Answer:
A) r•s = -21.99 units
B) rXs = 22.77 units (In the direction of k)
Completed question
Two vectors, r and s, lie in the xy plane. Their magnitudes are 4.25 and 7.45 units, respectively, and their directions are 312° and 86.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?
A) r•s =
B) (r)X(s)=
Explanation:
A) dot product of r and s can be expressed mathematically as;
r•s = |r||s|cos(A) .......1
Where A is the angle between the two vectors.
|r| and |s| are the magnitude of vector r and s.
r•s = 4.25×7.45 ×cos(86-312) = -21.99 units
B) cross product of r and s can be expressed mathematically as;
rXs = |r||s|sin(A) .......2
Where A is the smallest angle between the two vectors
|r| and |s| are the magnitude of vector r and s.
A = 312-86 = 226
A = 360-226 = 134 smallest
rXs = 4.25 × 7.45 × sin134.
rXs = 22.77 units
In the direction of k
(a) -22.00 units
(b) 22.77 units
Explanation:Given vectors are r and s
Where;
r = |r| = 4.25 and ∠r = 312° measured anticlockwise
s = |s| = 7.45 and ∠s = 86° measured anticlockwise
First, let's calculate the angle between vectors r and s by representing them in the figure below;
y | /s
| /
| /
| /
|/ )86° x
|\) 48°
| \
| \
| \
| \ r
To get the acute angle between r and the +x axis, subtract the reflex angle of r (312°) from 360° as follows;
360 - 312 = 48°
As shown in the diagram, the angle between vectors r and s is 48° + 86° = 134°
Now,
(a) The (r)(s) represents the dot or scalar product of the two vectors and it is given as;
(r) (s) = r x s cos θ ---------------------------(i)
Where;
r = magnitude of vector r = 4.25
s = magnitude of vector s = 7.45
θ is the angle between the two vectors r and s = 134°
Substitute these values into equation (i) as follows;
(r) (s) = 4.25 x 7.45 cos 134°
(r) (s) = 4.25 x 7.45 x -0.6947
(r) (s) = 31.66 x -0.6947
(r) (s) = -22.00 units
(b) The (r) X (s) represents the vector product of the two vectors and it is given as;
(r) (s) = r x s sin θ ---------------------------(ii)
Where;
r = magnitude of vector r = 4.25
s = magnitude of vector s = 7.45
θ is the angle between the two vectors r and s = 134°
Substitute these values into equation (ii) as follows;
(r) (s) = 4.25 x 7.45 sin 134°
(r) (s) = 4.25 x 7.45 x 0.7193
(r) (s) = 31.66 x 0.7193
(r) (s) = 22.77 units
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature of 30°C and a pressure of 101 kPa. Find the density change, the temperature change, and the velocity change across this wave
Answer:
Density change, Δρ = 2.4 × 10⁻⁴ kg/m³
Temperature Change, ΔT = 0.0258 K
Velocity Change, Δc = 0.0148 m/s
Explanation:
For sound waves moving through the air,
Pressure and Temperature varies thus
(P₀/P) = (T₀/T)^(k/(k-1))
Where P₀ = initial pressure of air = 101KPa = 101000 Pa
P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa
T₀ = initial temperature of air = 30°C = 303.15 K
T = final temperature of air = ?
k = ratio of specific heats = Cp/Cv = 1.4
(101000/101030) = (303.15/T)^(1.4/(1.4-1))
0.9990703 =(303.15/T)^(3.5)
Solving This,
T = 303.1758 K
ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K
Density can be calculate in two ways,
First method
Δρ = ρ - ρ₀
P₀ = ρ₀RT₀
ρ₀ = P₀/RT₀
R = gas constant for air = 287 J/kg.k
where all of these are values for air before the wave propagates
P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K
ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³
ρ = P/RT
P = 101030 Pa, T = 303.1758K
ρ = 101030/(287×303.1758) = 1.1611115 kg/m³
Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³
Second method
(ρ₀/ρ) = (T₀/T)^(1/(k-1))
Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.
Velocity Change
c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s
c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s
Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s
QED!
In this exercise we have to use the pressure knowledge to calculate the velocity, temperature and density so we have:
Density: [tex]\Delta \rho = 2.4 * 10^{-4} kg/m^3[/tex]Temperature: [tex]\Delta T = 0.0258 K[/tex]Velocity: Δc = 0.0148 m/sThe variation of temperature and pressure is given by the formula of:
[tex](P_0/P) = (T_0/T)^{(k/(k-1))}[/tex]
From the formula given above we can identify:
P₀ = initial pressure of airP = final pressure of air due to the change brought about by the moving sound wave T₀ = initial temperature of air T = final temperature of airk = ratio of specific heatsSolving the formula for temperature we find:
[tex](101000/101030) = (303.15/T)^{(1.4/(1.4-1))}\\0.9990703 =(303.15/T)^{(3.5)}\\T = 303.1758 K\\\Delta T = T - T_0 = 303.1758 - 303.15 = 0.0258 K[/tex]
They are using the formula for density is:
[tex]\rho_0 = 101000/(287 * 303.15) = 1.1608655 kg/m^3\\\rho = P/RT\\\rho = 101030/(287*303.1758) = 1.1611115 kg/m^3\\\Delta \rho = 2.4 * 10^{-4} kg/m^3[/tex]
Calculating the speed we find that:
[tex]c_0 = \sqrt{kRT_0} = 349.00669 m/s\\c = \sqrt{kRT} = 349.0215415 m/s\\\Delta c = 0.0148 m/s[/tex]
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Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the string?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Answer:
Explanation:
=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Here, the spatial and temporal parts of the wave function are A sin(kx) and sin(ωt), respectively. The displacement of the string at x=0 is 0. Although the string is straight at time t=π/2ω, there is still energy stored in it due to parts of the string having nonzero velocity.
Explanation:Part C: The spatial part of the wave function, ye(x), can be expressed as A sin(kx), and the temporal part, yt(t), is represented by sin(ωt). These expressions are valid for a wave of unit amplitude.
