Answer:
Yes
Explanation:
The velocity measured by Jennifer and Johnny is
[tex]v_m = 34.87 m/s[/tex]
The actual velocity is
[tex]v=34.15 m/s[/tex]
We can calculate the % error of the students measurement as follows:
[tex]Err = \frac{v_m - v}{v}\cdot 100 = \frac{34.87 m/s-34.15 m/s}{34.15 m/s}\cdot 100 =0.021 \cdot 100 = 2.1 \%[/tex]
Which is lower than the 2.5% maximum error required, so the two students will pass the test.
Final answer:
Jennifer and Johnny's measurement has a percentage error of approximately 2.11%, which is less than the maximum allowed error of 2.5%. Therefore, they will pass their final project.
Explanation:
To determine if Jennifer and Johnny passed their final project, we need to calculate the percentage error of their measured velocity. The percentage error is calculated using the formula:
Percentage Error = |(Actual Value - Experimental Value) / Actual Value| × 100%
First, let's find the absolute difference between the actual velocity (34.15 m/s) and the measured velocity (34.87 m/s):
|34.15 m/s - 34.87 m/s| = |(-0.72 m/s)| = 0.72 m/s
Now, we calculate the percentage error:
Percentage Error = (0.72 m/s / 34.15 m/s) × 100% ≈ 2.11%
Since the percentage error they obtained (2.11%) is less than the maximum allowed error of 2.5%, Jennifer and Johnny will pass their final project.
Question Part Points Submissions Used The difference in potential between the accelerating plates of a TV set is about 218 V. If the distance between these plates is 1.29 cm, find the magnitude of the uniform electric field (N/C) in this region. Consider entering your answer using scientific notation.
Answer:
[tex]1.69\cdot 10^4 N/C[/tex]
Explanation:
The relationship between electric field strength and potential difference is:
[tex]E=\frac{V}{d}[/tex]
where
E is the electric field strength
V is the potential difference
d is the distance between the plates
Here we have
V = 218 V
d = 1.29 cm = 0.0129 m
So, the electric field is
[tex]E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C[/tex]
The minimum distance required to stop a car moving at 35.6 mi/h is 40.8 ft. What is the minimum stopping distance, in ft, for the same car moving at 71.5 mi/h, assuming the same rate of acceleration? (60.0 mi/h = 88.0 ft/s).
Answer:
Minimum stopping distance = 164.69 ft
Explanation:
Speed of car = 35.6 mi/hr = 15.82 m/s
Stopping distance = 40.8 ft = 12.44 m
We have equation of motion
v² = u² + 2as
0²=15.82²+ 2 x a x 12.44
a = -10.06 m/s²
Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.
Speed of car = 71.5 mi/h = 31.78 m/s
We have
v² = u² + 2as
0² = 31.78² - 2 x 10.06 x s
s = 50.2 m = 164.69 ft
Minimum stopping distance = 164.69 ft
The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on an electron with kinetic energy of 1 keV? Compare with the gravitational force on the electron.
Answer:
[tex]F = 1.5 \times 10^{-16} N[/tex]
this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force
Explanation:
Kinetic Energy of the electron is given as
[tex]KE = 1 keV[/tex]
[tex]KE = 1 \times 10^3 (1.6 \times 10^{-19}) J[/tex]
[tex]KE = 1.6 \times 10^{-16} J[/tex]
now the speed of electron is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
now we have
[tex]v = \sqrt{\frac{2 KE}{m}}[/tex]
[tex]v = 1.87 \times 10^7 m/s[/tex]
now the maximum force due to magnetic field is given as
[tex]F = qvB[/tex]
[tex]F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})[/tex]
[tex]F = 1.5 \times 10^{-16} N[/tex]
Now if this force is compared by the gravitational force on the electron then it is
[tex]\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}[/tex]
[tex]\frac{F}{F_g} = 1.68 \times 10^{13}[/tex]
so this force is [tex]1.68 \times 10^{13}[/tex] times more than the gravitational force
Final answer:
The maximum magnetic force on an electron in Earth's magnetic field is calculated using the charge of the electron, its velocity derived from kinetic energy, and the field strength. This force is much larger than the gravitational force acting on the electron due to its minuscule mass.
Explanation:
The maximum possible magnetic force on an electron moving through Earth's magnetic field can be found using the formula F = qvB sin(\theta), where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. Since the maximum force occurs when the angle is 90 degrees (sin(\theta) = 1), we first need to calculate the velocity from its kinetic energy, 1 keV. Given the kinetic energy E = 1 keV = 1 × 103 eV and using the relation E = \frac{1}{2}mv2, we can find v. The electron charge is q = -1.6 × 10-19 C, and the magnetic field is B = 0.5 G = 5 × 10-5 T.
