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swat32
3 years ago
15

An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and

its volume is 1.08×1012km3. What is the earth’s average density?
Physics
1 answer:
ladessa [460]3 years ago
5 0

Answer:

The answer to your question is 5.5 x 10¹² kg/km³

Explanation:

Data

density = rho = X

mass = 5.94 x 10²⁴ kg

volume = 1.08 x 10¹² km³

Process

To solve this problem use the density formula and simplify it to find the result.

Formula

density = \frac{mass}{volume}

Substitution

density = \frac{5.94 x 10^{24} }{1.08 x 10^{12}}

Simplification and result

density = 5.5 x 10¹² kg/km³

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The critical angle for total internal reflection occurs when:___________.
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Answer:

b) the refracted ray has an angle of 90 degrees

Explanation:

The angle of incidence is measured with respect to the normal separation of the media. The critical angle for total internal reflection occurs is:

\theta=arcsin(\frac{n_2}{n_1})

Here n_2 and n_1 are the refractive index of the mediums. This equation is an application of Snell's law, for the case where  the refracted ray has an angle of 90^\circ.

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2 years ago
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3 years ago
How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci
kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

Q=m*C_{p}*(T_{final}-T_{initial})

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

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T_initial = initial temperature = 27 [°C]

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Now replacing:

Q=450*0.47*(32-27)\\Q=1057.5[cal]

5 0
2 years ago
Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis
solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

v^{2}=u^2+2as

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 39.66m/s

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