Answer:
D=1.54489 m
Explanation:
Given data
S=6.10 mm= 0.0061 m
To find
Depth of lake
Solution
To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation
[tex]S=v_{1}t+(1/2)gt^{2} \\ 0.0061m=(0m/s)t+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{0.0061m}{4.9m/s^{2} }\\ t=\sqrt{1.245*10^{-3} }\\ t=0.035s[/tex]
So ball takes 0.035sec to hit the water
As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as
[tex]v_{f}=v_{i}+gt\\v_{f}=0+(9.8m/s^{2} )(0.035s)\\ v_{f}=0.346m/s[/tex]
Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake
So the depth of lake given as:
[tex]D=|vt|\\D=|0.346m/s*4.465s|\\D=1.54489m[/tex]
Answer: d = 1.54m
The depth of the lake is 1.54m
Explanation:
The final velocity of the ball just before it hit the water can be derived using the equation below;
v^2 = u^2 + 2as ......1
Where ;
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance travelled.
Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:
v^2 = 2gs
v = √2gs ......2
g = 9.8m/s^2
s = 6.10mm = 0.0061m
substituting into equation 2
v = √(2 × 9.8× 0.0061)
v = 0.346m/s
The time taken for the ball to hit water from the time of release can be given as:
d = ut + 0.5gt^2
Since u = 0
d = 0.5gt^2
Making t the subject of formula.
t = √(2d/g)
t = √( 2×0.0061/9.8)
t = 0.035s
The time taken for the ball to reach the bottom of the lake from the when it hits water is:
t2 = 4.5s - 0.035s = 4.465s
And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;
depth d = velocity × time = 0.346m/s × 4.465s
d = 1.54m
The depth of the lake is 1.54m
If the length of the air column in the test tube is 14.0 cm, what is the frequency of this standing wave?
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :
[tex]f=\dfrac{nv}{4l}[/tex]
[tex]f=\dfrac{1\times 344}{4\times 0.14}[/tex]
f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then load up the pickup and pump up itstires so that its total weight increases by 42% whilethe coefficient of rolling friction decreases by19%.
a) Now what horizontal force will you need to move the pickupalong the same road at the same speed? The speed is low enough thatyou can ignore air resistance.
Answer:
[tex]F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
[tex]F_H_n=F_f\\F_h=mg*u[/tex]
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
[tex]200=mg*u[/tex] Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
[tex]F_H=(1.42m*g*.81u)[/tex] Eq (2)
Divide Eq(2) and Eq (1)
[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N
After a parallel-plate capacitor has been fully charged by a battery, the battery is disconnected and the plate separation is increased. Which of the following statements is correct? Please explain in detail why the staement is correct!
A) The energy stored in the capacitor increases.
B) The charge on the plates increases.
C) The charge on the plates decreases.
D) The potential difference between the plated decreases.
E) The energy stored in the capacitor decreases.
Answer:
A) The energy stored in the capacitor increases.
Explanation:
For a capacitor fully charged by battery, when disconnected from battery and the plate separation is increased. The charge on the plate remain constant because there is no where for it to go( it has been disconnected from battery). But the capacitance would decrease, while also the potential difference would increase.
Q = CV ....1
Q is the charge, C is capacitance, V is the potential difference.
The energy stored in a capacitor is given by:
E = 1/2 CV^2
E = 1/2 QV .......2
E is the energy stored in the capacitor,
Therefore since Q remain constant and V increases when the distance between the plates is increased, then according to the equation 2 above the energy stored in the capacitor increases.
What is the ratio of the intensities and amplitudes of an earthquake P wave passing through the Earth and detected at two points 27 km and 13 km from the source?
(a) I 27 / 13 =__________
(b) A 27 / 13 =_____________-
For both cases we will use the proportional values of the distance referring to the amplitude and intensity. Theoretically we know that the intensity is inversely proportional to the square of the distance, while the amplitude is inversely proportional to the distance, therefore,
PART A ) Intensity is inversely proportional to the square of the distance
[tex]Intensity \propto \frac{1}{distance^2}[/tex]
Therefore the intensity of the two values would be
[tex]\frac{I_{27}}{I_{13}} = \frac{(13km)^2}{(27km)^2}[/tex]
[tex]\mathbf{\therefore \frac{I_{27}}{I_{13}} = 0.232 }[/tex]
PART B) Amplitude is inversely proportional to the distance
[tex]Amplitude \propto \frac{1}{distance}[/tex]
[tex]\frac{A_{27}}{A_{13}}= \frac{(13km)}{(27km)}[/tex]
[tex]\mathbf{\therefore\frac{A_{27}}{A_{13}}= 0.4815}[/tex]
The intensity ratio of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the amplitude ratio.
Explanation:The ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the ratio of their amplitudes.
Let's assume the amplitudes of the earthquake P wave at the two points are given by A27 and A13.
The intensity of a wave is given by the square of its amplitude. Therefore, the ratio of the intensities I27/I13 is equal to the square of the ratio of the amplitudes A27/A13.
So, the answer is:
(a) I27/I13 = (A27/A13)2
(b) A27/A13
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Light from the star Betelgeuse takes 640 years to reach Earth. How far away is Betelgeuse in units of light-years? Name any historical event that was occurring on Earth at about the time the light left Betelgeuse. Is the distance to Betelgeuse unusual compared with other stars?
Answer:
The distance is 641.8207 light years, and the star Betelgeuse is further away when compared to other stars
Historical event: Benedict XI succeeds Boniface VIII as pope (1302)
Explanation:
the solution is in the attached Word file
A charged comb often attracts small bits of dry paper that then fly away when they touch the comb. Explain why that occurs.
Explanation:
Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.
If a scuba diver fills his lungs to full capacity of 5.7 L when 8.0 m below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume he dives in the sea, thus the water is salt. Express your answer using two significant figures
To calculate the pressure in the body we will use the definition of the hydrostatic pressure for which the pressure of a body at a certain distance submerged in a liquid is defined. After calculating this relationship we will apply the equations of the relationship between the volume and the pressure to calculate the volume in state 2,
[tex]P = P_{atm} + \rho gh[/tex]
Here,
[tex]\rho[/tex]= Density of the Fluid (Water)
g = Acceleration due to gravity
h = Height
[tex]P = P_{atm} + 10^3*9.8*8[/tex]
[tex]P = 1.01*10^{5} +10^3*9.8*8[/tex]
[tex]P = 179400Pa[/tex]
Applying the equations of relationship between volume and pressure we have
[tex]P_1V_1 = P_2 V_2[/tex]
[tex]179400*5.7 = 101000*V_2[/tex]
[tex]V_2 = 10.12L[/tex]
Therefore the volume that would his lungs expand if he quickly rose to the surface is 10.12L
Calculate the percent weight reduction of the plane as it moves from the sea level to 11,719.342 m.
Answer: %(∆W) = 0.37%
Explanation:
According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
F = Gm1m2/r^2
Where
F = force between the masses
G universal gravitational constant
m1 and m2 = mass of the two particles
r = distance between the centre of the two mass
Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth
W₁/W₂ = r₂²/r₁² .....1
W₂ = W₁r₁²/r₂²
At sea level the weight of the plane is W1 and at distance r₁ from the centre of the earth which is equal to the radius of the earth.
The radius of the earth is = 6378.1km
r₁ = radius of the earth = 6 378.1km = 6,378,100m
r₂ = r₁ + 11,719.342m = 6,378,100m + 11,719.342m
r₂ = 6,389,819.342m
W₂ = W₁r₁²/r₂²
W₂ = W₁[(6378100)²/(6,389,819.342)²]
W₂ = W₁[0.996335234422]
W₂/W₁ = 0.9963
fraction reduction of the weight is
ΔW/W₁ = 1 - W₂/W₁ = 1 - 0.9963 = 0.0037
percentage change :
%(∆W) = 0.0037 × 100% = 0.37%
Therefore, the percentage weight loss is 0.37%
Vector vector A has a magnitude A and is directed at an angle theta measured with respect to the positive x-axis. What is the magnitude of vector A sub x, the x-component of vector A?
Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]
Explanation:
A vector in this situation have two components
1) The X-component
2) The Y-component
So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be
[tex]A_x[/tex] = Acos[tex]\theta[/tex]
I hope this will answer your question,
An image is also provided please have a look at that.
Thank you.
Answer: Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
Explanation:
The attached image is a pictorial representation of the question:
From the attached image,
Cos(theta) = adjacent/hypothenus
Cos(theta) = Ax/A
Making Ax the subject of formula,
Ax = Acos(theta)
Therefore, the x component of A: Ax = Acos(theta)
A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the velocity of the skater?
a. -0.07 m/s
b. 0.15 m/s
c. 0.30 m/s
d. 0.07 m/s
Answer:
(d) 0.07 m/s
Explanation:
Given Data
Snowball mass m₁=0.15 kg
Ice skater mass m₂=65.0 kg
Snowball velocity v₁=32.0 m/s
To find
Velocity of Skater v₂=?
Solution
From law of conservation of momentum
[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]
So Option d is correct one
The velocity of skater is [tex]0.07m/s[/tex]
Option d is correct.
Law of conservation of momentum:For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.
The expression is given as,[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
Given that, mass of ice skater [tex]m_{1}=65kg[/tex], mass of snow ball [tex]m_{2}=0.15kg[/tex] Velocity of snow ball [tex]v_{2}=32m/s[/tex]Substitute values in above expression.[tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]
The velocity of skater is [tex]0.07m/s[/tex]
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According to the U.S. Green Building Council, what percentage of the world’s energy use and greenhouse gas emissions can be attributed to buildings?
According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions
Answer:
40%
Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.
A student who grew up in a tropical country and is studying in the United States may have no experience with static electricity sparks and shocks until their first American winter. Explain.
Explanation:
Water in the air (humidity) helps to dissipate static charge that builds up. If the air is very dry, the charge can't dissipate, so it builds up until there is enough to spark.
Tropical countries are typically more humid than the United States, but I guess that depends on where you are in the US.
A student's first experience with static electricity in an American winter is due to the dry air, which allows for greater accumulation and discharge of electrical charges, unlike in humid tropical climates.
This is because static electricity is more prevalent in cold, dry environments.
In tropical countries, the air tends to have higher humidity, which allows electrical charges to dissipate more easily. Moist air is a better conductor of electricity, thereby reducing the likelihood of a significant buildup of static charge.In contrast, during an American winter, the air is typically cold and dry. Dry air is a poor conductor, meaning that electrical charges are more likely to accumulate on surfaces and objects, leading to more frequent static electricity shocks. When you walk on a rug or take off a woollen sweater in such conditions, electrons can be transferred between your body and the surfaces, creating an imbalance.As a result, when you touch a metal object like a doorknob, the built-up charge is suddenly released, creating a spark and a shock. Even though the spark may carry a few hundred watts of power, it happens so quickly and involves such a small amount of current that it doesn't cause injury. The brief duration and low current ensure that the shock is felt but not harmful.In summary, the primary reason a student may experience static electricity for the first time in an American winter is due to the dry air conditions that favour the accumulation and sudden discharge of electric charges.
An equilateral triangle with side lengths of 0.50 m has a 5.0 nC charge placed at each corner. What is the magnitude of the electric field at the midpoint of one of the three sides? (A) 240 N/C (B) 180 N/C (C) 720 N/C (D) 480 N/C (E) 120 N/C
To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,
[tex]x = \frac{\sqrt{3}}{2}a[/tex]
[tex]x = \frac{\sqrt{3}}{2}(0.5)[/tex]
[tex]x = 0.43301m[/tex]
The magnitude of the electric field is
E=\frac{kQ}{x^2}
Here,
k = Coulomb's Constant
Q = Charge
x = Distance
[tex]E=\frac{(9*10^9)(5*10^{-9})}{(0.43301)^2}[/tex]
[tex]E = 240.002N/C \approx 240N/C[/tex]
Therefore the correct answer is A.
An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. Suppose that a typical car rounds the curve with a speed of 11.7m/s and that the radius of the curve is 50.0m. At what angle should the curve be banked?
To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,
[tex]T_x = nsinA = \frac{mv^2}{r}[/tex]
[tex]T_y = ncosA = mg[/tex]
Dividing both.