Part E: At position x=0, the displacement of the string is 0. This occurs because, for the wave equation, when inputting x=0 into the spatial part, the value for y becomes 0.
Part G: The correct answer is option c. Although the string is straight at time t=π/2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string. Option d is incorrect as the total mechanical energy in a perfectly straight string does not oscillate, but remains constant.
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What is the resistance of a 4.4-m length of copper wire 1.5 mm in diameter? The resistivity of copper is 1.68×10−8Ω⋅m.
Answer:
R = 4.18 * 10^8ohms
Explanation:
R=resistance in ohms=?
ρ=resistivity of material in ohms meters = 1.68*10^-8 oh ohm meters
L= length of the object (m) = 4.4m
A = cross-sectional area of the object in square meters (m^2)= πr^2
r = (1.5mm/1000)/2= 0.00075m
A=π*(0.00075)^2 = 1.76714586764426*10^-6
Approximately, A = 1.767m^2
R = ρL/A= (1.68*10^8Ω⋅m) * (4.4m)/(1.767m^2)
R = 418336162.988115 ohms
Approximately, R = 4.18 * 10^8ohms.
The resistivity of copper will be "4.18 × 10⁸ Ω (ohms)".
ResistanceAccording to the question,
Resistivity of ohms, ρ = 1.68 × 10⁻⁸ Ω.m
Object's length, L = 4.4 m
Radius, r = [tex]\frac{\frac{1.5}{1000} }{2}[/tex]
= 0.00075 m
We know,
The cross-sectional area,
A = πr²
By substituting the values,
= π × (0.00075)²
= 1.767 × 10⁻⁶
= 1.767 m²
hence,
The resistivity will be:
→ R = [tex]\frac{\rho L}{A}[/tex]
= [tex]\frac{1.68\times 10^{-8}\times 4.4}{1.767}[/tex]
= 4183316162.9 Ω or,
= 4.18 × 10⁸ Ω
Thus the above answer is correct.
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A tetrahedron has an equilateral triangle base with 27.0-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 280 N/C passes upward through the tetrahedron.What is the electric flux through each of the three sides?
Answer:
The electric flux through each of the 3 sides is 2.95 Wb
Solution:
As per the question:
Length of the edges, l = 27.0 cm = 0.27 m
Strength of the electric field, E = 280 N/C
To calculate the electric field through each of the three sides:
Area of the equilateral triangle is given by:
[tex]A = \frac{\sqrt{3}}{4}l^{2}[/tex]
[tex]A = \frac{\sqrt{3}}{4}\times (0.27)^{2} = 0.0316\ m^{2}[/tex]
Now, the electric flux that passes through the base is given by:
[tex]\phi = - E\cdot A = - EA[/tex]
[tex]\phi = 280\times 0.0316 = - 8.85\ Wb[/tex]
Now, the overall flux that passes through the surface and the base of the tetrahedron is zero.
[tex]\phi_{total} = \phi + 3phi_{surface}[/tex]
[tex]0 = \phi + 3phi_{surface}[/tex]
[tex]phi_{surface} = -\frac{\phi }{3}[/tex]
[tex]phi_{surface} = -\frac{- 8.85}{3} = 2.95\ Wb[/tex]
Two 2.00 cm * 2.00 cm plates that form a parallel-plate capacitor are charged to { 0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) 1.00 mm and (b) 2.00 mm.
The electric field strength between parallel conducting plates can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. For a spacing of 1.00 mm, the electric field strength is 7.08 × 10^-8 N/C. For a spacing of 2.00 mm, the electric field strength is 3.54 × 10^-8 N/C.
Explanation:The electric field strength between parallel conducting plates can be calculated using the formula:
E = V/d
Where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. In this case, the potential difference is given as 0.708 nC, or 0.708 × 10-9 C.
To find the electric field strength, we need to convert the charge to coulombs and divide it by the spacing between the plates. For a spacing of 1.00 mm (or 0.01 cm), the electric field strength is:
E = (0.708 × 10-9 C) / (0.01 cm) = 7.08 × 10-8 N/C
For a spacing of 2.00 mm (or 0.02 cm), the electric field strength is:
E = (0.708 × 10-9 C) / (0.02 cm) = 3.54 × 10-8 N/C
The electric field strength inside the capacitor is 2000 N/C. The potential difference is 2 V for a 1.00 mm separation and 4 V for a 2.00 mm separation between the plates.
To find the electric field strength (E) and potential difference (V) across a parallel-plate capacitor, we use the following steps:
Calculate the area (A) of the plates: A = 2.00 cm * 2.00 cm = 4.00 cm² = 4.00 * 10-4 m².Determine the charge (Q) on the plates: Q = 0.708 nC = 0.708 * 10-9 C.(a) For plate separation (d) of 1.00 mm:Compute the electric field: E = Q / (ε₀ * A) = (0.708 * 10-9 C) / (8.854 * 10-12 C²/N∙m² * 4.00 * 10-4 m²) ≈ 2000 N/C.Potential difference: V = E * d = 2000 N/C * 1.00 * 10-3 m = 2 V.(b) For plate separation of 2.00 mm:Electric field remains the same: E ≈ 2000 N/C.Potential difference: V = E * d = 2000 N/C * 2.00 * 10-3 m = 4 V.Therefore, the electric field strength inside the capacitor is 2000 N/C, the potential difference is 2 V for 1.00 mm separation, and 4 V for 2.00 mm separation.
