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2 years ago

7

A women who normally weighs 400 N stands on top of a ladder so high she is one full additional earth radius above the surface of

the earth. How much does she weigh now

1 answer:

AveGali [126]2 years ago

3 0

The woman weighs 100N now.

Explanation:

Given:

F is given as 400N

one full additional earth radius formed

To Find:

Weight of the woman=?

Solution:

We know that gravitational force is inversely proportional to the square of radius.

$f_g=\frac{1}{r^2}$

where

$f_g$ is the gravitational force

$r^2$ is the radius square

So when radius is doubled, the force becomes one fourth of the original.

Therefore $\frac{1}{4} \times400 = 100N$

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An arrow, starting from rest, leaves the bow with a speed of25

xz_007 [3.2K]

Answer:

The speed will be $v_f= 43.30 \frac{m}{s}$

Explanation:

We can use the following kinematics equation

$v_f^2-v_i^2=2 \ a \ d$

where $v_f$ is the final speed, $v_i$ its the initial speed, a is the acceleration, and d the distance.

The force will be tripled, the force is:

$\vec{F} = m \vec{a}$

in 1D

$F = m a$

Now, for the original problem, we have

$(25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'$

$(25 \frac{m}{s}))^2=2 \ a' \ d'$

$625 \frac{m^2}{s^2}=2 \ a' \ d'$

For the second problem, we have

$(v_f})^2-(v_i)^2=2 \ a'' \ d''$

starting from the rest, we have the same initial velocity.

$(v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''$

$(v_f})^2=2 \ a'' \ d''$

As the force is tripled, we have:

$F'' = 3 F'$

$m'' \ a'' = 3 m' \ a'$

But the mass its the same, so

$m' \ a'' = 3 m' \ a'$

$a'' = 3 \ a'$

So the acceleration its also tripled.

$(v_f})^2=2 \ (3 * a') \ d''$

$(v_f})^2=3 ( 2 \ ( a' \ d'' )$

As the distance traveled by the arrow must also be the same, we have:

$(v_f})^2=3 ( 2 \ ( a' \ d' )$

$(v_f})^2=3 (625 \frac{m^2}{s^2})$

$v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}$

$v_f= \ sqrt{3} 25 \frac{m}{s}$

$v_f= 43.30 \frac{m}{s}$

And this will be the speed from the arrow leaving the bow.

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