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Anettt [7]
3 years ago
6

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

a) At what angleθis therock thrown? b) In terms of its original rangeR, what is the rangeRmaxtherock can attain if it is launched at the same speed but at the optimal anglefor maximum range? c) Would your answer to part a) be different if the rockis thrown with the same speed on a different planet? Explain.
Physics
1 answer:
Snowcat [4.5K]3 years ago
5 0
A) The vertical component of velocity v is taking the rock to a height

Vertical component = vsin\theta
The time taken to reach maximum height = \frac{vsin\theta}{g}
So total time of rocks flight = \frac{2vsin\theta}{g}
Range of rock is due to the horizontal component of velocity = vcos\theta
Range = \frac{2*v*sin\theta*v*cos\theta}{g} = \frac{2*v^2*sin\theta*cos\theta}{g}
Maximum height = \frac{g*t^2}{2} = \frac{v^2*sin^2\theta}{2*g}
Since range = maximum height
We have \frac{2*v^2sin\theta*cos\theta}{g} = \frac{v^2*sin^2\theta}{2*g}
tan\theta = 4
\theta = 75.96^0
So when angle of projection is \theta = 75.96^0 range is equal to maximum height reached.
b) We have range = \frac{2*v^2*sin\theta*cos\theta}{g} =\frac{2*v^2*sin2\theta}{g}
Maximum of range is reached when \theta = 45^0
Maximum range = \frac{2*v^2}{g}
c) For range to be equal to maximum height only condition is tan\theta = 4, it does not depend upon acceleration due to gravity and velocity. That angle is a constant.
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An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,
Taya2010 [7]

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}

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A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a
s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

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