A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude of 86.0 m and points in a direction 42.0° above the negative x-axis. a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C
Answer with Step-by -step explanation:
We are given that
b.[tex]\mid A\mid=46 m[/tex]
[tex]\theta=20^{\circ}[/tex] below the positive x-axis
Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=[tex]x=360-20=340^{\circ}[/tex]
x-component of vector A=[tex]A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24[/tex]
y-Component of vector A=[tex]A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64[/tex]
Magnitude of vector B=86 m
The vector B makes angle with positive x- axis=[tex]x'=42^{\circ}[/tex]
x-component of vector B=[tex]B_x=86cos42=63.64[/tex]
y-Component of vector B=[tex]B_y=86sin42=57.62[/tex]
Vector A=[tex]A_xi+A_yj=43.24i-15.64j[/tex]
Vector B=[tex]B_xi+B_yj=63.64i+57.62j[/tex]
Vector C=A+B
Substitute the values
[tex]C=43.24i-15.64j+63.64i+57.62j[/tex]
[tex]C=106.88i+41.98j[/tex]
c.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}[/tex]
The direction of the vector C=21.5 degree
To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, start by drawing the x and y axes. Use the component method of vector addition to find the magnitude of C⃗. Use the inverse tangent function to find the direction of C⃗.
Explanation:To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, we start by drawing the x and y axes. Vector A⃗ has a magnitude of 46.0 m and points 20.0° below the positive x-axis, so we draw A⃗ starting from the origin and make an angle of 20.0° with the positive x-axis in a downward direction. Vector B⃗ has a magnitude of 86.0 m and points 42.0° above the negative x-axis, so we draw B⃗ starting from the origin and make an angle of 42.0° with the negative x-axis in an upward direction. To find the vector C⃗=A⃗+B⃗, we add the x-components and the y-components of A⃗ and B⃗ separately. Then we use the Pythagorean theorem to find the magnitude of C⃗ and the inverse tangent function to find the direction of C⃗ in relation to the positive x-axis.
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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cmcm . The explorer finds that the pendulum completes 103 full swing cycles in a time of 132 s.What is the value of the acceleration of gravity on this planet?
Answer:
13.01 m/s²
Explanation:
The period of a simple pendulum is given as
T = 2π√(L/g) .......................... Equation 1
Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.
Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14
Substitute into equation 1
1.28 = (2×3.14)√(0.54/g)
1.28 = 6.28√(0.54/g)
Making g the subject of the equation,
√(0.54/g) = 1.28/6.28
√(0.54/g) = 0.2038
0.54/g = (0.2038)²
0.54/g = 0.0415
g = 0.54/0.0415
g = 13.01 m/s²
Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²
A rock is thrown horizontally from a tower 4.9 m above the ground, and the rock strikes the ground after travelling 20 m horizontally. What was the rock's initial launch speed (in m/s)?
We have that the the rock's initial launch speed is mathematically given as
u=20m/sFrom the question we are told
A rock is thrown horizontally from a tower 4.9 m above the ground, and the rock strikes the ground after travelling 20 m horizontally. What was the rock's initial launch speed (in m/s)?SpeedGenerally the equation for the Motion is mathematically given as
[tex]H=ut+0.5gt^2\\\\Therefore\\\\4.9=0*t+0.5*9.8*t^2\\\\t=1s\\\\Therefore\\\\sx=uxt\\\\20=ux*1\\\\[/tex]
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To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, and it travels 20 m horizontally in 1 second, the initial launch speed can be calculated as 20 m/s.
Explanation:To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, its initial vertical velocity is zero. The only force acting on the rock in the vertical direction is gravity, which causes it to accelerate downward at a rate of 9.8 m/s². Using the formula:
-yf = yi + vit - 1/2gt²
where yi represents the initial height and yf represents the final height, we can solve for the initial downward velocity. Since the rock reaches the ground (yf = 0) and starts at a height of 4.9 m, the equation becomes:
0 = 4.9 + 0 - 1/2(9.8)t²
Simplifying the equation gives:
t² = 4.9/4.9
t = 1 second.
Since the horizontal distance traveled by the rock is 20 m, and it takes 1 second to reach the ground, the horizontal speed (initial launch speed) of the rock can be calculated using the formula:
v = d/t = 20 m/1 s = 20 m/s.
Therefore, the rock's initial launch speed is 20 m/s.
Borrow soil is used to fill a 75,000m3 depression. The borrow soil has the following characteristics. Density: 1540kg/m3, water content: 8%, specific gravity of the solids: 2.66. The final in-place dry density should be 1790 kg/m3 and the final water content should be 13%. a) How many m3 of borrow soil are needed to fill the depression? b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?
Answer
given,
Volume of Depression, V = 75000 m³
borrow soil
Density of soil,γ = 1540 Kg/m³
water content,w = 8 % = 0.08
Specific gravity of solid,G = 2.66
Final in-place
dry density = 1790 kg/m³
water content = 13 % = 0.13
a) Volume of borrow soil require to fill the depression
Mass of solid solid,m= dry density of inplace soil x Volume of depression
= 1790 x 75000
m = 1.3425 x 10⁸ Kg
dry density of the borrow pit
[tex]\gamma_d = \dfrac{\gamma}{1 + w}[/tex]
[tex]\gamma_d = \dfrac{1540}{1 + 0.08}[/tex]
[tex]\gamma_d = 1425.93\ kg/m^3[/tex]
Volume of borrow soil required
[tex]V = \dfrac{m}{\gamma_d}[/tex]
[tex]V = \dfrac{1.3425\times 10^8}{1425.93}[/tex]
V = 94149 m³
b) Water required
[tex]W = 1790\times 75000\times 0.13 - 1425.93\times 94149\times 0.08[/tex]
W = 6.71 x 10⁶ Kg
Water required to achieve 13% moisture is equal to W = 6.71 x 10⁶ Kg
Final answer:
To fill a 75,000 m³ depression, 93,590.46 m³ of borrow soil is required when accounting for the dry density change. To achieve a final water content of 13%, an additional 17,452,500 kg of water mass is needed.
Explanation:
To answer this question, we must first understand the relationship between the initial and final states of the borrow soil in terms of their volumes, densities, and moisture contents. The question is divided into two parts: calculating the volume of borrow soil needed and determining the water mass addition required to achieve the desired moisture content.
a) How many m³ of borrow soil are needed to fill the depression?
Since we want to fill a depression with a volume of 75,000 m³, we need to consider the change in density from the borrow state to the in-place dry density. The in-place dry density is given as 1790 kg/m³. The dry density of the borrow soil can be calculated by subtracting the mass of water from the total mass.
