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2 years ago

A shot-putter exerts a force of 0.142 kN on a shot, accelerating it to 22.75 m/s2. What is the mass of the shot?

Physics

1 answer:

Svetach [21]2 years ago

8 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Force and Power.

Since, according to the Newton's law,

Force = mass * Acceleration.

hence, here

Force = 142 N, accelration = 22.75 m/s2

hence, mass = 142/22.75

===> Mass = 6.24 Kg

hence the mass of the shot is 6.24 Kg

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Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co

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2 years ago

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice

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remains the same, but the apparent brightness is decreased by a factor of four.

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2 years ago

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

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A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$