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3 years ago

Which term describes the amount of water vapor in the air?

Physics

2 answers:

Reika [66]3 years ago

6 0

Answer:

Humidity

Explanation: its 100% verified

loris [4]3 years ago

4 0

That would be humidity

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What happens if you drop a chicken wing from the top of the Eiffel Tower

horsena [70]

Answer:

It either goes WEEEEEEE. Or it just breaks apart.

Explanation:

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2 years ago

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Humans obtain energy in a number of different ways. About half of the electrical energy in the U.S. is generated by burning coal

ivanzaharov [21]

Answer:

C. Burning coal tends to harm the environment more than using solar panels.

Explanation:

When coal is burned, it reacts with the oxygen in the air. This reaction converts the stored potential energy, which turns into thermal energy, which is released as heat. But it also produces methane and carbon dioxide which is released into the air.

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3 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho

NISA [10]

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u = 50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

v = u + at

0 = 25 - 9.81 x t

t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u = 50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

5 0

3 years ago

A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on

Juliette [100K]

Answer:

k = $5\times 10^{4}\ N/m$

b = $0.707\times 10^{3}$

t = $7.1\times 10^{- 5}\ s$

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

$v^{2} = u^{2} + 2gh$

where

u = initial velocity = 0

g = 10 $m/s^{2}$

Thus

$v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s$

Force on the mass is given by:

F = mg = $1000\times 10 = 10000 N = 10\ kN$

Also, we know that the spring force is given by:

F = - kx

Thus

$k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m$

Now, to find the damping constant b, we know that:

F = - bv

$b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}$

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

$t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s$

4 0

3 years ago