First, we convert the water content to a decimal by dividing by 100:
Water content = 8% / 100 = 0.08
Using the formula for dry density (
dry density = total density / (1 + water content)), we find:
Dry density of borrow soil = 1540 kg/m³ / (1 + 0.08) = 1425.93 kg/m³
The volume of the borrow soil needed can be calculated by the volume of the depression divided by the ratio of the final in-place dry density to the dry density of the borrow soil:
Volume of borrow soil needed = 75,000 m³ x (1790 kg/m³ / 1425.93 kg/m³) = 93,590.46 m³
b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?
For the final water content of 13%, we'll use the in-place dry density to determine the total mass, then calculate the water mass needed:
Total in-place mass = volume x in-place dry density
Total in-place mass of fill = 75,000 m³ x 1790 kg/m³= 134,250,000 kg
Convert 13% to decimal:
Water content = 13% / 100 = 0.13
We want 13% of the total mass to be water, so the water mass required is:
Water mass = total mass x water content
Water mass needed = 134,250,000 kg x 0.13 = 17,452,500 kg
This is the additional water mass required to achieve the desired moisture content of 13%.
If it takes about 8 minutes for light to travel from the Sun to Earth, and Pluto is 40 times this distance from us, how long does it take light to reach Earth from Pluto?
Light takes 320 minutes to reach Earth from Pluto
Explanation:
We can solve this problem by applying the rule of three. We can do as follows:
- We call [tex]x[/tex] the distance between the Sun and the Earth
- Then the distance between the Earth and Pluto is 40 times this distance, so [tex]40x[/tex]
- Light takes about 8 minutes to cover the distance x between Sun and Earth
Therefore, calling T the time that light takes to cover distance between Earth and Pluto (40x), we can write:
[tex]\frac{x}{8}=\frac{40x}{T}[/tex]
And solving for T,
[tex]T=\frac{8\cdot 40 x}{x}=320 min[/tex]
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A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ∘ incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?
Final answer:
A 26-N force must act down and parallel to the incline for the bin to move down at constant velocity since this force would balance out the 26-N frictional force acting up the incline.
Explanation:
The subject of this question is Physics, specifically the concepts related to mechanics and forces on inclines with friction. The student is in High School level, most likely studying the fundamentals of Newtonian mechanics.
Given that the bin is already being pushed up the incline with a 26-N force and moves at a constant speed, this means the net force on the bin is zero, therefore the force of friction must also be 26 N but acting downwards along the incline. When the bin moves down at a constant velocity, the force of friction still acts up the incline (opposite to the bin's movement) and therefore, to maintain a constant velocity, the force applied must be equal in magnitude to the frictional force but directed down the incline. Hence, a 26-N force must be applied down and parallel to the incline for the bin to move down the incline at constant velocity.
A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horizontal component of the velocity.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
A 238 92U nucleus is moving in the x-direction at 5.0 × 105 m/s when it decays into an alpha particle 4 2He and a 234 90Th. If the alpha particle moves off at 25.4° above the x axis with a speed of 1.4 × 107 m/s, what is the recoil velocity of the thorium nucleus? Assume the uranium-thorium-alpha system is isolated; you may also assume the particles are pointlike.
Answer:
v_th = 3.1 * 10^5 m/s
Explanation:
Given:
- mass of 238-Uranium m_u = 3.952 *10^-25 kg
- mass of alpha particle m_a = 6.64 * 10^-27 kg
- mass of thorium particle m_th = 3.885*10^-25 kg
- velocity of 238-Uranium v_u = 5.0 *10^5 m/s
- velocity of alpha particle v_a = 1.4 *10^7 m/s
Find:
- The recoil velocity of the thorium particle.
Solution:
- To solve this problem we will use conservation of momentum in both x and y direction.
- Momentum conservation in x-direction:
P_i = P_f
m_u*v_u = m_a*v_a*cos(Q) + m_th*v_th,x
where v_th,x is the x component of thorium velocity:
P_i = 3.952 *10^-25 * 5.0 *10^5 = 1.976*10^-19
P_f = 6.64 * 10^-27*1.4 *10^7*cos(25.4) + 3.885*10^-25*v_th,x
P_f = 8.3974*10^-20 + 3.885*10^-25*v_th,x
Hence,
1.976*10^-19 = 9.296*10^-20 + 3.885*10^-25*v_th,x
v_th,x = 2.92473 * 10^5 m/s
- Momentum conservation in y-direction:
P_i = P_f
0 = m_a*v_a*sin(Q) + m_th*v_th,y
where v_th,x is the x component of thorium velocity:
v_th,y = m_a*v_a*sin(Q) / m_th
v_th,y = 6.64 * 10^-27*1.4 *10^7*sin(25.4) / 3.885*10^-25
Hence,
v_th,y = 1.02635 * 10^5 m/s
- The magnitude of recoil velocity:
v_th = sqrt ( v_th,x ^2 + v_th,y ^2 )
v_th = sqrt ( (2.92473 * 10^5)^2 + (1.02635 * 10^5)^2 )
v_th = 3.1 * 10^5 m/s
An astronaut takes her bathroom scales to the moon, where g = 1.6 m/. On the moon, compared to at home on earth:
(A) Her weight is less, and her mass is less.
(B) Her weight is the same, and her mass is less.
(C) Her weight is the same, and her mass is the same.
(D) Her weight is less, and her mass is the same.
(E) Her weight is zero, and her mass is the same.
Answer: (D) Her weight is less, and her mass is the same.
Explanation:
Mass is the total amount of matter contained in the body.
Weight is the force exerted by gravity on a mass.
[tex]weight=mass\times gravity[/tex]
When the value of g is more, weight is more.
For moon g= [tex]1.6m/s^2[/tex] , whereas for earth [tex]g=9.8m/s^2[/tex]
Thus weight is more on earth as compared to moon.
On the moon, compared to at home on earth: Her weight is less, and her mass is the same.
Answer:
Explanation:
Weight is defined as the force with which a planet pulls the object towards its centre.
