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Diano4ka-milaya [45]
3 years ago
8

A 10 kg rock is suspended 320 m above the ground. Calculate the approximate speed with which the rock hits the ground.

Physics
2 answers:
Ket [755]3 years ago
8 0

Answer:

B. 79 m/s

Explanation:

Since we have information about the distance of the fall, we can use the following formula to find the final velocity of the object:

v_{f}^2=v_{i}^2+2gh

clearing for v_{f}

v_{f}=\sqrt{v_{i}^2+2gh}

where v_ {f} is the final velocity (when it hits the ground), v_ {i} the initial velocity, g is the acceleration of gravity (g=9.8m/s^2), and h is the height from where the rock is dropped, in this case: h=320m

Assuming that the rock was only dropped without adding an initial velocity, v_ {i} is zero.

Substituting all the values in the formula:

v_{f}=\sqrt{(0)^2+2(9.8m/s^2)(320m)}

v_{f}=\sqrt{6272}

v_{f}=79.2m/s

The value that best approximates from the options is 79 m/s,  the answer is B.

Rama09 [41]3 years ago
4 0

The answer for this is actually b. 79 m/s

I submitted it to odyssey-ware and i got the question correct

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Explanation:

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3 years ago
Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far
marta [7]

a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

Explanation:

a)

The angular speed of Andrea and Chuck is the same.

Let's call \omega the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

\omega= \frac{2\pi}{T}

where

2 \pi is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

v=\omega r

where

\omega is the angular speed

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Here let's call r_c the distance at which Chuck is rotating, so his tangential speed is

v_c = \omega r_c

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

r_a = 2 r_c

So his tangential speed is

v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

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2 years ago
What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s
nika2105 [10]

Answer:

<em>An average net force of 5200 N is needed to stop the car</em>

Explanation:

<u>Cinematics and Dynamics</u>

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

\displaystyle a=\frac{v_f-v_o}{t}

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

\displaystyle a=\frac{0-28}{10.5}

a = -2.67~m/s^2

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

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F = 5200 N

An average net force of 5200 N is needed to stop the car

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