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yKpoI14uk [10]
3 years ago
5

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

d a mass of 130kg, how much would you weigh when standing on the surface of this asteroid?B. If you could walk on the surface of this asteroid, what minimum speed would you need to launch yourself into an orbit just above the surface of the asteroid?
Physics
1 answer:
WITCHER [35]3 years ago
3 0

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

F=\frac{GMm}{R^2}

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

\frac{GMm}{R^2}=\frac{mv^2}{R}

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s

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the same with that of products

Explanation:

In a chemical reaction, the total charge of the reactants must be the same with that of products.

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Learn more:

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calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.​
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25%
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How many items are present in the compound (NH4)2Cr2O7
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Or speed is distance over time

So...

40min / 60min = 0.6667 or 2/3 --> Finding what proportion 40 minutes is to an hour or 60 minutes as we need the units of hours to match up

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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
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Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

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