Answer:
1.74 m/s
Explanation:
From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.
Therefore, using the formula below, we can calculate the speed of the can, V(can);
===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).
Since the question says the collision was elastic, we use the formula below
Slotting in the given values into the equation (1) above, we have;
1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).
Therefore, final velocity of the can= 2M1V1/M1+M2.
==> 2×2.7×1.1/ 2.7 + 0.72.
The velocity of the can after collision = 1.74 m/s
Answer:
Explanation:
As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time
So we can say
now we have
here since magnetic field is constant so we have
now we have
now we have
A
~~~hope this helps~~~
~~have a beautiful day~~
~davatar~
Answer:
The value of impulse of the net force on the ball during its collision with wall is 
The value of average horizontal force that the wall exerts on the ball during the impact is F = 15000 N
Explanation:
Mass of the ball = 0.5 kg
Horizontal velocity 
= 20 
Velocity after collision 
= - 10
(A). Impulse of the net force on the ball during its collision with wall is
I = 0.5 × (20 + 10)
This is the value of impulse of the net force on the ball during its collision with wall.
(B). The magnitude of average horizontal force
Where F = Force
I = impulse & t = time interval = 0.001 sec
F = 15000 N
This is the value of average horizontal force that the wall exerts on the ball during the impact.
Explanation:
Given:
v₀ₓ = 15 m/s cos 20° = 14.10 m/s
aₓ = 0 m/s²
v₀ᵧ = 15 m/s sin 20° = 5.13 m/s
aᵧ = -9.8 m/s²
t = 1.5 s
Find: Δx and Δy
Δx = v₀ₓ t + ½ aₓ t²
Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²
Δx = 21.1 m
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²
Δy = -3.33 m
