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Rzqust [24]
3 years ago
15

3. A train has broken through the wall of a train station. During the collision, what can be said about the force exerted by the

train on the wall?
a. The force exerted by the train on the wall was larger than the force the wall could exert on the train.
b. The wall cannot be said to "exert" a force; after all, it broke.
c. The force exerted by the train on the wall was the same in magnitude as the force exerted by the wall on the train.
d. The force exerted by the train on the wall was less than the force exerted by the wall on the train.
Physics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

c. The force exerted by the train on the wall was the same in magnitude as the force exerted by the wall on the train.

Explanation:

Newton's Third Law states that if a body exerts a force on another body, this body exerts a force of equal magnitude but in the opposite direction on the body that produces it. So, one body cannot exert a force on another without experiencing a force in itself.

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Help me please Am I correct???​
trasher [3.6K]

Answer:

They both experienced the same force as they weigh the same amount

Newtonian physics

Explanation:

5 0
3 years ago
A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

3 0
3 years ago
A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent hea
Sunny_sXe [5.5K]

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

7 0
4 years ago
Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8
Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

V2=\frac {M1v1}{m2}

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s

4 0
3 years ago
Hanging from a horizontal beam are nine simple pendulums of the following lengths:
LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

L =\frac{g}{w^{2} }

where g=9.8m/s

         ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

L= \frac{9.8}{2^{2} }

L = 2.45m

when  ω= 4rad/sec

L=\frac{9.8}{4^{2} }

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

8 0
4 years ago
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