Weight = mass x acceleration due to gravity of the planet
W = m x g
The amount of matter contained in the substance is called mass.
the mass of the substance remains same at every planet but the weight is changed as the value of acceleration due to gravity is different for all the planets.
As the acceleration due to gravity on moon is 1.6 m/s^2 and it is less than the acceleration due to gravity on earth. So, the weight of the astronaut is less than the weight on earth and mass remains same.
Weight is less but the mass is same.
Option (D) is true.
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.97 MeV, how close (in m) to the gold nucleus (79 protons) could it come before being deflected?
Answer:
3.8 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) = 9.564 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 9.564 × 10⁻¹³ = 3.8 × 10 ⁻¹⁴ m
A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of argon gas at 130°C and 20 bar on the other. If the partition is removed and the gases mix adiabatically and completely, what is the change in entropy? Assume nitrogen to be an ideal gas with CV = (5/2)R and argon to be an ideal gas with CV = (3/2)R.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C in kelvin t1=75+273
t1=348K
T2=130°C in kelvin t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
putting values:
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
by simplifying:
Tfinal=363K
To determine the change in entropy when the gases mix adiabatically and completely, we'll follow these steps:
1. Calculate the initial entropy of each gas before mixing.
2. Calculate the final entropy of the combined gases after mixing.
3. Calculate the change in entropy.
Given information:
For nitrogen gas:
- Moles of nitrogen gas, [tex]n_N2[/tex] = 4 mol
- Temperature of nitrogen gas, [tex]T_N2[/tex] = 75°C = 75 + 273.15 K
- Pressure of nitrogen gas, [tex]P_N2[/tex] = 30 bar = 30 * 100 kPa
For argon gas:
- Moles of argon gas, [tex]n_Ar[/tex] = 2.5 mol
- Temperature of argon gas, [tex]T_Ar[/tex] = 130°C = 130 + 273.15 K
- Pressure of argon gas, [tex]P_VR[/tex] = 20 bar = 20 * 100 kPa
Given specific heat capacities :
- [tex]CV_N2[/tex] = (5/2)R for nitrogen gas
- [tex]CV_Ar[/tex] = (3/2)R for argon gas
The change in entropy, ΔS, can be calculated using the equation:
[tex]\[ΔS = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_f}{T_{\text{N2}}}\right) + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_f}{T_{\text{Ar}}}\right)\][/tex]
where [tex]\(T_f\)[/tex] is the final temperature after mixing.
1. Calculate the initial entropy of each gas:
[tex]\[S_{\text{N2}} = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_{\text{N2}}}{T_0}\right)\][/tex]
[tex]\[S_{\text{Ar}} = n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_{\text{Ar}}}{T_0}\right)\][/tex]
where [tex]\(T_0\)[/tex] is the reference temperature (usually taken as 298.15 K).
2. Calculate the final temperature after mixing using the adiabatic process equation:
[tex]\[T_f = \left(\frac{n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot T_{\text{N2}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot T_{\text{Ar}}}{n_{\text{N2}} \cdot C_{V_{\text{N2}}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}}}\right)\][/tex]
3. Calculate the change in entropy using the formula above.
Let's plug in the values and calculate the change in entropy. First, we'll calculate the initial entropies of each gas:
[tex]\[S_{\text{N2}} = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{75 + 273.15}{298.15}\right)\][/tex]
[tex]\[S_{\text{Ar}} = 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{130 + 273.15}{298.15}\right)\][/tex]
Now, calculate [tex]\(T_f\)[/tex] using the adiabatic process equation:
[tex]\[T_f = \left(\frac{4 \times \left(\frac{5}{2}\right)R \times 75 + 2.5 \times \left(\frac{3}{2}\right)R \times 130}{4 \times \left(\frac{5}{2}\right)R + 2.5 \times \left(\frac{3}{2}\right)R}\right)\][/tex]
Finally, calculate the change in entropy:
[tex]\[ΔS = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{T_f}{75 + 273.15}\right) + 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{T_f}{130 + 273.15}\right)\][/tex]
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A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. What is the fundamental frequency if the tube is 0.203-m long, by taking air temperature to be 37.0ºC?
Answer:
435.467980296 Hz
Explanation:
T = Temperature = 37.0ºC
L = Length of the tube = 0.203 m
Speed of sound at a specific temperature is given by
[tex]v=331.4+0.6\times T\\\Rightarrow v=331.4+0.6\times 37\\\Rightarrow v=353.6\ m/s[/tex]
Frequency is given by (one end open other end closed)
[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{353.6}{4\times 0.203}\\\Rightarrow f=435.467980296\ Hz[/tex]
The fundamental frequency is 435.467980296 Hz
The fundamental frequency of a tube, considered as an approximation for the vocal apparatus and closed at one end, with a length of 0.203 meters and at an air temperature of 37°C, is calculated using the formula for frequency f=Vw/(4L). Substituting in the values, the fundamental frequency is about 432 Hz.
Explanation:The given question involves a concept from physics, specifically sound waves and resonance. When considering the breathing passages and mouth as a resonating tube closed at one end, the fundamental resonant frequency can be calculated using the formula Vw = fa, where Vw is the speed of sound, f represents frequency, and a is the wavelength.
For a tube closed at one end, the wavelength (λ) is four times the length of the tube. So, λ = 4L. We can rearrange the formula to solve for fundamental frequency: f = Vw / a = Vw / (4L).
If the air temperature is 37°C, the speed of sound (Vw) in air is approximately 351 m/s. Substituting these values in, you get: f = 351m/s / (4*0.203m) ≈ 432 Hz. So, the fundamental frequency of this tube (our vocal apparatus approximation) at 37°C would be about 432 Hz.
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F1 = (3.3,-0.5) and F2 = (-3.8,-0.3) where all components are in newtons. What angle does the vector F1 + F2 make with the positive x-axis? The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees.
Answer:
238 Degree
Explanation:
Data given
F1=(3.3,-0.5) and F2=(-3.8,-0.3).
To determine the angle F1+F2 makes with the positive x-axis, we need to determine the magnitude of the force F1+F2.
