Help me answer this ASAP

The Earth orbits the Sun with a period of 3.16 x 107 seconds (1 year!) at a distance of 1.5 x 1011 m. The mass of the Earth is 6

SVETLANKA909090 [29]

Answer:

a) M = 2 10³⁰ kg , b) $T_{m}$ = 5.81 10⁷ s

Explanation:

a) For this exercise let's use Newton's second law where force is the law of universal gravitation and acceleration is centripetal

G m M / R² = m a

a = v² / R

G M / R = V²

The orbit of the two planets is approximately circular, therefore the velocity module (speed) is constant

v = d / t

The distance is the length of the circular orbit

d = 2π R

G M / R = 4π² R² / T²

G M T² = 4π² R³

Let's write this equation for each planet

For the earth

The period is T = 3.16 10⁷ s and the radius of the orbit R = 1.5 10¹¹ m, let's calculate the mass of the sun

M = 4π² R³ / G T²

M = 4π² (1.5 10¹¹)³ / (6.67 10⁻¹¹ (3.16 10⁷)²)

M = 133.24 10³³ / 66.60 10³

M = 2 10³⁰ kg

b) For this part we write this equation for the two points

For the earth

$T_{E}$ ² = (4π² / G M) $R_{E}$ ³

For mars

$T_{m}$ ² = (4π² / G M) $R_{m}$ ³

Let's divide the two expressions

$T_{m}$ ² / $T_{E}$ ² = $R_{m}$ ³ / $T_{E}$ ³

They indicate that the orbit of Mars is $R_{m}$ = 1.5 $R_{E}$

$T_{m}$ ² / $T_{E}$ ² = (1.5 $R_{E}$ / $R_{E}$ )³

$T_{m}$ ² = $T_{E}$ ² 1.5³

$T_{m}$ ² = (3.16 10⁷)² 1.5³

$T_{m}$ = √ (33.70 10¹⁴)

$T_{m}$ = 5.81 10⁷ s

5 0

3 years ago

Problem:

kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a = -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

7 0

3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

4 years ago

Read 2 more answers

Beryllium has a charge of 2, and bromine has a charge of –1. which is the best name for the ionic bond that forms between them?

Otrada [13]

The best name for the ionic bond that forms between them is Beryllium Bromide.

We have been provided with data,

Beryllium charge, q = 2

Bromine charge, q = -1

As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.

Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.

So, the best name for the ionic bond that forms between them is Beryllium Bromide.

Learn more about ionic bonds here:

brainly.com/question/21464719

#SPJ4

8 0

3 years ago

What is the direction of the electric field at the corner? What is the direction of the electric field at the corner? along the

ehidna [41]

Answer:

jsjsj isso isso isms sim

Explanation:

jjed djdn djms disse sjsm disse mism

7 0

3 years ago