[tex]tan A = \frac{v^2}{rg}[/tex]
[tex]tan A = \frac{11.7^2}{50*9.8}[/tex]
[tex]A = tan^{-1} (0.279367)[/tex]
[tex]A = 15.608\°[/tex]
Therefore the angle that should the curve be banked is 15.608°
What is the correct answer
Answer:
D and compound
Explanation:
because N2 is = to a compound
A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 93 ft long. When unloaded its draft (depth of submergence) is 6 ft, and with the load of grain the draft is 9 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.
Answer:
a) [tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
b) [tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
Explanation:
Part a
For this case we have the situation illustrated on Figure 1. We will have two forces involved in equilibrium the weight [tex] W_B[/tex] and the Bouyance force[tex] F_B[/tex], and since the system is on equilibrium we have:
[tex] \sum F_{vertical}=0[/tex]
So then we have:
[tex] W_B = F_B = \gamma_{w} V_s[/tex]
Where [tex] V_s[/tex] represent the submerged volume. [tex]\gamma_w[/tex] represent the specific weight for the fluid. So we can replace and we have:
[tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
Part b
As we can see on figure 2 attached we have the illustration for this case. We add the weight for the grain and now the depth is 9ft.
W can do the balance of forces in the vertical and we got again:
[tex] W_B +W_g = F_B[/tex]
Where [tex] W_g[/tex] represent the weight for the grain.
And if we solve for [tex] W_g[/tex] we got:
[tex] W_g = F_B -W_B[/tex]
[tex] W_g =\gamma_w V_S -W_B[/tex]
Where [tex] \gamma_w[/tex] represent the specific weight of rthe water and [tex] V_s[/tex] the submerged volume. If we replace we got:
[tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
The weight of the unloaded barge is 975,769.6 lbs, and the weight of the grain is 486,572.8 lbs. These calculations were made based on Archimedes' principle, which states that the buoyant force (the weight of water displaced) is equal to the weight of the object.
Explanation:This question is about calculating the weight of a river barge and its load based on the principles of fluid mechanics. Here, we are examining the fact that the weight of the water displaced by the barge equals the weight of the barge according to Archimedes' principle.
Firstly, we calculate the unloaded weight of the barge. The volume of water displaced by the barge when it is empty is the volume of a rectangular prism with dimensions 28ft x 93ft x 6ft, which gives us 15,624 cubic feet. Given that the density of water is 62.4 lbs/ft³, the weight of this water, which is equal to the weight of the empty barge, would be (Volume x Density) = 15,624ft³ x 62.4 lbs/ft³ = 975,769.6 lbs.
Secondly, let's calculate the weight of the grain. The volume of water displaced when the barge is loaded is the volume of a rectangular prism with dimensions 28ft x 93ft x 9ft, which equals 23,436 cubic feet. The weight of this water, which is equal to the weight of the loaded barge, is (Volume x Density) = 23,436ft³ x 62.4 lbs/ft³ = 1,462,342.4 lbs. Hence, the weight of the grain is the weight of the loaded barge minus the weight of the barge itself = 1,462,342.4 lbs - 975,769.6 lbs = 486,572.8 lbs.
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The work function of palladium is 5.22 eV.
(a) What is the minimum frequency of light required to observe the photoelectric effect on Pd?
(b) If light with a 200 nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons?
a) The minimum frequency of the light must be [tex]1.26\cdot 10^{15} Hz[/tex]
b) The maximum velocity of the electrons is [tex]5.93\cdot 10^5 m/s[/tex]
Explanation:
a)
The photoelectric effect is a phenomenon that occurs when electromagnetic radiation hits the surface of a metal causing the release of electrons from the metal's surface.