Since force is a vector quantity, we add the vectors component by component
[tex]F_{1}+F_{2}=<3.3,-0.5> +<-3.8,-0.3>\\F_{1}+F_{2}=<3.3+(-3.8), -0.5+(-0.3)>\\ F_{1}+F_{2}=<-0.5,-0.8>\\F_{1}+F_{2}=(-0.5,-0.8)[/tex]
To determine the angle, we use
[tex]F=(x,y)\\\alpha=arctan(\frac{y}{x} )\\Hence for \\F_{1}+F_{2}=(-0.5,-0.8)\\\alpha=arctan(\frac{-0.8}{-0.5} )\\\alpha=58^{0}[/tex]
Since the component of the force F1+F2 is a negative y and negative x which are located in the 3rd quadrant, the angle can be calculated as
∝=58+180=238 degree
Hence The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees is 238 Degree
The angle measured counterclockwise from the positive x-axis is θ = 57.99°
Finding the direction of the force.
Here we know that:
F1 = (3.3, -0.5)F2 = (-3.8, -0.3)First, we need to add the forces, we will get:
F1 + F2 = (3.3, -0.5) + (-3.8, -0.3) = (3.3 - 3.8, -0.5 - 0.3))
F1 + F2 = (-0.5, -0.8)
Now, the angle measured counterclockwise from the positive x-axis of a vector
(a, b) is given by:
θ = Atan(b/a).
Where Atan(x) is the inverse tangent function.
So in this case the angle will be:
θ = Atan(-0.8/-0.5) = 57.99°
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After a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)?There is positive charge on end B and negative charge on end A.There is negative charge spread evenly on both ends.There is negative charge on end A with end B remaining neutral.There is positive charge on end A with end B remaining neutral.The conducting rod is not grounded, so a negative charge accumulates on one end, but a charge cannot remain at a single end in a conductor, so it will spread to both ends:There is negative charge spread evenly on both ends
Answer:
The correct option is that both the ends will remain neutral.
Explanation:
As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.
The correct option is: There is negative charge spread evenly on both ends.
When a conducting rod is brought into contact with a charged ball, charge transfer occurs due to the process of induction and conduction. Initially, when the charged ball (let's assume it is positively charged) is brought close to the rod, it will induce a negative charge on the end of the rod nearest to the ball (end A) and a positive charge on the far end (end B) due to the separation of charges within the conductor. This is because the positive charge on the ball will repel the positive charges within the rod, pushing them away, and attract the negative charges, pulling them towards the ball.
Once the rod is actually touched to the ball, the negative charges on the rod (which were induced by the ball's positive charge) will be neutralized by the positive charges from the ball, leaving the rod with a net negative charge after the ball is moved away. This negative charge will be evenly distributed along the rod initially.
However, since the rod is not grounded, the negative charge will not have a path to escape and will remain on the rod. Due to the repulsion between like charges (negative-negative), these charges will repel each other and spread out as far as possible from each other along the length of the rod. This means that the negative charge will be spread evenly on both ends of the rod, as this arrangement minimizes the repulsive forces between the charges and results in the lowest potential energy configuration.
Therefore, the final arrangement of any charge on the rod, when the charged ball is far away, will be an even distribution of negative charge on both ends A and B of the rod. This is consistent with the principle of electrostatic equilibrium, where charges within a conductor will redistribute themselves to minimize repulsion and achieve the lowest energy state.
A glow-worm of mass 5.0 g emits red light (650nm) with a power of 0.10W entirely in the backward direction. To what speed will it have accelerated after 10 y if released into free space and assumed to live
Answer:
The speed the glow-worm would have accelerated after 10 years if released into free space and assumed to live is 3.09 X 10³⁴ m/s
Explanation:
Energy associated with photon of lights, is given as
E = hc/λ
Where;
h is Planck's constant = 6.626 x 10⁻³⁴ m2 kg/s
C is the speed of light = ?
λ is the light's wavelength = 650nm = 650 X10⁻⁹ m
Also Energy = Power X time
Time given = 10 years
Time in seconds = 10 yrs X 365 days X 24 hrs X 60mins X 60 secs
Time in seconds = 315360000 seconds
P*t = hc/λ
c = (P*t*λ)/h
c = (0.1 X 315360000 X 650 X 10⁻⁹)/(6.626 x 10⁻³⁴)
c = 3.09 X 10³⁴ m/s
The speed the glow-worm would have accelerated after 10 y if released into free space and assumed to live is 3.09 X 10³⁴ m/s
In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 m above the ground and has a speed of 64.0 m/s (143 mi/h), at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
Answer:
[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.
Explanation:
Given:
height of the plane, [tex]h=90\ m[/tex]horizontal speed of plane, [tex]v_x=64\ m.s^{-1}[/tex]Time taken by the canister to hit the ground:
using equation of motion
[tex]h=u_y.t+\frac{1}{2} g.t^2[/tex]
where:
[tex]u_y=[/tex] initial vertical velocity of the canister = 0 (since the the object is dropped from a horizontally moving plane)
[tex]t=[/tex] time taken to hit the ground
[tex]90=0+0.5\times 9.8\times t^2[/tex]
[tex]t=4.2857\ s[/tex]
Now the horizontal distance travelled by the canister after dropping:
[tex]s=v_x\times t[/tex]
[tex]s=64\times 4.2857[/tex]
[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.
Spring #1 has spring constant 61.0 N/m. Spring #2 has an unknown spring constant, but when connected in series with Spring #1, the connected springs have an effective spring constant of 20.0 N/m. What is the spring constant for Spring #2?
Answer:
29.79 N/m
Explanation:
A spring connected in series behaves like a capacitor connected in series.
Note: Spring and capacitor are alike because they both store energy, While the former store mechanical energy, the later store electrical energy.
From the above, the effective spring connected in series is given by the formula below
1/Kt = 1/K1 + 1/K2 ........................ Equation 1
Where Kt = effective spring constant, K1 = spring constant of spring 1, K2 = spring constant of spring 2
Making K2 the subject of the equation,
K2 = KtK1/(K1-Kt)..................... Equation 2
Given: K1 = 61 N/m, Kt = 20 N/m.
