If an astronaut has a mass of 80kg on earth, what is their mass on the moon?

Find the gravitational force that Earth (mass = 5.97 × 10²⁴ kg) exerts on the moon (mass = 7.35 × 10²² kg) when the distance bet

alexandr402 [8]

Explanation:

mass of earth (m1)=5.97×10^24

mass of moon (m2)=7.35×10^22

distance between their center (d)= 3.84×10^8

G=6.67×10^-11

now,

gravitational force =(F)= G(m1×m2)/d²

6.67×10^-11(5.97×10^24×7.35×10^22)/(3.84×10^8)
19.84×10^19

<h3>stay safe healthy and happy...</h3>

3 0

3 years ago

An object is inside a room that has a constant temperature of 292 K. Via radiation, the object emits three times as much power a

maw [93]

Answer: Maybe he has the heater on

Explanation: If there is that much temperature then it has to be from something heated.

8 0

3 years ago

Given: 3x + y = 1.<br><br> Solve for y.<br><br> y = -3 x - 1<br> y = -3 x + 1<br> y = 3 x - 1

maw [93]

Answer:

y = -3x + 1

Explanation:

Isolate the variable, y. Note the equal sign, what you do to one side, you do to the other. Subtract 3x from both sides of the equation:

3x + y = 1

3x (-3x) + y = 1 (-3x)

y = -3x + 1

y = -3x + 1 is your answer.

~

7 0

2 years ago

A tennis racquet swung with an angular velocity of 12 rad/s strikes a motionless ball at a distance of 0.5 m from the axis of ro

Anestetic [448]

Answer:

The linear velocity of the racquet at the point of contact with the ball is 6 m/s.

Explanation:

Given;

angular velocity of the racquet, ω = 12 rad/s

distance of strike, r = 0.5 m

The linear velocity of the racquet at the point of contact is given by;

V = ωr

V = (12)(0.5)

V = 6 m/s

Therefore, linear velocity of the racquet at the point of contact with the ball is 6 m/s.

8 0

3 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

3 years ago