For the gravitational force, we use F = mg, where m is the mass of the electron and g is the acceleration due to gravity, approximately 9.8 N/kg on the surface of Earth. The mass of the electron is about 9.11 × 10-31 kg. Comparing these forces, the magnetic force is several orders of magnitude greater than the gravitational force acting on an electron due to its negligible mass.
A soccer player hits a ball with a velocity of 22 m/s at an angle of 36.9 above the horizontal. Air resistance can be ignored. a. b. c. (3) What are the x and y components of the balls initial velocity? (3) How high does the ball go? (4) How long (time) does it take to get to the maximum height?
An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 3.40 rad/s2. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)?
(a) 4.03 s
The initial angular velocity of the wheel is
[tex]\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s[/tex]
The angular acceleration of the wheel is
[tex]\alpha = -3.40 rad/s^2[/tex]
negative since it is a deceleration.
The angular acceleration can be also written as
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 0[/tex] is the final angular velocity (the wheel comes to a stop)
t is the time it takes for the wheel to stop
Solving for t, we find
[tex]t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s[/tex]
(b) 27.6 rad
The angular displacement of the wheel in angular accelerated motion is given by
[tex]\theta= \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where we have
[tex]\omega_i=13.7 rad/s[/tex] is the initial angular velocity
[tex]\alpha = -3.40 rad/s^2[/tex] is the angular acceleration
t = 4.03 s is the total time of the motion
Substituting numbers, we find
[tex]\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad[/tex]
The grinding wheel takes 38.5 seconds to stop, covering an angular displacement of 105 radians.
(a) To calculate the time it takes for the grinding wheel to stop, we can use the equation: t = ωf / α where ωf is the final angular velocity (0), and α is the angular acceleration (-3.40 rad/s²). Solving gives t = 38.5 seconds.
(b) The total angular displacement can be found using the equation: θ = ωi*t + 0.5*α*t^2 where ωi is the initial angular velocity (1.31 × 10² rev/min converted to rad/s), t is the time found in part (a), and α is the angular acceleration. This gives θ = 105 radians.
An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car travel before stopping?
Answer:
The car traveled the distance before stopping is 90 m.
Explanation:
Given that,
Mass of automobile = 2000 kg
speed = 30 m/s
Braking force = 10000 N
For, The acceleration is
Using newton's formula
[tex]F = ma[/tex]
Where, f = force
m= mass
a = acceleration
Put the value of F and m into the formula
[tex]-10000 =2000\times a[/tex]
Negative sing shows the braking force.
It shows the direction of force is opposite of the motion.
[tex]a = -\dfrac{10000}{2000}[/tex]
[tex]a=-5\ m/s^2[/tex]
For the distance,
Using third equation of motion
[tex]v^2-u^2=2as[/tex]
Where, v= final velocity
u = initial velocity
a = acceleration
s = stopping distance of car
Put the value in the equation
[tex]0-30^2=2\times(-5)\times s[/tex]
[tex]s = 90\ m[/tex]
Hence, The car traveled the distance before stopping is 90 m.
A 33.1 g copper object is launched from a 1.5 m 30° steel incline positioned on the floor by being pulled up a string attached to a 50.0 g mass suspended vertically over a pulley. The object is projected towards a glass table where it lands when it is at the point along its trajectory with the lowest speed. It comes to a halt when it clears the opposite edge of the table. It then falls and lands on 9 physics texts each 5 cm thick that are stacked on the floor on the opposite side of the table. Assume that the table does not have a ledge and the rectangular object experiences 0.05 N of air resistance as it falls towards the books. How far and how long did the object travel and how fast does it hit the books. Assume that at the moment the copper object leaves the incline, the massless string and ideal pulley break off
Explanation:
To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass. We also need to know the kinetic coefficient of friction between the copper object and the steel incline. This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.
Let's divide the path into four sections to stay organized:
a) object moves up incline
b) object is in free fall
c) object slides across table
d) object falls onto books
a)
First, the object slides up the incline. There are four forces acting on the object. Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.
Sum of the forces normal to the incline:
∑F = ma
N - W cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
T - F = ma
T - Nμ = ma
T - mgμ cos θ = ma
Now sum of the forces in the y direction on the hanging mass:
∑F = ma
T - W = M(-a)
T = Mg - Ma
Substituting:
Mg - Ma - mgμ cos θ = ma
Mg - mgμ cos θ = (m + M) a
a = g (M - mμ cos θ) / (m + M)
Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:
a = 4.71 m/s²
So the velocity it reaches at the top of the incline is:
v² = v₀² + 2a(x - x₀)
v² = 0² + 2(4.71)(1.5)
v = 3.76 m/s
The time to reach the top of the incline is:
x = x₀ + v₀ t + ½ at²
1.5 = 0 + (0) t + ½ (4.71) t²
t = 0.798 s
The horizontal distance traveled is:
x = 1.5 cos 30°
x = 1.299 m
And the vertical distance traveled is:
y = 1.5 sin 30°
y = 0.750 m
b)
In the second stage, the object is in free fall. v₀ = 3.76 m/s and θ = 30°. The object lands on the table at the point where its speed is a minimum. This is at the highest point of the trajectory, when the vertical velocity is 0.
v = at + v₀
0 = (-9.8) t + (3.76 sin 30°)
t = 0.192 s
The horizontal distance traveled is:
x = x₀ + v₀ t + ½ at²
x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²
x = 0.625 m
The height it reaches is:
v² = v₀² + 2a(y - y₀)
(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)
y = 0.180 m
c)
The object is now on the glass table. As it slides, there are three forces acting on it. Normal force up, gravity down, and friction to the left.