Substitute into equation 2
K2 = (61×20)/(61-20)
K2 = 1220/41
K2 = 29.76 N/m.
Hence the spring constant in the second spring = 29.79 N/m
The spring constant for Spring #2 is 369.17 N/m.
Explanation:To find the spring constant of Spring #2, we need to use the formula for the effective spring constant of springs in series. The formula is:
1/keff = 1/k1 + 1/k2
Given that k1 = 61.0 N/m and the effective spring constant (keff) is 20.0 N/m, we can plug these values into the formula and solve for k2:
1/20 = 1/61 + 1/k2
Now we can solve for k2:
1/k2 = 1/20 - 1/61
k2 = 1/(1/20 - 1/61)
k2 = 369.17 N/m
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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.30 kN, and the radius of the circle is 12.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 3.60 m/s? (b) What is FB if v = 14.0 m/s? Use g=9.80 m/s2.
Explanation:
As the force is given as 5.30 kN or [tex]5.30 \times 1000 N[/tex]. Hence, mass will be calculated as follows.
[tex]F_{w}[/tex] = mg
[tex]5.30 \times 10^{3} = m \times 9.8 m/s^{2}[/tex]
m = 540.816 kg
(a) At the top, centripetal force [tex]F_{c}[/tex] is acting upwards and the weight of the riders and car, [tex]F_{w}[/tex] will be acting downwards.
Therefore, force on the car by the boom will be calculated as follows.
[tex]F_{B} = F_{w} - F_{c}[/tex]
or, [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]
= [tex]5.30 \times 1000 N - \frac{540.816 kg \times (3.60)^{2}}{12}[/tex]
= 4715.919 N
Hence, the force [tex]F_{B}[/tex] on the car from the boom is 4715.919 N. This means that the car will be hanging on the boom and the boom will exert an upward force.
(b) Now at the top, centripetal force [tex]F_{c}[/tex] will be acting upwards and the weight of cars and car riders will be acting in the downwards direction.
Hence, we will calculate the force on car by the boom as follows.
[tex]F_{B} = F_{w} - F_{c}[/tex]
or, [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]
= [tex]5.30 \times 1000 N - \frac{540.816 kg \times (14.0)^{2}}{12}[/tex]
= -3533.33 N
Therefore, car will be pushing on the boom and the boom will exert a downward force.
Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
0.12959085 J
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = 1.55 μC
d = Distance between charge = 0.5 m
Electric potential energy is given by
[tex]U=k\dfrac{q^2}{d}[/tex]
In this system with three charges which are equidistant from each other
[tex]U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}[/tex]
[tex]\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J[/tex]
The potential energy of the system is 0.12959085 J
"Sarah is out playing fetch with her dog, and throws a tennis ball as far as she can. At the moment the ball reaches its maximum height, which of the following are true. You may neglect air resistance."a-You may neglect air resistance.
b-The ball's vertical acceleration is downwards
c-The ball's vertical acceleration is zero
d-The ball's vertical acceleration is upwards
e-The ball's horizontal acceleration is zero
f-The ball's horizontal acceleration is in the direction of the throw
g-The ball's horizontal acceleration is oppoite the direction of the throw
h-The ball's vertical acceleration is upwards
Answer:
a.The ball's vertical acceleration is downwards.e.The ball's horizontal acceleration is zeroExplanation:
We are given that Sarah throws a tennis ball as far as she can.
At the moment the ball reaches its maximum height.
We have to find the true statement if air resistance is neglect.
When air resistance is negligible then the force act on the ball is force due to gravity.
The ball throw vertically then the acceleration act on the ball is acceleration due to gravity.
The value of g=-9.8 m/square sec
It acts on the ball in downward direction .
Therefore, the ball's vertical acceleration is downwards.
The horizontal acceleration is zero because the ball reaches at maximum height then there is no force which act in horizontal direction on the ball.
Therefore, horizontal acceleration of the ball is zero.
Hence, option a and e are true.
The statement B is correct which states that "The ball's vertical acceleration is downwards".
The statement E is correct which states that "The ball's horizontal acceleration is zero".
How do you find out which statement is true?The given condition is that Sarah throws a tennis ball as far as she can. Also given that the air resistance is negligible.
Let consider that the ball is thrown vertically. In this case, the force on the ball is the force due to gravitational acceleration on the ball.
Hence the vertical acceleration on the ball is equivalent to the gravitational acceleration of 9.8 m/s2. The vertical force acting on the ball will be in a downward direction because this is the force due to gravity.
Hence the statement B is correct which states that "The ball's vertical acceleration is downwards".
When the ball reaches its maximum height, the horizontal acceleration will be zero on the ball, as there is no force acting on the ball in the horizontal direction.
Hence the statement E is correct which states that "The ball's horizontal acceleration is zero".
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Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
About how long does it take a seed launched at 90° at the highest possible initial speed to reach its maximum height? Ignore air resistance. (a) 0.23 s; (b) 0.47 s; (c) 1.0 s; (d) 2.3 s.
Final answer:
The seed takes approximately 0.47 seconds to reach its maximum height when launched at 90° with the highest possible initial speed of 4.6 m/s.
Explanation:
When a seed is launched with an initial speed of 4.6 m/s and at an angle of 90°, the vertical component of the initial velocity is 4.6 m/s. The time it takes for a projectile to reach its maximum height can be calculated using the equation for vertical motion:
t = (Vf - Vi) / g
Where:
t is the time
Vf is the final velocity (which is 0 m/s at the maximum height)
Vi is the initial velocity
g is the acceleration due to gravity (9.8 m/s²)
Using the given information, we can calculate the time it takes for the seed to reach its maximum height:
t = (0 - 4.6) / -9.8 = 0.469 s
Therefore, the seed takes approximately 0.47 seconds to reach its maximum height.