The equation of the photoelectric effect is:
[tex]hf = \phi +K_{max}[/tex]
where :
[tex]hf[/tex] is the energy of the incoming photons, where
[tex]h[/tex] is the Planck's constant
[tex]f[/tex] is the frequency of the incoming photons
[tex]\phi[/tex] is the work function of the metal, the minimum energy that the photons must have in order to be able to free electrons from the metal
[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted electrons
In order to free electrons, the minimum energy of the photons must be at least equal to the work function (so that the kinetic energy of the electrons is zero, [tex]K_{max}=0[/tex]. Therefore,
[tex]h f_0 = \phi[/tex]
In this case,
[tex]\phi = 5.22 eV \cdot (1.6\cdot 10^{-19})=8.35\cdot 10^{-19} J[/tex]
Therefore, the minimum frequency of the photons must be
[tex]f_0 = \frac{\phi}{h}=\frac{8.35\cdot 10^{-19}}{6.63\cdot 10^{-34}}=1.26\cdot 10^{15} Hz[/tex]
b)
In this case, the wavelength of the incoming light is
[tex]\lambda = 200 nm = 200 \cdot 10^{-9} m[/tex]
We can find the frequency by using the wave equation:
[tex]f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{200\cdot 10^{-9}}=1.5\cdot 10^{15} Hz[/tex]
Now we can use the equation of the photoelectric effect to find the maximum kinetic energy of the electrons:
[tex]K_{max} = hf-\phi = (6.63\cdot 10^{-34})(1.5\cdot 10^{15})-8.35\cdot 10^{-19}=1.60\cdot 10^{-19} J[/tex]
And therefore, we can find their velocity by using the equation for the kinetic energy:
[tex]K_{max} = \frac{1}{2}mv^2[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electrons
v is their speed
Solving for v,
[tex]v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})}{9.11\cdot 10^{-31}}}=5.93\cdot 10^5 m/s[/tex]
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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where tis in s. Its initial position is x0 = 2.3m at t0 = 0 s
A)
At 2.2s , what is the particle's position?
Express your answer with the appropriate units.
B)
At 2.2s , what is the particle's velocity?
Express your answer with the appropriate units.
C)
At 2.2s , what is the particle's acceleration?
Answer:
A) At 2.2 s the position of the particle is 9.4 m.
B) At t =2.2 s the velocity is 9.7 m/s.
C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²
Explanation:
Hi there!
A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:
dx/dt = 2 · t²
Separating variables, we can find the equation of position as a function of time:
dx = 2 · t² · dt
Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:
∫ dx = 2 ∫ t² · dt
x - 2.3 m = 2/3 · t³
x = 2.3 m + 2/3 m/s³ · t³
Replacing t = 2.2 s:
x = 2.3 m + 2/3 m/s³ · (2.2 s)³
x = 9.4 m
At 2.2 s the position of the particle is 9.4 m
B) Now, let´s evaluate the velocity function at t = 2.2 s:
v = 2 · t²
v = 2 m/s³ · (2.2 s)²
v = 9.7 m/s
At t =2.2 s the velocity is 9.68 m/s
C) The acceleration is the variation of the velocity over time (the derivative of the velocity):
dv/dt = a
a = 4 · t
At t = 2.2 s:
a = 4 m/s³ · 2.2 s
a = 8.8 m/s²
At t = 2.2 s the acceleration of the particle is 8.8 m/s²
(A) The particle's position at time, t = 2.2 s is 7.1 m.
(B) The velocity of the particle at 2.2 s is 9.68 m/s.
(C) The acceleration of the particle at 2.2 s is 8.8 m/s².
The given parameters:
Velocity, Vx = 2t² m/sInitial position of the particle, X₀ = 2.3 mThe particle's position at time, t = 2.2 s is calculated as follows;
[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]
The velocity of the particle at 2.2 s is calculated as follows;
[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]
The acceleration of the particle at 2.2 s is calculated as follows;
[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]
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The magnitude of the electric force between two protons is 2.30 x 10^-26 N. How far apart are they?
a) 0.0220 mb) 0.100 mc) 0.480 md) 0.000570 me) 3.10 m
To solve this problem we will use the concepts given by Coulomb's law defined for force, said law is mathematically described as
[tex]F = \frac{kq_1 q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex]= Charge of two protons
r = Distance between them
F = Force
Our values are given as,
[tex]F =2.3*10^{-26} N[/tex]
[tex]q_1 =q_2 = 1.6*10^{-19} C[/tex]
[tex]k =9*10^9 Nm^2/C2[/tex]
Rearrenging to find the distance and replacing we have that
[tex]r^2=\frac{(9*10^9 )(1.6*10^{-19})^2 }{2.3*10^{-26} }[/tex]
[tex]r^2=10.01*10^{-3} m^2[/tex]
[tex]r =\sqrt{10.01*10^{-3} m^2}[/tex]
[tex]r = 0.100m[/tex]
Therefore the correct option is B.
A railroad freight car, mass 15 000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50 000-kg loaded second car, initially at rest and with brakes released. What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision
Answer:
23.0760769 %
Explanation:
[tex]m_1[/tex] = Mass of freight car = 15000 kg
[tex]m_2[/tex] = Mass of second car = 50000 kg
[tex]v_1[/tex] = Velocity of freight car = 2 m/s
[tex]v_2[/tex] = Velocity of second car = 0
v = Combined mass velocity
As the linear momentum of the system is conserved we have
[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]
The initial kinetic energy
[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]
Final kinetic energy
[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]
The percentage is given by
[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]
The change in percentage of initial kinetic energy is 23.0760769 %
There are great similarities between electric and gravitational fields. A room can be electrically shielded so that there are no electric fields in the room by surrounding it with a conductor. Can a room be gravitationally shielded? Explain.
Answer:
Can a room be gravitationally shielded? No, it can't.
Explanation:
the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.
Suppose that 10 moles of an ideal gas have a gauge pressure of 2 atm and a temperature of 200 K. If the volume of the gas is doubled and the pressure dropped to a gauge pressure of 1 atm, what is the new temperature?
Select one:
a.
267 K
b.
300 K
c.
400 K
d.
200 K
The new temperature is: d. 200 K
The Ideal Gas Law is given by:
[tex]PV = nRT[/tex]
where:
[tex]P[/tex] = pressure
[tex]V[/tex] = volume
[tex]n[/tex] = number of moles
[tex]R[/tex] = universal gas constant
[tex]T[/tex] = temperature
Given initial conditions:
[tex]P_1 = 2[/tex] atm (gauge pressure)
[tex]T_1 = 200[/tex] K
[tex]V_1 = V[/tex]
[tex]n = 10[/tex] moles
Final conditions:
[tex]P_2 = 1[/tex] atm (gauge pressure)
[tex]V_2 = 2V[/tex]
[tex]T_2 = ?[/tex]
We can use the combined gas law equation to relate the initial and final states of the gas:
[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]
Substituting in the given values:
[tex]\frac{2V}{200} = \frac{1 \cdot 2V}{T_2}[/tex]
Solving for [tex]T_2[/tex]:
[tex]\frac{2V}{200} = \frac{2V}{T_2}[/tex]
By simplifying, we get:
[tex]\frac{2}{200} = \frac{2}{T_2}[/tex]
Cross-multiplying gives:
[tex]2 \cdot T_2 = 400[/tex]
Dividing both sides by 2 gives:
[tex]T_2 = 200[/tex] K
Therefore, the new temperature [tex](T_2)[/tex] after the volume is doubled and the pressure is halved is 200 K.
In the microscopic view of electrical conduction in a copper wire, electrons are accelerated by an electric field and then collide with metal atoms after traveling about4.2×10−8m.
If an electron begins from rest and is accelerated by a field of 0.080 N/C, what is its speed when it collides with a metal atom?
Answer:
34.35464 m/s
Explanation:
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
E = Electric field = 0.08 N/C
s = Displacement = [tex]4.2\times 10^{-8}\ m[/tex]
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Electrical force is given by
[tex]F=qE[/tex]
Work done is given by
[tex]W=Fs\\\Rightarrow W=qEs[/tex]
Work done is also given by the kinetic energy
[tex]\dfrac{1}{2}mv^2=qEs\\\Rightarrow v=\sqrt{\dfrac{2qEs}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 0.08\times 4.2\times 10^{-8}}{9.11\times 10^{-31}}}\\\Rightarrow v=34.35464\ m/s[/tex]
The velocity of the electron is 34.35464 m/s
A water gun fires 5 squirts per second. The speed of the squirts is 15 m/s.
By how much distance is each consecutive squirt
separated?
Final answer:
Each consecutive squirt from the water gun is separated by 3 meters, calculated by multiplying the speed of the squirts (15 m/s) by the interval between each squirt (0.2 seconds).
