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jeyben [28]
3 years ago
11

What happens to jetstream’s as they get closer to the equator

Physics
2 answers:
MAVERICK [17]3 years ago
6 0

Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

Semmy [17]3 years ago
6 0

Answer:

Hey!

Jet Streams don't really tend to move towards the Equator...

But when the area they are located becomes warm, the tend to STOP SHIFTING and deliver a constant temperature of air (most commonly it is warm air)

They follow boundaries formed when air MOVES AWAY FROM THE EQUATOR and the EARTHS ROTATION!

Explanation:

HOPE THIS HELPS!!

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aev [14]

Answer:

C

Explanation:

through the desk....here desk is the student's medium to hear the sound. its oblivious because when he lifts his head away from the desk he hears nothing else

6 0
3 years ago
On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2 )/6 = 1.63 m/s2 .
MAVERICK [17]
The pendulum would differ from 300 inches
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3 years ago
Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni
nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0
3 years ago
Read 2 more answers
When using a micron gauge, if the pressure gauge falls frim 1,300 to 1,000 instantly, the vehicle needs to?
julsineya [31]
If the micron gauge pressure falls instantly, it means that the moisture of the air-conditioning is frozen, therefore you need to move the vehicle to a warmer place in order to unfreeze the moisture.
5 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
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