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miv72 [106K]
3 years ago
8

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

he surface of the moon (i.e., x = 6r)?
Physics
1 answer:
Cloud [144]3 years ago
7 0

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

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"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro
pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0
3 years ago
What type of change accurs when a substance stays the same
Thepotemich [5.8K]
This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.

The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.
3 0
3 years ago
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa
bearhunter [10]

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

8 0
3 years ago
Does a light bulb with a greater wattage have a greater brightness
Tamiku [17]
Only within the same technology. / / / If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
8 0
3 years ago
Read 2 more answers
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