a)
We use the formula :
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the values in:
4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f
Calculating this we get:
32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f
Rearrange for v2f:
v2f = 
This gives us 8.0 m/s as the final velocity of the second ball.
b)
Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.
This means we use the formula:
Ek = 
+ = + 
Substituting in values:
Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2
This simplifies to:
Ek= 128J + 0J = 0J + 128J
This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.
Answer:
P = 86956.52 Pa
Explanation:
Data:
- F = 800 N
- A = 0.0092 m²
- P = ?
Use the formula:
Replace and solve:
The pressure it exerts on the ground is <u>86956.52 Pascal.</u>
Greetings.
Velocity probably well that’s what i would put
We need the frequency of the photon, it is v = c/ λ
Where c is 3 x 10^8 ms^-1 and λ
is the wave length
We also need the expression of
connecting frequency to energy of photon
which is E = hv where h is Planck’s
constant
Combining the two equations
will give us:
E = h x c/λ
Inserting the values, we will
have:
E = 6.626 x 10^-34 x 3 x 10^8 /
0.126
E = 1.578 x 10^ -24 J
Answer:
B. Increases
Explanation:
Momentum is given as 
. Since 
is maintained, as the velocity decreases, the momentum decreases, and similarly, as the velocity increases, the momentum increases.
