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3 years ago

How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?

Physics

1 answer:

Anestetic [448]3 years ago

3 0

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

$d=vt$

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

$d=(2.1)(10)=21 m$

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3 years ago

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

$F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}$

Now, if the body is on the surface of the Earth, its weight (w) would be:

$F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}$

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

$\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\$

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3 years ago