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Arada [10]
3 years ago
8

When a car is weighed, it is driven slowly on a horizontal floor over a scale that records a reading as the front wheels go over

the scale, and then records a second reading as the rear wheels go over the scale. The weight of the car is equal to
A) the weight under the rear wheels.

B) the average of the two weights.

C) the sum of the two weights.

D) the weight under the front wheels.

E) the difference of the two weights.
Physics
1 answer:
Ymorist [56]3 years ago
7 0

Answer: C

Explanation:

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Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis
WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{Mm}{r^{2}} (1)

Where:

F is the gravitational force between Earth and Moon

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M=5.972(10)^{24} kg is the mass of the Earth

m=7.349(10)^{22} kg is the mass of the Moon

r=3.9(10)^{8} m is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

F=m.a_{C} (2)

Where a_{C} is the centripetal acceleration given by:

a_{C}=\frac{V^{2}}{r} (3)  

Being V the orbital velocity of the moon

Making (1)=(2):

m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

a_{C}=G\frac{M}{r^{2}} (5)

Making (5)=(3):

\frac{V^{2}}{r}=G\frac{M}{r^{2}} (6)  

Finding V:

V=\sqrt{\frac{GM}{r}} (7)

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}} (8)

Finally:

V=1010.92 m/s

5 0
3 years ago
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LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0
3 years ago
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