Physics, 13.07.2019 21:30 bigmouth804

Use this information to answer the next 2 questions: a 0.32 kg ball is thrown with a speed of 11.5 m/s. 18. how much kinetic energy does the ball have? 42.3 j 1.84 j 35.9 j 21.2 j 19. how much work would be needed to stop the ball?

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Answer from: lloydlunsford1685

Given:

m = 0.32 kg

v = 11.5 m/s

To find:

1) Kinetic energy = ?

2) Work needed to stop the ball = ?

Formula used:

1) Kinetic energy = $\frac{1}{2} m v^{2}$

2) Work = kinetic energy

Solution:

Kinetic energy is given by,

Kinetic energy = $\frac{1}{2} m v^{2}$

K.E. = $\frac{1}{2} 0.32 \times 11.5 \times 11.5$

K.E. = 21.16 Joule

Thus, option (d) is the correct answer.

Work needed to stop the ball is same as the kinetic energy because energy is the capacity to do work. Since, work is opposing the movement of the ball. Thus, Work = - Kinetic energy

Work = -21.16 Joule

Answer from: Quest

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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