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kirza4 [7]
3 years ago
8

How does wind from? nnnnnnnnnn

Physics
1 answer:
irina [24]3 years ago
6 0

Answer:

It comes from the sun unevenly heating the earth. Creating warm and cool spots.

Explanation:

You might be interested in
Will give brainliest! A team of engineers is asked to evaluate several different
Levart [38]

Answer:

the answer if im not mistaken is A

Hope this helps:)

6 0
2 years ago
Who was the first scientist to propose that an object could emit only certain amounts of energy?
Rzqust [24]
It was Niels Bohr who proposed it
3 0
3 years ago
Planet x has a mass of 4x1022 kg and a radius of 6x105 m What lise the grav ritational field strength in the surface of planet X
SCORPION-xisa [38]

Answer:

  g ≈ 7.4 m/s²

Explanation:

The acceleration due to gravity on planet XX is ...

  g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

  g ≈ 7.4 m/s²

6 0
2 years ago
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

7 0
3 years ago
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?
Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u =  6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0
2 years ago
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