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3 years ago

Help is requested. Will give brainliest to anyone who answers correctly.

Physics

1 answer:

sp2606 [1]3 years ago

4 0

The answer should be d because they are constantly rotating

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The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea

djverab [1.8K]

Answer:

Acceleration, $a=9.39\ m/s^2$

Explanation:

Given that,

Mass of the planet Krypton, $m=8.8\times 10^{23}\ kg$

Radius of the planet Krypton, $r=2.5\times 10^{6}\ m$

Value of gravitational constant, $G=6.6726\times 10^{-11}\ Nm^2/kg^2$

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}$

$a=9.39\ m/s^2$

So, the value of acceleration of an object in free fall near the surface of Krypton is $9.39\ m/s^2$

5 0

3 years ago

Pleases help with science ASAP!!

sattari [20]

Yes I am subscribed to coryxkenshin but are you subscribed to me jp

3 0

3 years ago

Read 2 more answers

A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000

allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

Mass of the car, m = 1700 kg
Velocity of the car, v = 21 m/s
Time of motion, t = 10 s
Additional energy provided by the engine, E₁ = 22,000 J
Energy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

$\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J$

The final speed of the car is calculated as follows;

$\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s$

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

4 0

3 years ago

A car travels from point A to Be in 3 hours and returns back to point A in 5 hours .Points A and B are 150 miles apart along str

Rainbow [258]

Explanation:

Given that,

The time taken from A to B is 3 hours

The time from B to A = 5 hours

The total distance between A and B is 150 miles

The average velocity from A to B,

$v=\dfrac{150}{3}=50\ mph$

The average velocity from B to A,

$v=\dfrac{150}{5}=30\ mph$

As the velocity of an object is a vector quantity. It means when the car reaches the initial point, the displacement is 0. So, the average velocity is equal to 0.

3 0

3 years ago

A sprinter can go from a state of resting to running at 7 m/s in just 3 seconds. What is the sprinter's average acceleration?

liberstina [14]

The formula for this equation is:
change in velocity/time taken = acceleration

7/3 = 2.3...

so the answer is 2.3 m/s
(I am not positive with the answer but I do know that is the formula for the equation)

4 0

3 years ago