All categories

3 years ago

If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

1 answer:

7 0

·The acceleration of gravity is proportional to

   1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

   = 490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

   (9.8 m/s²) · (1/5)² = 0.39 m/s² .

The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)

= 19.6 newtons .

Just as we expected, his weight at that distance is

   (19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.

You might be interested in

Stars, like our Sun, are massive luminous balls of plasma held together by their own gravity, while galaxies are made of billion

Fudgin [204]

Galaxies are comprised of groups of stars circling a common gravitational point (which is theorized in the Milky Ways case to be a black hole). The answer is True.

5 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the y-component of velocity we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago

Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig

MissTica

Answer:

v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

Em_f = K = ½ m v_f²

Emo = Em_g

½ m v₀² + m g y₀ = ½ m v_f²

v_f² = v₀² + 2 g y₀

let's calculate

v_f = √(10² + 2 9.8 25)

v_f = 24.3 m / s

6 0

3 years ago

!!!HELP ASAP!!! in witch situation is work being done? theres a photo added.

Arturiano [62]

C..............................

8 0

3 years ago

Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ?

NeTakaya

Explanation:

A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.

the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.

3 0

3 years ago