In the y direction:
∑F = ma
N - W = 0
N = mg
In the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
The time it takes to reach the end of the table is:
v = at + v₀
0 = (-9.8×0.35) t + (3.76 cos 30°)
t = 0.949 s
And the distance it travels is:
v² = v₀² + 2a(x - x₀)
(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)
x = 1.546 m
d)
Finally, the object falls onto the books. There are 9 books, each 5 cm thick, so the height is 45 cm. As the book falls, there are two forces acting on it: drag up and gravity down.
∑F = ma
D - W = ma
0.05 - (0.0331)(9.8) = 0.0331 a
a = -8.29 m/s²
Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m. So the time it takes the object to land is:
y = y₀ + v₀ t + ½ at²
0.45 = 0.93 + (0) t + ½ (-9.8) t²
t = 0.313 s
Adding up the times from all four parts, the total time is:
t = 0.798 + 0.192 + 0.949 + 0.313
t = 2.25 s
Adding up the horizontal distances, the total distance traveled is:
x = 1.299 + 0.625 + 1.546
x = 3.47 m
The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50
Answer:
Liquid's index of refraction, n₁ = 1.27
Explanation:
It is given that,
The critical angle for a liquid in air is, [tex]\theta_c=52^o[/tex]
We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.
Using Snell's law as :
[tex]n_1sin\theta_c=n_2sin\theta_2[/tex]
Here, [tex]\theta_2=90[/tex]
[tex]sin\theta_c=\dfrac{n_2}{n_1}[/tex]
Where
n₂ = Refractive index of air = 1
n₁ = refractive index of liquid
So,
[tex]n_1=\dfrac{n_2}{sin\theta_c}[/tex]
[tex]n_1=\dfrac{1}{sin(52)}[/tex]
n₁ = 1.269
or n₁ = 1.27
Hence, the refractive index of liquid is 1.27
A block slides on a horizontal surface with an initial speed of 10 m/s. If the block takes 2s to come to a stop, what is the coefficient of kinetic friction between the block and the surface? (The answer does not depend on the mass of the block.) (A) 0.1 (B) 0.2 (C) 0.4 (D) 0.5
Answer:
The coefficient of kinetic friction between the block and the surface is 0.5
Explanation:
It is given that,
Initial velocity of the block, u = 10 m/s
Time taken by the block to come to rest, t = 2 s
So, final velocity, v = 0
We need to find the coefficient of kinetic friction between the block and the surface. According to second law of motion :
F = ma
And friction force F = -μmg
i.e.
[tex]\mu mg=ma[/tex]
[tex]\mu=\dfrac{a}{g}[/tex]...........(1)
Firstly, we will find the value of a i.e. acceleration
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-10\ m/s}{2\ s}[/tex]
a = -5 m/s²
So, equation (1) becomes :
[tex]\mu=\dfrac{5\ m/s^2}{9.8\ m/s^2}[/tex]
[tex]\mu=0.5[/tex]
So, the coefficient of kinetic friction between the block and the surface is 0.5. hence, this is the required solution.
A reservoir has a surface area of 30 km^2 and an average depth of 49 m. What mass of water is held behind the dam?
Answer:
Mass of water, [tex]m=1.47\times 10^{12}\ kg[/tex]
Explanation:
Given that,
Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²
Average depth, d = 49 m
The volume of the reservoir is V such that,
[tex]V=A\times d[/tex]
[tex]V=3\times 10^7\ m^2\times 49\ m[/tex]
[tex]V=1.47\times 10^9\ m^3[/tex]
We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³
Density, [tex]d=\dfrac{m}{V}[/tex]
[tex]m=d\times V[/tex]
[tex]m=1000\ kg/m^3\times 1.47\times 10^9\ m^3[/tex]
[tex]m=1.47\times 10^{12}\ kg[/tex]
Hence, this is the required solution.
A 40000-Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 225 kN. (Round the final answer to the nearest whole number.)
Answer:
197.2 s
Explanation:
First of all, let's convert the mass of the ocean liner into kilograms:
[tex]m=40,000 Mg = 40,000,000 kg = 4\cdot 10^7 kg[/tex]
and the initial velocity into m/s:
[tex]u=4 km/h =1.11 m/s[/tex]
The force applied is
[tex]F=-225 kN = -2.25\cdot 10^5 N[/tex]
So we can find first the deceleration of the liner:
[tex]a=\frac{F}{m}=\frac{-2.25\cdot 10^5 N}{4\cdot 10^7 kg}=-5.63\cdot 10^{-3} m/s^2[/tex]
And now we can use the following equation:
[tex]a=\frac{v-u}{t}[/tex]
with v = 0 being the final velocity, to find t, the time it takes to bring the liner to rest:
[tex]t=\frac{v-u}{a}=\frac{0-1.11 m/s}{-5.63\cdot 10^{-3} m/s^2}=197.2 s[/tex]
An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the primary, what voltage and current are present in the secondary?
Explanation:
It is given that,
Number of turns in primary coil, [tex]N_p=60[/tex]
Number of turns in secondary coil, [tex]N_s=300[/tex]
Voltage in primary coil, [tex]V_p=120\ V[/tex]
Current in primary coil, [tex]I_p=2\ A[/tex]
We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]
[tex]\dfrac{60}{300}=\dfrac{120}{V_s}[/tex]
[tex]V_s=720\ Volts[/tex]
Now, calculating the current present in the secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}[/tex]
[tex]\dfrac{60}{300}=\dfrac{I_s}{2\ A}[/tex]
[tex]I_s=0.4\ A[/tex]
Hence, this is the required solution.
Final answer:
The voltage present in the secondary is 600 V and the current is 0.4 A.
Explanation:
The question involves an ideal transformer where the primary coil has 60 turns and the secondary coil has 300 turns, with 120 V at 2.0 A applied to the primary. To find the voltage and current present in the secondary, we can use the transformer equations:
Vp/Vs = Np/Ns (where V is voltage and N is the number of turns in the primary (p) and secondary (s) coils)
According to the conservation of energy in an ideal transformer, Pp = Ps (where P is power), and since P = VI, we get:
VpIp = VsIs (where I is current)
Using these equations:
Is = (Np/Ns) \\times Ip = (60/300) \\times 2.0 A = 0.4 A
Thus, the voltage present in the secondary is 600 V and the current is 0.4 A.
A soft drink manufacturer claims that a new diet drink is "low Joule." The label indicates the available energy per serving is 6300 J. What is the equivalent of this energy in Calories (1 Calorie = 1000 cal)?
Answer:
6300 J = 1.51 Calories
Explanation:
We have 1 Calorie = 1000 cal
1 cal = 4.184 J
That is 1 Calorie = 1000 cal = 1000 x 4.184 = 4184 J
Here we have 6300 J
That is
[tex]1Calorie=4184J\\\\1J=\frac{1}{4184}Calorie\\\\6300J=\frac{6300}{4184}Calorie=1.51Calorie[/tex]
6300 J = 1.51 Calories
A person walks first at a constant speed of 4.89 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.95 m/s. What is the average speed over the entire trip?
Answer:
approximately 3.68
Explanation:
Explained in the attached picture.
A nuclear power plant operates at 40.0% efficiency with a continuous production of 1192 MW of usable power in 1.00 year and consumes 1.22×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?
Final answer:
To calculate the energy released by fission of a single U-235 atom, convert the power output to total energy in joules, adjust for plant efficiency, and divide by the number of fissioned U-235 atoms, using Avogadro's number and U-235's molar mass.
Explanation:
The question involves calculating the energy released from the fission of a single uranium-235 (U-235) atom, given the power output and efficiency of a nuclear power plant. First, determine the total energy produced by the power plant in a year by converting megawatts to joules. Since 1 watt equals 1 joule per second (J/s), and there are 3.1536×107 seconds in a year, the total energy output in joules is 1192 MW × 3.1536×107 s/year × 106 W/MW. Taking into consideration the 40.0% efficiency of the plant, the total energy from fission is obtained by dividing the calculated energy output by 0.40. Finally, to find the energy per fission event, divide the total fission energy by the number of grams of U-235 consumed, converted to number of atoms using Avogadro's number, which is approximately 6.022×1023 atoms/mol. Using the atomic mass of U-235, which is roughly 235 grams per mole, we can calculate the energy per fission event.
A 93.5 kg snowboarder starts from rest and goes down a 60 degree slope with a 45.7 m height to a rough horizontal surface that is 10.0 m long and a coefficient of kinetic friction of 0.102. The then snowboarder goes up a slope with angle of 30 degrees up coming to a stop and then sits to stay in the same place. be considered frictionless. a. What is the speed of the snowboarder at the bottom of the 60 degree slope? b. What is the speed of the snowboarder at the bottom of the 30 degree slope? c. What is the final height of the snowboarder above the horizontal surface? until Both slopes are smooth and can
Answer:
a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m
Explanation:
This can be answered with either force analysis and kinematics, or work and energy.
a) Using force analysis, we can draw a free body diagram for the snowboarder. There are two forces: normal force perpendicular to the slope and gravity down.
