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musickatia [10]
3 years ago
15

What should be the speed of an artificial satellite moving on a circular orbit around the Earth at a distance of 400 km from the

surface of the Earth? Express your answer in km / sec and also miles per hour. This exercise tells you that objects in space are typically moving very fast and gives you an idea of why the impact of a meteorite or comet on Earth can cause extinction level events. Just for fun, watch the movie "Deep impact" as a follow up to this exercise.
Physics
1 answer:
Drupady [299]3 years ago
5 0

Answer:

7.67001846 km/s or 17157.38529 mph

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth =  5.972 × 10²⁴ kg

m = Mass of satellite

v = Velocity of satellite

The distance between the Earth's center and the satellite is

r = 6371000+400000 = 6771000 m

As the centripetal force balances the force of gravity we have

\frac{mv^2}{r}=\frac{GMm}{r^2}\\\Rightarrow v=\sqrt{\frac{GM}{r}}\\\Rightarrow v=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{6771000}}\\\Rightarrow v=7670.01846\ m/s=7.67001846\ km/s

Converting to mph

7670.01846\times \frac{3600}{1609.34}=17157.38529\ mph

The velocity of the satellite is 7.67001846 km/s or 17157.38529 mph

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how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance
Sergio039 [100]

Answer:

W = F * s    

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ)     since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g    if applied force is in the horizontal direction

8 0
2 years ago
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
Can a relative velocity of two bodies be greater than the absolute velocity of
steposvetlana [31]

Yes, eg., when 2 bodies move in opposite directions

, the relative velocity of each is greater than the individual velocity of either

8 0
3 years ago
Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides
alukav5142 [94]

Answer:

Bob's angular speed is the same as that of lily

Explanation:

Because for a carousel the angular speed remains the same since velocity at center and edge are the same

6 0
3 years ago
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
3 years ago
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