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.
C. located in front of the lens
Distance = speed* time
10x10^-3 *u + 10x10^-3 * 2u = 125
0.01u + 0.02u = 125
0.03u = 125
u = 4166.66666.... s
acceleration = (final velocity - initial velocity)/time
(10x10-3 - 0)/4167 = 2.3998 * 10^-6 m s-2
The crate is in equilibrium. Newton's second law gives
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where
• <em>n</em> = magnitude of the normal force
• <em>mg</em> = weight of the crate
• <em>p</em> = mag. of push exerted by movers
• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>
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It follows that
<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N
so that the movers perform
<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J
of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)
Answer:
metre (m) - unit of length
kilograms (kg) - unit of mass
second (s) - unit of time
ampere (A) - unit of electrical current
kelvin (K) - unit of temperature
mole (mol) - unit of the amount of substance
Explanation:
