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3 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]3 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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In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an

kirill115 [55]

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

$I_1\omega_1 = I_2\omega_2$

now we have

$I_1 = mr^2$

$I_1 = (1kg)(0.05^2)$

$I_1 = 25\times 10^{-4} kg m^2$

similarly let the final distance is "r"

so now we have

$I_2 = mr^2$

$I_2 = 1r^2$

now from above equation we have

$(25\times 10^{-4})37 = (r^2)(58)$

$r = 0.04 m$

so final distance is 0.04 m between them

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3 years ago

Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq

andrew-mc [135]

The force on each balloon is 2×10^−3 N.

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

$\sum F_{y} = Tcos10\textdegree - mg = 0\\\\T = \frac{mg}{cos10\textdegree } \\\\\sum F_{y} = Tsin10\textdegree - mg = 0\\$

So,

$F_{e} = \frac{mg}{cos10\textdegree }sin10\textdegree = mgtan10\textdegree \\\\= (0.00100kg)(9.8m/s^{2})tan10\textdegree \\\\F_{e} = 2 \times 10^{-3}N$

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).

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1 year ago

A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic

valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

$E=\frac{1}{2}kx^2$

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k is the spring constant

x is the elongation of the spring with respect to its equilibrium position

For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(84.08)}{342.25}}=0.701 m$

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3 years ago

A cannon fired horizontally at 20 m/s from the top of a cliff lands 80m away. how tall is the cliff

Scorpion4ik [409]

Answer:

The height of the cliff is, h = 78.4 m

Explanation:

Given,

The horizontal velocity of the projectile, Vx = 20 m/s

The range of the projectile, s = 80 m

The projectile projected from a height is given by the formula

<em> S = Vx [Vy + √(Vy² + 2gh)] / g </em>

Therefore,

h = S²g/2Vx²

Substituting the values

h = 80² x 9.8/ (2 x 20²)

= 78.4 m

Hence, the height of the cliff is, h = 78.4 m

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3 years ago

If the distance between two asteroids is halved, the gravitational force they exert on each other will

mamaluj [8]

Answer:

e) Be four times greater

Explanation:

Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses ( $m_{1}$ & $m_{2}$ ) and the distance between them ( $r$ ) in the next way:

$F=G\frac{m_{1}m_{2}}{r^{2}}$ (2)

Now if distance between asteroids is halved:

$F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}}$

$F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}}$

Note that $G\frac{m_{1}m_{2}}{r^{2}}$ because (1) is F so:

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It's four times greater!

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3 years ago