Sum of the forces parallel to the slope:
∑F = ma
mg sin θ = ma
a = g sin θ
Therefore, the velocity at the bottom is:
v² = v₀² + 2a(x - x₀)
v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)
v = 29.9 m/s
Alternatively, using energy:
PE = KE
mgh = 1/2 mv²
v = √(2gh)
v = √(2×9.8×45.7)
v = 29.9 m/s
b) Drawing a free body diagram, there are three forces on the snowboarder. Normal force up, gravity down, and friction to the left.
Sum of the forces in the y direction:
∑F = ma
N - mg = 0
N = mg
Sum of the forces in the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Therefore, the snowboarder's final speed is:
v² = v₀² + 2a(x - x₀)
v² = (29.9)² + 2(-9.8×.102) (10 - 0)
v = 29.6 m/s
Using energy instead:
KE = KE + W
1/2 mv² = 1/2 mv² + F d
1/2 mv² = 1/2 mv² + mgμ d
1/2 v² = 1/2 v² + gμ d
1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)
v = 29.6 m/s
c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.
∑F = ma
-mg sin θ = ma
a = -g sin θ
Therefore, the final height reached is:
v² = v₀² + 2a(x - x₀)
(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)
h = 44.7 m
Using energy:
KE = PE
1/2 mv² = mgh
h = v² / (2g)
h = (29.6)² / (2×9.8)
h = 44.7 m
A boxcar traveling at 12 m/s approaches a string of 5 identical boxcars sitting stationary on the track. The moving boxcar collides and links with the stationary cars and they all move off together along the track. What is the final speed of the cars immediately after the collision? (You may take the mass of each boxcar to be 18,537 kg.)
Answer:
2 m/s
Explanation:
Momentum is conserved:
mv = MV
where m is mass of the boxcar, v is its initial velocity, M is the mass of all six box cars, and V is the final velocity.
v = 12 m/s, and M = 6m, so:
m (12 m/s) = 6m V
12 m/s = 6V
V = 2 m/s
Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.7-V battery, the current from the battery is 1.81 A. When the resistors are connected in parallel to the battery, the total current from the battery is 9.06 A. Determine R1 and R2. (Enter your answers from smallest to largest.)
In the case of resistors in series and in parallel to a battery, the resistances can be determined by applying Ohm's law and using equations for total resistance in both series and parallel configurations. The calculations yield the values R1 = 1.53 ohms and R2 = 5.48 ohms.
Explanation:
The problem involves applying Ohm's law and equations for series and parallel resistors. When the resistors are in series, we add the resistances so R1 + R2 = 12.7 V / 1.81 A = 7.01 ohms. In a parallel connection, the total resistance R is given by 1/R = 1/R1 + 1/R2 or 12.7 V / 9.06 A = 1.40 ohms. Solving this pair of equations we get R1 = 1.53 ohms and R2 = 5.48 ohms, giving the pair of resistances asked for.
Learn more about Ohm's law here:https://brainly.com/question/36009177
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The distance between two successive minima of a transverse wave is 3.10 m. Five crests of the wave pass a given point along the direction of travel every 14.7 s. (a) Find the frequency of the wave. Hz (b) Find the wave speed. m/s
Answer:
The frequency is 0.34 Hz and the wave speed is 1.05 m/s.
Explanation:
Given that,
The distance between two successive minima of a transverse wave is wavelength.
Wavelength = 3.10 m
Time = 14.7 s
(I). We need to calculate the frequency
[tex]f=\dfrac{number\ of\ wave\ propagated\ per\ second }{times}[/tex]
[tex]f=\dfrac{5}{14.7}[/tex]
[tex]f=0.34\ Hz[/tex]
(II). We need to calculate the wave speed
Formula of wave speed
[tex]v= \lambda\times f[/tex]
Where,
[tex]\lambda = wavelength[/tex]
[tex]f = frequency[/tex]
Put the value into the formula
[tex]v=3.10\times0.34[/tex]
[tex]v=1.05\ m/s[/tex]
Hence, The frequency is 0.34 Hz and the wave speed is 1.05 m/s.
The frequency of the wave is 0.34 Hz and the wave speed is 1.054 m/s.
Explanation:The distance between two successive minima in a transverse wave is the same as the wavelength (λ). Given that the question provided this distance as 3.10 m, we can use this as the wavelength. On the other hand, the frequency (f) of the wave is the number of wave cycles that pass a point per unit time. The information from the question gives us five crests (complete cycles) over 14.7 seconds. Thus, we can calculate the frequency by dividing the total cycles by the total time, i.e., f = 5 cycles / 14.7 seconds = 0.34 Hz.
The speed of the wave (v) can be obtained from the equation v = λf. Substituting the values we have, v = (3.10 m) * (0.34 Hz) = 1.054 m/s. Therefore, the frequency of the wave is 0.34 Hz, and the wave speed is 1.054 m/s.
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If the work required to stretch a spring 3 ft beyond its natural length is 15 ft-lb, how much work is needed to stretch it 27 in. beyond its natural length?
Answer:
8.4 ft-lb
Explanation:
Work = change in energy
W = ½ kx²
When x = 3 ft, W = 15 ft-lb:
15 ft-lb = ½ k (3 ft)²
k = 30/9 lb/ft
When x = 27 in = 2.25 ft:
W = ½ kx²
W = ½ (30/9 lb/ft) (2.25 ft)²
W = 8.4375 ft-lb
Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.
Final answer:
The work required to stretch the spring 27 in. beyond its natural length is 12.728 N-m.
Explanation:
To find the work required to stretch the spring 27 in. beyond its natural length, we can use the work done formula:
Work = (1/2)kx²
Where k is the spring constant and x is the displacement of the spring. We know that the work required to stretch the spring 3 ft beyond its natural length is 15 ft-lb. Let's convert the given measurements into consistent units:
3 ft = 36 in.
15 ft-lb = 12.728 N-m
Now, we can solve for the work required to stretch the spring 27 in. beyond its natural length:
(1/2)k(27)² = (1/2)k(729) = 12.728 N-m
We can rearrange the equation to solve for k:
k = (2 * 12.728) / 729 = 0.0349 N/m
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass M. The escape velocity is a function of the distance of the object from the center of the planet R, but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?"
Answer: [tex]V=\sqrt{\frac{2GM}{R}}[/tex]
Explanation:
Taking into account what is stated in this problem and considering there is no friction during the takeoff of the rocket of the planet, the rocket will escape the gravitational attraction of the massive body when its kinetic energy [tex]K[/tex] and its potential energy [tex]P[/tex] are equal in magnitude.
Written mathematically is:
[tex]K=P[/tex] (1)
Where:
[tex]K=\frac{1}{2}mV^{2}[/tex] (2)
Being [tex]m[/tex] the mass of the rocket
And:
[tex]P=-\frac{GMm}{R}[/tex] (3)
Being [tex]M[/tex] the mass of the planet, [tex]G[/tex] the gravitational constant and [tex]R[/tex] the radius of the planet.
Substituting (2) and (3) in (1):
[tex]\frac{1}{2}mV^{2}=-\frac{GMm}{R}[/tex] (4)
Finding [tex]V[/tex], which is the escape velocity:
[tex]V=\sqrt{\frac{2GM}{R}}[/tex] this is the velocity the rocket must have in order to escape from the surface of the planet
A particle leaves the origin with an initial velocity of 4.90 m/s in the x direction, and moves with constant acceleration ax = -1.50 m/s^2 and ay = 3.00 m/s^2. How far does the particle move in the x direction before turning around?
Answer:
it will move by d = 8.00 m in x direction before it will turn back
Explanation:
Here the initial velocity of particle is along +x direction
it is given as
[tex]v_i = 4.90 m/s[/tex]
now its acceleration is given as
[tex]a_x = -1.50 m/s^2[/tex]
[tex]a_y = 3.00 m/s^2[/tex]
now when it turns back then the velocity in x direction will become zero
so we will say
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 4.90^2 = 2(-1.50) d[/tex]
[tex] d = 8.00 m[/tex]
so it will move by d = 8.00 m in x direction before it will turn back
Final answer:
The particle moves 8.01 m in the x direction before it turns around, determined by using kinematic equations with the given initial velocity and constant acceleration.
Explanation:
To determine how far the particle moves in the x direction before turning around, we need to find the point at which its velocity in the x direction is zero. Since the particle starts with an initial velocity of 4.90 m/s in the x direction and has a constant acceleration of ax = -1.50 m/s², we can use the kinematic equation v = u + at to find the time when the velocity becomes zero:
v = 0 m/s (the velocity at the turnaround point)
u = 4.90 m/s (initial velocity)
a = -1.50 m/s² (constant acceleration)
Solving for the time (t) gives:
0 = 4.90 + (-1.50)t
t = 4.90 / 1.50
t = 3.27 s
Now using the equation s = ut + (1/2)at² to find the displacement in the x direction:
s = 4.90 m/s ⋅ t + (1/2)(-1.50 m/s²) ⋅ t²
s = 4.90 ⋅ 3.27 + (0.5) ⋅ (-1.50) ⋅ (3.27)²
s = 8.01 m
The particle moves 8.01 m in the x direction before turning around.
What is a plank in science?
