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3 years ago

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A 360-N child is in a swing that is attached to a pair of ropes 1.90 m long. Find the gravitational potential energy of the chil

d–Earth system relative to the child's lowest position at the following times.

1 answer:

Igoryamba3 years ago

7 0

Answer:

684J

Explanation:

So basically the formula for gravitational potential energy is Mass X Gravity X height. That is G.p.e = mgh

We don't have the mass but since we have the height, we multiply directly with the height since the quantity of weight is already given.

so G.p.e = 360 X 1.9 = 684J

Note that; The answer is in joules because g.p.e is work done.

Hope that was helpful!!

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The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever

hammer [34]

Answer:

Approximately $5.19 \times 10^{-5}\; \rm N$ .

Explanation:

Let $G$ denote the gravitational constant. ( $G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}$ .)

Let $M$ and $m$ denote the mass of two objects separated by $r$ .

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

$\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}$ .

In this question: $M = 7 \times 10^{22}\; \rm kg$ is the mass of the moon, while $m = 1\; \rm kg$ is the mass of the water. The two are $r = 3\times 10^{5}\; \rm km$ apart from one another.

Important: convert the unit of $r$ to standard units (meters, not kilometers) to reflect the unit of the gravitational constant $G$ .

$\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m$ .

$\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}$ .

5 0

3 years ago

Bonus Points. A ball was thrown straight up in the air from 1.2 meter above the grouind. After 3 seconds the ball returnes to th

nekit [7.7K]

Answer:

Explanation:

given that

Distance above the ground, s = 1.2 m

Time taken by the ball, t = 3 s

Velocity of the ball, v = 1.2/3 = 0.4 m/s

Maximum height reached by the ball is then given by the formula

H = v² / 2g

H = 0.4² / 2 * 9.8

H = 0.16 / 19.6

H = 0.0082 m or rather, 0.82 cm

5 0

3 years ago

If a small motor does 520 J of work to move a toy car 260 m, what force does the engine exert on the car?

ohaa [14]

Answer:

<h2>2 N</h2>

Explanation:

The force engine exert on the car can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{520}{260} = 2 \\$

We have the final answer as

<h3>2 N</h3>

Hope this helps you

4 0

3 years ago

What is true when an object is moved farther from a plane mirror?

musickatia [10]

Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

8 0

3 years ago

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have:

$\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}$

or:

$\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}$

which for our values is:

$\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s$

6 0

3 years ago