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2 years ago

If we increase the force applied to an object and all other factors remain the same that amount of work will

Physics

2 answers:

AlekseyPX2 years ago

7 0

It is c. increase................

worty [1.4K]2 years ago

5 0

Hello there.

<span>If we increase the force applied to an object and all other factors remain the same that amount of work will

</span><span>C. Increase
</span>

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Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the char

Kamila [148]

Answer:

B. 3.1 × 10^5 V

Explanation:

8 0

2 years ago

Read 2 more answers

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

Read more about the flows rate

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5 0

1 year ago

Is James Charles a boy or a girl ? I am so confused

eimsori [14]

Answer:

both

Explanation:

and I'm serious he is literally both

7 0

2 years ago

While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?

sergiy2304 [10]

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

3 0

2 years ago