Explanation:
The question is asking to calculate the distance at which each consecutive squirt from a water gun is separated when the water gun fires squirts at a certain rate and speed. Given that the water gun fires 5 squirts per second and the speed of the squirts is 15 m/s, we can use the formula distance = speed × time to find the separation distance between squirts.
Since there are 5 squirts per second, each squirt is separated by 1/5 of a second, or 0.2 seconds. To find the separation distance, we multiply the speed of the squirts by the time interval between each squirt:
Distance = Speed × Time
Distance = 15 m/s × 0.2 s
Distance = 3 meters
Therefore, each consecutive squirt is separated by 3 meters.
A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a long, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Answer:
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
Explanation:
If we define the magnitude of the field as B, then we have:
Total magnitude of the field [tex]B_{t}[/tex] = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.
B_loop = (μ0)*([tex]I_{2}[/tex])/2*R
B_wire = (μ0)*([tex]I_{1}[/tex])/2*π*D
Thus:
B_loop = B_wire at the center of the loop.
(μ0)*([tex]I_{2}[/tex])/2*R = (μ0)*([tex]I_{1}[/tex])/2*π*D
[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R
The magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
The given parameters;
radius of the loop = Rcurrent in the loop, I = I₂distance of the loop from the wire, = DThe magnetic field at the center of the loop is calculated as follows;
[tex]B_o = \frac{\mu_o I_2}{2R}[/tex]
The magnetic field at the distance of the wire is calculated as follows;
[tex]B_o = \frac{\mu_o I_1}{2\pi D}[/tex]
The magnitude of current at the center of the loop is calculated as follows;
[tex]\frac{\mu_o I_2 }{2R} = \frac{\mu_o I_1}{2\pi D} \\\\I_1 = \frac{\pi D I_2}{R}[/tex]
Thus, the magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].
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By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 3.0 N ?
To solve this problem we will apply the definition of electrostatic force. From the variables present there and explained later we will find the value of the distance reorganizing said expression, that is
[tex]F = \frac{k q_1 q_2}{d^2}[/tex]
Here
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object
d = Distance
Replacing our values we have that
[tex]3 = \frac{(9*10^9)(1)(1)}{d^2}[/tex]
Rearranging and solving for the distance we have
[tex]d = 54772.25m[/tex]
Therefore the distance between the two objects is 54772.25m
A rubber ball with a mass of 0.30 kg is dropped onto a steel plate. The ball's velocity just before impact is 4.5 m/s and just after impact is 4.2 m/s and just after impact is 4.2 m/s. What is the change in the ball's momentum?
Answer:
Change in momentum will be -2.61 kgm/sec
Explanation:
We have given mass of the rubber ball m = 0.30 kg
Velocity of the ball before the impact [tex]v_1=4.5m/sec[/tex]
Velocity of ball after impact [tex]v_2=-4.2m/sec[/tex] ( negative sign is due to opposite direction of motion )
Change in momentum is given by [tex]m(v_2-v_1)=0.3\times (-4.2-4.5)=0.3\times =0.3\times -8.7=-2.61kgm/sec[/tex] ( negative sign shows the direction of change in momentum )
Answer:
-0.09 kg m/s
Explanation:
2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.
The specific kinetic energy of a mass (in kilojoules per kilogram) moving at a velocity of 40 m/s is calculated using the kinetic energy formula, resulting in 0.8 kJ/kg.
Explanation:The specific kinetic energy of a mass moving with a velocity can be determined using the formula for kinetic energy, K = 0.5*m*v². In this case, where the velocity 'v' is given as 40 m/s, and we want to solve the kinetic energy per kilogram, we can consider the mass 'm' as 1 kg. Hence the specific kinetic energy would be K = 0.5*(1 kg)*(40 m/s)² = 800 J = 0.8 kJ/kg, because 1 kilojoule (kJ) = 1000 joules (J).
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Two charges are located in the x – y plane. If q 1 = − 3.65 nC and is located at ( x = 0.00 m , y = 0.600 m ) , and the second charge has magnitude of q2 = 4.20 nC and is located at ( x = 1.10 m , y = 0.800 m ) , calculate the x and y components, Ex and Ey , of the electric field, → E , in component form at the origin, ( 0 , 0 ) . The Coulomb force constant is 1/(4πϵ0 ) = 8.99 × 10^9 N ⋅ m^2 /C^2.
Answer:
Ex = -16.51 N/C Ey = 79.14 N/C
Explanation:
As the electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.
So, we can first the field due to q1.
Due to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward,(like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows.
E₁ = k*(3.65 nC) / r₁²
If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:
E₁ₓ = 0 E₁y = 8.99*10⁹*3.65*10⁻⁹ / (0.600)²m² = 91.15 N/C (1)
For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.
It will be just the pythagorean distance between the points located at the coordinates (1.10 m, 0.800 m) and (0,0), as follows:
r₂² = 1.10²m² + (0.800)²m² = 1.85 m²
The magnitude of the electric field due to q2 can be found as follows:
E₂ = k*q₂ / r₂² = 8.99*10⁹*(4.2)*10⁹ / 1.85 = 20.41 N/C (2)
Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.
In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:
E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ
the cosine of θ, is just, by definition, the opposite of x/r₂:
⇒ cos θ =- (1.10 m / √1.85 m²) =- (1.10 / 1.36) = -0.809
By the same token, sin θ can be obtained as follows:
sin θ = - (0.800 m / 1.36 m) = -0.588
⇒E₂ₓ = 20.41 N/C * (-0.809) = -16.51 N/C (pointing to the left) (3)
⇒E₂y = 20.41 N/C * (-0.588) = -12.01 N/C (pointing downward) (4)
The total x and y components due to both charges are just the sum of the components of Ex and Ey:
Ex = E₁ₓ + E₂ₓ = 0 + (-16.51 N/C) = -16.51 N/C
From (1) and (4), we can get Ey:
Ey = E₁y + E₂y = 91.15 N/C + (-12.01 N/C) =79.14 N/C
Answer:
Ex = 15.505 N/C
Ey = 79.144 N/C
Explanation:
Particle 1
[tex]E_{1} = \frac{k*q_{1} }{r^2_{1}} \\\\r^2_{1} = 1.1^2+0.8^2\\\\r^2_{1} = 1.85 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.8}{1.1}) = 36.03 degree\\\\ E_{1} = \frac{(8.99*10^9)*(4.2*10^(-9)) }{1.85}\\\\E_{1} = 20.4096 N/C[/tex]
Away from the particle at (0,0) due to + charge
Particle 2
[tex]E_{2} = \frac{k*q_{2} }{r^2_{2}} \\\\r^2_{2} = 0.36 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.6}{0}) = 90 degree\\\\ E_{2} = \frac{(8.99*10^9)*(3.65*10^(-9)) }{0.36}\\\\E_{2} = 91.149 N/C[/tex]
Towards from the particle at (0,0) due to - charge
Resultant Electric field in y direction
[tex]E_{res,y} = E_{2} -E_{1}*cos(Q)\\E_{res,y} = (91.149) - (20.4096)*sin(36.03)\\\\E_{res,y} = 79.144 N/C[/tex]
Resultant Electric field in x direction
[tex]E_{res,x} = E_{1}*cos(Q)\\E_{res,y} =(20.4096)*cos(36.03)\\\\E_{res,x} = 16.505 N/C[/tex]
Describe the difference between technology based effluent standards and water quality based effluent standards under the Clean Water Act. Also, indicate which of these two different standards is likely to be controlling on a small stream designated as a cold water fishery and why.
Explanation:
Technology-based:
1.As the name implies technology, no technology will be clarified about it. It only depends on the variables, which describes them.
2. That is based on a single facility's findings.
3. It takes into account the contaminants type and volume, and their equations to monitor them.
4. This is reserved for city or urban wastewater treatment plants only.
5. It takes into account the pH, need for oxygen and the suspended solids.
Water quality based on:
1.This is enforced if there is a need to apply stricter limits to pollutants that are not pleased with the limits of technology.
2. All basing on water quality was risk-based.
3. They placed some less importance on the technologies which is used in the technology based limit.
Water quality dependent restrictions are used as cool water fishery for streams.
The act on clean water will also include bodies of water belonging to wildlife, agriculture and others. The law also included that the physical chemical and biological variables of all the state water bodies must be controlled by these water quality based limits.