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White raven [17]
3 years ago
14

10dm/s=...m/phút=......km/h

Physics
1 answer:
Lynna [10]3 years ago
8 0
150kmp miles to the second power add all together them together then you get 46 to the 3rd ad 104
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WORTH 47 POINTS!!! It's EASY too!
Vikentia [17]
The Factors are Temperature and kinetic energy, the temperature is because the particles are going to move fast which means the particles in a solid container or in solid pattern they will actually vibrate and they will expand , when temperature increase, more kinetic energy between the particles .
examples is that in the steel when you heat a steel the particles inside it will vibrate then the particles speed up because the vibration increases, therefore the temperature increases so a a thermal expansion occurs that the vibration of the particles will take up more space so the steel bar expands slightly in all Direction if the temperature Falls the reverse happens and the material or steel contracts which means get smaller .
another example is the thermometer, the thermometer has a liquid inside it which is Mercury or alcohol this liquid expands when the temperature rises, the tube is made narrow so that a small increase in volume of the liquid produces a large movement along the tube.

5 0
3 years ago
True or False:
Leto [7]
True
False
True
My answers
4 0
3 years ago
A man has a mass of 66 kg on earth. What’s his weight in pounds?
Vlad1618 [11]

Answer:

646.8 N

Explanation:

3 0
2 years ago
A student is performing experiments on a particular substance. Which
boyakko [2]

Answer: A.

Explanation:

6 0
2 years ago
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that
NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0
3 years ago
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