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3 years ago

What is the on ohooke benden

1 answer:

6 0

Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

You might be interested in

Which of the following are capital cities in the Caribbean? Lima La Habana San Juan San Jos Guinea Ecuatorial

mars1129 [50]

Caribbean:
Habana ... Cuba
San Juan ... Puerto Rico
San Jose ... Costa Rica

Other:
Lima ... capital of Peru in South America
Equatorial Guinea ... country in Africa

5 0

3 years ago

Help? its due in like 12 minutes lolzz

Gnom [1K]

Answer:

Question 1: the plates are moving toward one another.

Question 2: The Himalayan Mountains in India

Question 3: Because mountains are formed instead.

Explanation:

The paragraph explains that the plates continue to move closer to one another while forming multiple mountains.

The paragraph explains, " a well-known example of this is the formation of the Himalayan Mountains in India,"

The area of the Himalayan Mountains are better suited for the formation of mountains rather than volcanoes.

Have a nice day!! Good Luck!! Brainliest would be appreciated!!!

4 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

Does the under water pressure depend on the molecules of the air or what does it depend on

miskamm [114]

Answer:the pressure depends on gas and it will be half as much underwater

Explanation:

Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.

This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .

4 0

3 years ago

65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight

Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

Weight of liquid displaced (u)=2.5N

Let weight of body in air = W₁

Solution,

U=W₁-W₂

W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0

3 years ago