A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car
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Answer:
Explanation:
mass of object, m = 3 kg
spring constant, K = 750 n/m
compression, x = 8 cm = 0.08 m
angle of gun, θ = 30°
(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.
(b) Let v be th evelocity of ball at the tim eof launch.
by using the conservation of energy
1/2 Kx² = 1/2 mv²
750 x 0.08 x 0.08 = 3 x v²
v = 1.265 m/s
By use of the formula of maximum height
h = 0.02 m
h = 2 cm
The center of mass of the two objects is the average position of the parts of the two object system
The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>
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Reason:
The given parameters are;
The position of block A and block B at t = 0 is x = 0
The mass of block A, m₁ = 8 kg
Mass of block B, m₂ = 16 kg
Speed of block <em>A</em> = 0 m/s
Speed of block <em>B</em>, v₂ = 10 m/s
Location of the center of mass of the two object at t = 3 s; Required
Solution;
The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)
The location of block <em>B</em>, = v₂ × t
The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m
The center of mass of two masses are given as follows;
The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)
Learn more about the center of mass here:
Answer:
<em>The velocity of the carts after the event is 1 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum </u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of the individual momentums:
If a collision occurs and the velocities change to v', the final momentum is:
Since the total momentum is conserved, then:
P = P'
In a system of two masses, the equation simplifies to:
If both masses stick together after the collision at a common speed v', then:
The common velocity after this situation is:
The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:
The velocity of the carts after the event is 1 m/s