Answer:
The impulse exerted by one cart on the other has a magnitude of 4 N.s.
Explanation:
Given;
mass of the first cart, m₁ = 2 kg
initial speed of the first car, u₁ = 3 m/s
mass of the second cart, m₂ = 4 kg
initial speed of the second cart, u₂ = 0
Let the final speed of both carts = v, since they stick together after collision.
Apply the principle of conservation of momentum to determine v
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 3 + 0 = v(2 + 4)
6 = 6v
v = 1 m/s
Impulse is given by;
I = ft = mΔv = m(
The impulse exerted by the first cart on the second cart is given;
I = 2 (3 -1 )
I = 4 N.s
The impulse exerted by the second cart on the first cart is given;
I = 4(0-1)
I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).
Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.
3 is the answer teeeeeeeeeheeeeeeeee
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.
Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.
Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
