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3 years ago

How does the theory of relativity explain the gravity exerted by massive objects?

Physics

1 answer:

Maslowich3 years ago

8 0

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

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1. What is the arrangement of the outer planets? 2. What effect does their placement have the planets?

Rasek [7]

the arrangement of the outer planets is
1. Mercury
2. Venus 
3. Earth 
4. Mars 
5. Jupiter 
6. Saturn 
7. Uranus 
8. Neptune
the inner most of the outer plannets is jupitor it is followed by saturn uranus and neptune

4 0

3 years ago

STUDY QUESTIONS

tankabanditka [31]

Answer:

The blade of sharpener is made up of iron. Iron is a magnetic material because of this pencil sharpener gets attracted by the poles of a magnet although the body is made up of plastic.

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3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

Determine the maximum r-value of the polar equation r =3+3 cos 0

AURORKA [14]

[r] =6

Solve for r by simplifying both sides of the equation, then isolating the variable.

I hope this makes sense

8 0

3 years ago

Read 2 more answers

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago