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padilas [110]
3 years ago
13

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,

the acceleration due to gravity is what factor times the value of g at the Earth's surface?
Physics
1 answer:
Taya2010 [7]3 years ago
7 0

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

F=\frac{GM_1M_2}{r^2}

Force F=mg'=\frac{GMm}{4R^2}-----1

Force on earth surface F=mg=\frac{GMm}{R^2}------2

Divide 1 and 2 we get

\frac{g'}{g}=\frac{R^2}{4R^2}

g'=\frac{g}{4}  

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Who pioneered the use of galvanoplastic compounds for preserving footprints and ballistics?
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Answer:

Option D (Alphonse Bertillon) is the correct response.

Explanation:

  • He seems to have been a policeman turned biometrics expert from France. Forensic techniques such as forensic record analysis were developed by Bertillon.
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All those other choices weren’t connected to the instance offered. So, the best one is the one described.

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Who was the first scientist to propose that an object could emit only certain amounts of energy?
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It was Niels Bohr who proposed it
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A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
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