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fredd [130]
3 years ago
9

. A 1.00 kg rock is thrown up into the air from ground level at a speed of 8.00 m/s. The ball travels up

Physics
1 answer:
Liula [17]3 years ago
5 0

Answer:

p = -8 kg-m/s

Explanation:

Given that,

Initial speed of the rock, u = 8 m/s

Mass of the rock, m = 1 kg

The ball travels up  to a maximum height, then returns to the ground.

We need to find the rock's momentum as it strikes the  ground. Let v be the final speed of the rock. Its final speed is as same as initial speed i.e. 8 m/s but in negative direction. So

p = mv

p = 1 kg × (-8 m/s)

= -8 kg-m/s

So, the rock's momentum as it strikes the  ground is (-8 kg-m/s).

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A ball is thrown off the top of a building and lands on the ground below.
natita [175]

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

K=\frac{1}{2} mv^2

where m is the mass of the object, and v is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

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  • velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball  before touching the gound can be determined.

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Read 2 more answers
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

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