At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le
1 answer:
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The cart travelled a distance of 14.4 m
Explanation:
The work done by a force when pushing an object is given by:
where:
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and the displacement
In this problem we have:
W = 157 J is the work done on the cart
F = 10.9 N is the magnitude of the force
, assuming the force is applied parallel to the motion of the cart
Therefore we can solve for d to find the distance travelled by the cart:
Learn more about work:
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Answer:
The velocity of the second ball is approximately 2.588 m/s
The angle direction of the second ball is 75° counterclockwise from the horizontal
Explanation:
The initial velocity of the first billiard ball = 5 m/s
The initial velocity of the billiard ball the first billiard ball strikes = 0 m/s
The final velocity of the first billiard ball = 4.35 m/s
The final direction of motion of the first billiard ball = 30° below its original motion
For perfectly elastic collision, whereby the target is at rest initially, by conservation of momentum, we have;
m₁ × = m₁·
+ m₂·
Which gives;
m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂·
∴ m₂· = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)
m₂· = m₁ × 5·(1 - √3/2)·i + m₁·2.5·j = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j
Therefore, given that the mass of both billiard balls are equal, we have, m₁ = m₂, which gives;
m₂· = m₁·
= m₁ × 2.5·(2 - √3)·i + m₁·2.5·j
∴ = 2.5·(2 - √3)·i + 2.5·j
The magnitude of the velocity of the second ball is = √((2.5·(2 - √3))² + 2.5²) ≈ 2.588 m/s
The direction of the second ball, θ = arctan(2.5/((2.5·(2 - √3))) = 75° counterclockwise from the horizontal.