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3 years ago

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A satellite encircles Mars at a distance above its surface equal to 3 times the radius of Mars. If g m is the acceleration due t

o gravity at the surface of Mars, what is the acceleration due to gravity at the location of the satellite?

1 answer:

Sav [38]3 years ago

6 0

Acceleration due to gravity at the surface of Mars is given by the formula

$g_m = \frac{GM}{R^2}$

here

M = mass of Mars

R = radius of Mars

Now if we take a point which is at height 3 times the radius of Mars from its surface

So the acceleration due to gravity is given as

$g = \frac{GM}{(3R+R)^2}$

now using this

$g = \frac{GM}{16R^2}$

$g = \frac{g_m}{16}$

so here the gravity is 16 times lesser than gravity at surface

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Three identical resistors have an equivalent resistance of 85 ω when connected in parallel. part a what is their equivalent resi

pishuonlain [190]

The equivalent resistance of the three resistors when connected in parallel is:
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}$
Since the three resistors in this problem are identical, we can call R their resistance, and we can rewrite the previous equation as
$\frac{1}{R_{eq}} = \frac{1}{R}+ \frac{1}{R}+ \frac{1}{R}= \frac{3}{R}$
And since we know the value of the equivalent resistance, $R_{eq}=85 \Omega$ , we can find the value of R:
$R=3 R_{eq} = 3 \cdot 85 \Omega=255 \Omega$

Now the problem asks us what is the equivalent resistance of the three resistors when they are connected in series. In this case, the equivalent resistance is just the sum of the three resistances, so
$R_{eq} = R+R+R=3 R= 3 \cdot 255 \Omega = 765 \Omega$

7 0

3 years ago

Read 2 more answers

As you run toward a source of sound, you perceive the frequency of that sound to

Nata [24]

As you run toward a source of sound, you perceive the frequency of that sound to decrease.

<u>Explanation:</u>

Doppler's effect is a principle used to describe the frequency and the intensity of sound and wavelengths of a source and observer with the two possibilities.

(i) Stationary sound source and moving observer.

(ii) Moving sound source and a stationary observer. It is a relative motion.

Consider when the observer is moving towards a source, the frequency of the sound will be higher and when moving away from the source, the frequency will decrease.

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4 years ago

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Two physics students, Ben and Bonnie, are in the weightlifting room. Bonnie lifts the 50 kg barbell over her head (approximately

KatRina [158]

Answer:

Both of them

Explanation:

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3 years ago

A 3 kg block is released from rest to slide down a ramp with friction. The length that the block slides is 2 meters and the angl

Arte-miy333 [17]

Answer:

(a). The acceleration is 3.20 m/s².

(b). The amount of work done by each force is 19.2 J.

(c). The total work done on the block is 19.2 J.

(d). The final speed of the block is 6.26 m/s.

Explanation:

Given that,

Mass of block = 3 kg

Distance = 2 m

Angle = 30°

Coefficient of kinetic friction = 0.20

(a). We need to calculate the acceleration

Using balance equation of force

$N = mg\co\theta$

$mg\sin\theta-f_{k}=ma$

$a = g\sin\theta-\mu g\cos30$

Put the value into the formula

$a=g(\sin30-0.20\cos30)$

$a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})$

$a=3.20\ m/s^2$

The acceleration is 3.20 m/s².

(b). We need to calculate the amount of work done by each force

Using formula of work done

Normal force is

$N = mg\cos30$

So due to this the net force is zero then the no work done by reaction force.

By another force,

$W= F\times d$

$W=ma\times d$

Put the value into the formula

$W= 3\times3.20\times2$

$W=19.2\ J$

The amount of work done by each force is 19.2 J.

(c). We need to calculate the total work done on the block

The total work done on the block is 19.2 J.

(d). We need to calculate the final speed of the block

Using equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=0+2\times9.8\times2$

$v=\sqrt{2\times9.8\times2}$

$v=6.26\ m/s$

The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

4 0

3 years ago

Introduction Team Problem 2 You jump into the deep end of Legion Pool and swim to the bottom with a pressure gauge. The gauge at

Kaylis [27]

Answer:

Depth of the pool, h = 4.004 cm

Explanation:

Pressure at the bottom, P = 39240 N/m²

The density of water, d = 1000 kg/m³

The pressure at the bottom is given by :

P = dgh

We need to find the depth of pool. Let h is the depth of the pool. So,

$h=\dfrac{P}{dg}$

$h=\dfrac{39240\ N/m^2}{1000\ kg/m^3\times 9.8\ m/s^2}$

h = 4.004 m

So, the pool is 4.004 meters pool. Hence, this is the required solution.

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4 years ago