Answer:
They both experienced the same force as they weigh the same amount
Newtonian physics
Explanation:
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 , 
mass of puck 2 , 
initial speed of puck 1 , 
initial speed of puck 2 , 
Final speed of puck 1 and puck 2 be
respectively
Apply conservation of linear momentum

=>
=>
-----(A)
Since collision is perfectly elastic , coefficient restitution e=1

=>
------(B)
From equation (A) and (B)

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

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

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;

where g=9.8m/s
ω= 2rad/s to 4rad/sec
when ω= 2rad/sec

L = 2.45m
when ω= 4rad/sec

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