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Inessa05 [86]
3 years ago
8

You drop two rocks. one rock has a mass of 8kg and the other a mass of 7kg. The 8kg rock falls no faster than the 7kg rock for w

hat reason. is it due to force of gravity for both being the same?
Physics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:

The gravitational force is significantly constant.

Explanation:

The gravitational force is expressed as:

F= G*M*m/d^2

G= Gravitational constant

M= in this case it is the mass of the planet

m= mass of the rock

d= distance of the rocks from the ground (suppose both at the same height for better comparison)

At the same time, we know that the force that each rock experiences is equal to the product of the mass due to acceleration:

F=m*a

We can match both expressions for F:

G*M*m/d^2 = m*a

we simplify m, and we obtain that the acceleration is independent of the mass of the attracted bodies:

a= G*M/d^2

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A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
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The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

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99.6t = pi/6 + 2pi

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The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

Read more on waves here

brainly.com/question/25699025

#SPJ4

complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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