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2 years ago

The atmosphere of earth does not escape into the space but remains attached to earth's surface why??

1 answer:

7 0

Just like any other gas or mixture of gases, the gas molecules are
zipping around in all different directions and with a whole range of
different speeds.

Those that happen to be moving at a speed greater than the Earth's
"escape velocity", AND are pointed away from Earth, AND don't hit
any other molecules before they escape, are lost.

With the combination of Earth's escape velocity, and the temperatures,
thickness, and density of the atmosphere, that process happens slowly
enough to have maintained an atmosphere around this planet until now.

Personally, I hope it hangs around for a while longer. But with the constant
increase in temperature that's been going on, you never know . . .

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The number ocean waves that pass a buoy in one second is _ of the wave

mr_godi [17]

The number of ocean waves that pass a buoy in one second is the frequency of the <span>wave. The crest of a transverse wave is its highest point. </span>

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2 years ago

Read 2 more answers

Marianne and her family are driving from West Virginia to California. While driving through the Rocky Mountains, they encounter

Elena-2011 [213]

Answer:

Point A

Explanation:

Point C is way to close to where the force would need to be.

Point B is close but not as close, the reasonable answer would be point A.

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2 years ago

3. The electric field of a sinusoidal electromagnetic wave has an amplitude of 5.0 V/m. How much radiation energy passes through

True [87]

Answer:

e) $179$ J

Explanation:

$E_{o}$ = Magnitude of electric field = 5 V/m

$A$ = Area of window = 1.5 m²

$c$ = speed of electromagnetic wave = 3 x 10⁸ m/s

$\Delta t$ = time interval = 1 h = 3600 sec

radiation energy is given as

$U = (0.5)\epsilon _{o}E_{o}^{2}cA\Delta t$

$U = (0.5)(8.85\times 10^{-12})(5)^{2}(3\times 10^{8})(1.5)(3600)$

$U = 179$ J

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2 years ago

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

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2 years ago

The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo

weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since $L=I \times \omega$ , the turning rate of the two-puck framework must stay unaltered.

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2 years ago