Answer:
It's a long explanation
Explanation:
The Planck mass is approximately the mass of the Planck particle, a hypothetical minuscule black hole whose Schwarzschild radius equals the Planck length. In physics, the Planck mass, denoted by mP, is the unit of mass in the system of natural units known as Planck units. It is approximately 0.02 milligrams. Plank length as the minimum distance at which space can be thought. In physics, the Planck length, denoted ℓP, is a unit of length that is the distance light travels in one unit of Planck time. It is equal to 1.616255×10⁻³⁵ m. It is a base unit in the system of Planck units, developed by physicist Max Planck...I miss you, Landon.
What is the escape speed (in km/s) from an Earth-like planet with mass 6.5e+24 kg and radius 55.0 × 105 m? Use the gravitational constant G = 6.67 × 10-11 m3 kg-1 s-2.
Answer:
Escape velocity, v = 12.6 km/s
Explanation:
It is given that,
Mass of earth like planet, [tex]m=6.5\times 10^{24}\ kg[/tex]
Radius of planet, [tex]r=55\times 10^5\ m[/tex]
The escape velocity of a planet is given by the following formula as:
[tex]v=\sqrt{\dfrac{2Gm}{r}}[/tex]
G = universal gravitational constant
[tex]v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\ m^3kg^{-1}s^{-2}\times 6.5\times 10^{24}\ kg}{55\times 10^5\ m}[/tex]
v = 12556.05 m/s
or
v = 12.6 km/s
Hence, this is the required solution.
A ball is whirled on the end of a string in a horizontal circle of radius R at const v. By which one of the following means can the centripetal acceleration of the b increased by a factor of two? 13. Keep the radius fixed and increase the period by a factor of two. Keep the radius fixed and decrease the period by a factor of two. Keep the speed fixed and increase the radius by a factor of two. Keep the speed fixed and decrease the radius by a factor of two. A) B) C) D)
Answer:
D
Explanation:
Centripetal acceleration is the square of velocity divided by radius:
a = v² / r
Velocity is equal to angular velocity times radius, so this can also be written as:
a = ω² r
And angular velocity is 2π divided by the period, so:
a = (2π / T)² r
We can use these two equations to determine which scenario results in double the centripetal acceleration.
A) Keep the radius fixed and increase the period by a factor of two.
(2π / 2T)² r
1/4 (2π / T)² r
1/4 a
B) Keep the radius fixed and decrease the period by a factor of two.
(2π / ½T)² r
4 (2π / T)² r
4a
C) Keep the speed fixed and increase the radius by a factor of two.
v² / 2r
1/2 a
D) Keep the speed fixed and decrease the radius by a factor of two.
v² / ½r
2a
Only D doubles the centripetal acceleration.
To double the centripetal acceleration of a ball moving in a circular path at a constant velocity, halve the radius of the path while keeping the speed constant. This is because centripetal acceleration is inversely proportional to the radius.
Explanation:This is possible because centripetal acceleration is inversely proportional to the radius (as described by the formula ac = v²/r). Conversely, increasing the radius or altering the period would not have the desired effect on the centripetal acceleration.
If the radius is halved and the speed remains constant, the force required to keep the object moving in a circular path increases, which is reflected in an increase in the centripetal acceleration. Hence, if you were to plot centripetal acceleration against radius, you would see it decrease as the radius increases and vice versa.
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Tom kicks a soccer ball on a flat, level field with initial speed 20 m/s at an angle 35 degrees above the horizontal. (1) How long will the ball stay in the air (2) Max height will ball attain (3) How far away from tom will the ball land (4) what speed will the ball have in the instant just before it lands
Explanation:
Given:
x₀ = 0 m
y₀ = 0 m
v₀ = 20 m/s
θ = 35°
aᵧ = -9.8 m/s²
1) Find t when y = 0.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
0 = 0 + (20 sin 35°) t + ½ (-9.8) t²
0 = t (20 sin 35° - 4.9 t)
t = 0, t = 2.34
The ball stays in the air 2.34 seconds.
2) Find y when vᵧ = 0.
vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)
0² = (20 sin 35)² + 2(-9.8) (y - 0)
y = 6.71 m
The ball reaches a maximum height of 6.71 meters.
3) Find x when y = 0.
x = x₀ + v₀ₓ t + ½ aₓ t²
x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²
x = 38.4 m
The ball lands 38.4 meters from Tom.
4) Find v when y = 0.
vₓ = aₓ t + v₀ₓ
vₓ = (0) (2.34) + 20 cos 35°
vₓ = 16.4 m/s
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8) (2.34) + 20 sin 35°
vᵧ = -11.5 m/s
v = √(vₓ² + vᵧ²)
v = √((16.4)² + (-11.5)²)
v = 20 m/s
The ball has a speed of 20 m/s just before it lands.
A soccer player kicks a ball horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air is 343 m/s. Determine the speed of the ball immediately before it reaches the pool of water.
the speed of the ball before it reaches the pool of water would be 9.91 m/s.
The initial speed given to the ball is 32.35m/s.
The speed of the ball immediately before it reaches the pool of water is 28m/s
What is Speed ?Speed is a rate of change of distance with respect to time. i.e. v=dx÷dt. Speed can also be defined as distance over time i.e. speed= distance ÷ time it is denoted by v and its SI unit is m/s. it is a scalar quantity. Speed shows how much distance can be traveled in unit time.
To find dimension for speed is, from formula
Speed = Distance ÷ Time
Dimension for distance is [L¹] ,
Dimension for Time is [T¹],
Dividing dimension of distance by dimension of time gives,
[L¹] ÷ [T¹] = [L¹T⁻¹]
Dimension for speed is [L¹T⁻¹].
In this problem,
Given,
Hight of the cliff y = 40m
speed of the sound = 343 m/s
vertical direction,
y = u(y)t + 1/2 at² where u(y) is initial velocity of ball along y direction which is zero.
-40 = 0t + 1/2 (-9.8)*t²
80 = 9.8t²
t = 2.85s
time taken by sound to hear is = 3-2.85 = 0.15s
distance covered by the sound,
343m/s × 0.15s = 51.45m
By applying Pythagoras theorem,
AC = √(x² + y²)
51.45² = x²+40²
x = 32.35
The initial velocity of the ball is,
u = 32.35m / 2.85s = 11.35m/s
The speed of the ball immediately before it reaches the pool of water is
v² = u² + 2as
v² = 0 +2(-9.8)40m
here initial velocity of the ball along y direction is 0. and distance along it is 40m.
v² = 784
v = 28m/s ......this is velocity along verticle.
Hence initial velocity of ball is 32.35m/s.
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In 1960 military pilot Joseph Kittinger jumped from a high altitude balloon in a fully pressurized jump suit about 100,000 feet, in the stratosphere, all the way to the ground. When in the stratosphere he felt like he was floating, even though he was free-falling in space. When he reached the troposphere he could feel the wind resistance and it no longer felt like he was floating. Why was this likely the case?
Answer:
because the troposphere has more oxygen than the stratosphere so when he felt like he was floating even though he was free falling it was due to the absence of air.
Which one of the following statements concerning superconductors is false? A constant current can be maintained in a superconducting ring for several years without an emf. Superconductors are perfect conductors. Critical temperatures for some superconductors exceed 100 K. All materials are superconducting at temperatures near absolute zero kelvin. Below its critical temperature, the resistivity of a superconductor is zero Ω-m
Answer:
All materials are superconducting at temperatures near absolute zero kelvin.
Explanation:
All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.
c. All materials are superconducting at temperatures near absolute zero kelvin.
What is superconductors?A superconductor is a material that achieves superconductivity, which is a state of matter that has no electrical resistance and does not allow magnetic fields to penetrate.
An electric current in a superconductor can persist indefinitely. Superconductivity can only typically be achieved at very cold temperatures.
Superconductivity is a phenomenon observed in several metals and ceramic materials.
When these materials are cooled to temperatures ranging from near absolute zero ( 0 degrees Kelvin, -273 degrees Celsius) to liquid nitrogen temperatures ( 77 K, -196 C), their electrical resistance drops with a jump down to zero.
Therefore,
All materials are superconducting at temperatures near absolute zero kelvin.
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A boat is traveling upstream at 10 km/h with respect to the water of a river. The water is flowing at 8.0 km/h with respect to the ground. What are the (a) magnitude and (b) direction of the boat’s velocity with respect to the ground? A child on the boat walks from front to rear at 4.0 km/h with respect to the boat.What are the (c) magnitude and (d) direction of the child’s velocity with respect to the ground?
Final answer:
The magnitude of the boat’s velocity with respect to the ground is 18.0 km/h downstream. The child’s velocity with respect to the ground is 14.0 km/h upstream.
Explanation:
To find the magnitude and direction of the boat’s velocity with respect to the ground, we can use vector addition. The boat's velocity with respect to the ground is the sum of its velocity with respect to the water and the water's velocity with respect to the ground. The magnitude of the boat's velocity with respect to the ground is the sum of its speed upstream and the speed of the water downstream: 10 km/h + 8.0 km/h = 18.0 km/h. The direction of the boat's velocity with respect to the ground is the same as the direction of the water's velocity, which is downstream.
To find the magnitude and direction of the child’s velocity with respect to the ground, we can also use vector addition. The child's velocity with respect to the ground is the sum of their velocity with respect to the boat and the boat's velocity with respect to the ground. The magnitude of the child's velocity with respect to the ground is the difference between their speed on the boat and the boat's speed with respect to the ground: 4.0 km/h - 18.0 km/h = -14.0 km/h (negative sign indicating direction opposite to the boat's velocity with respect to the ground). Therefore, the magnitude of the child's velocity with respect to the ground is 14.0 km/h. The direction is upstream, opposite to the direction of the boat's velocity with respect to the ground.