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3 years ago

What provides the centripetal force needed to keep Earth in orbit?

Physics

2 answers:

anyanavicka [17]3 years ago

8 0

The centripetal force is the gravitational attraction
between the Earth and the Sun.

kkurt [141]3 years ago

3 0

The centripetal force needed to keep earth in orbit is gravity.

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What will happen when the sun blows up?

RSB [31]

We will all chaotically burn to death!

Hope this helps!!

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3 years ago

Read 2 more answers

A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horiz

xxMikexx [17]

Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²

Explanation:

8 0

3 years ago

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

leva [86]

Answer:

$t = 2.52 billion \:years$

Explanation:

As we know by radioactivity law

$N = N_o e^{-\lambda t}$

so here we will have

$N = 3 mg$

$N_o = 12 mg$

now we will have

$3 = 12 e^{-\lambda t}$

$\lambda t = ln 4$

now we also know that

$\lambda = \frac{ln2}{1.26 \times 10^6 yrs}$

$t = 1.26\times 10^6\times \frac{ln4}{ln2}$

$t = 2.52 billion \:years$

7 0

3 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

3 years ago