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3 years ago

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Physics

2 answers:

Alenkasestr [34]3 years ago

8 0

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

$F=\frac{G\times m_1\times m_2}{r^2}$

G = gravitational constant

$m_1, m_2$ = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

$F\propto \frac{1}{r^2}$

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

Troyanec [42]3 years ago

8 0

Answer:

The correct answer is B.

Explanation:

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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) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

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 = 2.278 x 10^-5 volts

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

$I$ = E/R

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$I$ = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = $I$ E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

4 0

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Explanation:

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