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Bess [88]
4 years ago
8

At what distance above the surface of the earth is the gravitational field 4.9 m/s^2

Physics
1 answer:
san4es73 [151]4 years ago
5 0

That's 1/2 of what it is on the surface.

The distance between the center of the Earth and any object
on the surface is 1 Earth radius ... about 3960 miles.

Gravitational force is inversely proportional to the square of
the distance between the centers of the objects, so in order
to reduce the acceleration of gravity by 1/2, you increase the
distance by √2 .

           (3960 miles) x (√2) = 5,600 miles from the center

                                         = 1,640 miles above the surface.

                                          
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A type of energy embodied in oscillating electric and magnetic fields is called
yaroslaw [1]

Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

4 0
3 years ago
8. A tortoise moves a distance of 100 metres in 15 minutes. What is the average speed of
myrzilka [38]
Answer: 0.4 km/h



100 m = 0.1 km
15min./60min. = 0.25 hr

so, divide the total distance (100 m) by the total time (15 min)
0.1 km/0.25 hr = 0.4
8 0
3 years ago
Giving 12 points for this question.
Ratling [72]
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4 0
3 years ago
A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made
olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

B=\dfrac{\mu_oI}{2\pi r}

B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}

B = 0.000002 T

B=10^{-5}\ T

(2) Exact formula:

B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }

B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

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3 years ago
What is the typical treatment for Hodgkin's disease?
Evgesh-ka [11]

Answer:

Pretty sure its D.

radiation, chemotherapy, and bone marrow transplant

Explanation:

7 0
3 years ago
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