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balandron [24]
2 years ago
15

What is gravitonal force

Physics
1 answer:
zhuklara [117]2 years ago
8 0

Answer:

its something that hold the air for forceing liy by the exgen

Explanation:

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A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti
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B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.
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I believe it’s density and temperature
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A basic statement established by experiment or observation.
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c.Law

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i think you must learn and write however good luck

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On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters
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Answer:

d_1 = 16 d

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As we know that initial speed of the fall of the stone is ZERO

v_i = 0

also the acceleration due to gravity on Mars is g

so we have

d = v_i t + \frac{1}{2}gt^2

now we have

d = 0 + \frac{1}{2}g t^2

now if the same is dropped for 4t seconds of time

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A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

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