If Earth's clouds did not trap and release heat, how would most climate zones on Earth be affected?

(a) Check all of the following that are correct statements, where E stands for γmc2. Read each statement very carefully to make

Reil [10]

Answer:

V=9.60m/s

Attached is the solution

5 0

3 years ago

Please help with this one

olasank [31]

Answer:

OPTION A is the correct answer

6 0

3 years ago

The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths.less for long wavelengths a

kompoz [17]

<h2>Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>

Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

This means the smaller the slit or obstacle (diffracting object), the wider the resulting diffraction pattern, and the greater the obstacle, the narrower de resulting patter.

6 0

3 years ago

If the particle is located slightly to the left of point B, its acceleration is __________.

Igoryamba

(1) directed to the right

Explanation:

To the right of B, u(x) is a decreasing function & so its derivative is negative,this implies that the x component of the force on a particle at this position is positive,or that the force is directed towards right .Small deviations from equilibrium at point B causes a force to accelerate the particle away ,hence particle is in unstable equilibrium.

7 0

4 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago