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1 year ago

Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.

Physics

1 answer:

Nookie1986 [14]1 year ago

4 0

(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

M is mass of the planet
r is radius of the planet

v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 7,338.93 m/s

<h3>Escape velocity of the satellite</h3>

v = √2GM/r

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

<h3>Speed of the satellite at the given period </h3>

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

The radius of the synchronous orbit of a satellite is 69,801 km .

Learn more about minimum speed here: brainly.com/question/6504879

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Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

$R = \frac{u^{2}Sin2\theta }{g}$

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

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u = 31.55 m/s

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3 years ago

Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f

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Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m

μo = 4 π × 10^-7 Tm/A

The formula for the force per unit length between the two wires is given by

F = μo / 4π x (2 i1 x i2) / d

3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082

I = 0.0362 A = 36.22 mA

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3 years ago

Which of the following statements describe the transfer of energy?

Lady_Fox [76]

The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.

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2 years ago

Plz i need help for the 5 problems. plz show the work!!!

Artemon [7]

Answer:

1. 3 $m/s^{2}$

2. 1.5 $m/s^{2}$

3. 3 seconds

4. 0 $m/s^{2}$

5. 2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = $\frac {30-0}{10}=3 m/s^{2}$

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=10m/s, v=22m/s and t=8 seconds

a = $\frac {22-10}{8}=1.5 m/s^{2}$

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=0m/s since at rest, v=15m/s and a=5 $\frac {m}{s^{2}}$

= $\frac {15-0}{5}=3s$

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When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

$a=\frac {v-u}{t}$

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = $\frac {20-20}{10}=0 m/s^{2}$

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

$t=\frac {v-u}{a}$

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 $\frac {m}{s^{2}}$

= $\frac {0-9}{-4.1}=2.2s$

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2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

$s_2=\frac{32.4+0}{2}(5)=81 m$

So, the total displacement of the car is

$s=s_1+s_2=324+81=405 m$

Learn more about accelerated motion:

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3 years ago