The bullet starts at rest in the gun. An 8.6 g bullet leaves the muzzle of a rifle with a speed of 430.1 m/s. What constant force is exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle?
Answer:
The constant force exerted on the bullet is 1590.87 N.
Explanation:
It is given that,
Mass of the bullet, m = 8.6 g
Initial speed of the bullet, u = 0
Final speed of the bullet, v = 430.1 m/s
We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad[/tex]
[tex]a=\dfrac{v^2}{2d}[/tex]
[tex]a=\dfrac{(430.1)^2}{2\times 0.5}[/tex]
[tex]a=184986.01\ m/s^2[/tex]
Let F is the force exerted. It is given by :
[tex]F=ma[/tex]
[tex]F=8.6\times 10^{-3}\times 184986.01[/tex]
F = 1590.87 N
So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.
hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of the following statements could be an operational definition of electric current?The magnitude of the physical quantity of electric current I in a wire equals the magnitude of the electric charge q that passes through a cross section of the wire divided by the time interval Δt needed for that charge to pass.
The statement "The magnitude of the physical quantity of electric current \( I \) in a wire equals the magnitude of the electric charge q that passes through a cross-section of the wire divided by the time interval [tex]\( \Delta t \)[/tex] needed for that charge to pass" could be an operational definition of electric current.
This definition precisely defines electric current as the rate of flow of electric charge through a conductor over a specific time period.
By quantifying the amount of charge passing through a cross-section of the wire in a given time interval, this definition provides a measurable and practical way to understand and quantify electric current.
It aligns with the fundamental concept that electric current represents the flow of charge, making it a suitable operational definition in the context of electrical systems and circuits.
By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum speed during a vibration? A is the old amplitude and A′ is the new one.
Answer:
[tex]A'=2A[/tex]
Explanation:
According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:
[tex]E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}[/tex]
When the spring is in its equilibrium position, that is [tex]x=0[/tex], the object speed its maximum. So, we have:
[tex]\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}[/tex]
In order to double its maximum speed, that is [tex]v'{max}=2v_{max}[/tex]. We have:
[tex]A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A[/tex]
The factor we must increase the amplitude of the vibration to double its maximum speed is 2.
Apply the principle of conservation of energy to determine the amplitude;
[tex]U = K.E\\\\\frac{1}{2} KA^2 = \frac{1}{2}mv^2\\\\KA^2 = mv^2\\\\A^2 = \frac{m}{K} v^2\\\\\frac{A_1^2}{v_1^2 } = \frac{A_2^2}{v_2^2 } \\\\A_2 = \frac{v_2 A_1}{v_1} \\\\A_2 = \frac{2v_1 \times A_1}{v_1} \\\\A_2 =2A_1[/tex]
Thus, the factor we must increase the amplitude of the vibration to double its maximum speed is 2.
Learn more here:https://brainly.com/question/14494944
The space shuttle fleet was designed with two booster stages. if the first stage provides a thrust of 53 kilo-newtons and the space shuttle has an acceleration of 18,000 miles per hour squared, what is the mass of the spacecraft in units of pounds-mass?
Use Newton's second law to find the mass in kilograms, and finally convert it to pounds-mass, which is approximately 14.52 pounds-mass.
The question asks us to calculate the mass of the spacecraft given the thrust of the first booster stage and the acceleration of the space shuttle. To find the mass, we use Newton's second law of motion, which states that Force = [tex]mass imes acceleration (F = m imes a).[/tex] However, we first need to convert the given acceleration from miles per hour squared to meters per second squared, and then convert the mass obtained in kilograms to pounds-mass.
First, let's convert the acceleration: 18,000 miles/hour2 is approximately 8,046.72 m/s2 (using the conversion factor 1 mile = 1,609.34 meters and 1 hour = 3600 seconds). Next, we can calculate the mass (in kilograms) by rearranging the formula to m = F / a, which gives us 53,000 N / 8,046.72 m/s2
= approximately 6.59 kilograms. We then convert kilograms to pounds-mass (1 kilogram = 2.20462 pounds), resulting in the mass of the spacecraft being approximately 14.52 pounds-mass.
The mass of the spacecraft is approximately [tex]261,313.24\ pounds[/tex] mass.
To find the mass of the spacecraft, we can use Newton's second law of motion:
[tex]\[ F = ma \][/tex]
where:
[tex]- \( F \)[/tex] is the force (thrust) provided by the first stage ([tex]53 kN[/tex]),
[tex]- \( m \)[/tex] is the mass of the spacecraft,
[tex]- \( a \)[/tex] is the acceleration ([tex]18,000 \ mph^2[/tex]).
First, let's convert the thrust from kilo-newtons (kN) to newtons (N), since the unit of acceleration is meters per second squared (m/s²):
[tex]\[ 1 \text{ kN} = 1000 \text{ N} \][/tex]
So, [tex]53 kN[/tex] is equivalent to [tex]\(53 \times 1000 = 53000\) N.[/tex]
Now, let's convert the acceleration from miles per hour squared (mph²) to meters per second squared (m/s²).
[tex]\[ 1 \text{ mph} = \frac{1609.34}{3600} \text{ m/s} \][/tex]
[tex]1 mph^2 = \left(\frac{1609.34}{3600}\right)^2 \text{ m/s²}[/tex]
[tex]1 mph^2 = 0.447 m/s^2}[/tex]
We can rearrange Newton's second law to solve for the mass [tex]\( m \)[/tex]
[tex]\[ m = \frac{F}{a} \][/tex]
[tex]\[ m = \frac{53000 \text{ N}}{0.447 \text{ m/s²}} \][/tex]
[tex]\[ m = 118488 \text{ kg} \][/tex]
To convert kilograms to pounds-mass, we use the conversion factor:
[tex]\[ 1 \text{ kg} = 2.20462 \text{ pounds-mass} \][/tex]
So,
[tex]\[ \text{mass (pounds-mass)} = 118488 \text{ kg} \times 2.20462 \][/tex]
[tex]\[ \text{mass (pounds-mass)} = 261313.24 \text{ pounds-mass} \][/tex]