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Tamiku [17]
3 years ago
5

A steel ball and a piece of clay have equal mass. They are dropped from the same height on a horizontal steel platform. The ball

bounces back with nearly the same speed with which it hit. The clay sticks to the platform. Which object experiences the greater momentum change? (A) the ball (B) the clay change(C) Both experience the same momentum (D) there is no momentum change for either (E) more information is required
Physics
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer: The ball (option A)

Explanation: change in momentum is defined by the formulae m(v - u) where m = mass of object, v = final velocity and u = initial velocity.

For the ball, it hits the ground and bounces back with the same speed, that's final velocity equals initials (v = - u)

Change in momentum = m( -u- u) = m(-2u) = m(-2u) = -2mu

For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)

m(v - u) = m(0 - u) = - mu.

-2mu (change in momentum from the ball) is greater than - mu ( change in momentum of clay)

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Which choice has more thermal energy?
prohojiy [21]

Answer:

If thermal energy is the motion energy of the particles of a substance, which has more thermal energy—the cup of hot tea or a spoonful of hot tea? It makes sense that the more particles of a substance you have, then the more thermal energy the substance has. The cup of hot tea would have more thermal energy, even if the temperature of the tea is the same in the cup and in the spoon. But which cools down the quickest (has the highest rate of thermal energy transfer)—the tea in the cup or the tea in the spoon? If I have fewer particles of the same substance, then the rate of thermal energy transfer is faster. The tea in the spoon would lose thermal energy more rapidly. So the amount of a substance you have is one factor that affects the rate of thermal energy transfer.

Explanation:

7 0
3 years ago
A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p
Softa [21]

Answer:

20 Watts

Explanation:

Work = force × distance

W = (20 N) (10 m)

W = 200 Joules

Power = work / time

P = 200 J / 10 s

P = 20 Watts

8 0
3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

4 0
3 years ago
Physics is defined as:
nikitadnepr [17]

Answer:

B. Developed out of efforts of man trying to explain our physical environment.

Explanation:

This is because physics, as we all know of, is considered to be <em>crucial to understanding the world around us</em>, making it the most basic and fundamental of science.

5 0
3 years ago
Read 2 more answers
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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