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I am Lyosha [343]
3 years ago
8

Help please !!I need help with this!

Physics
1 answer:
belka [17]3 years ago
6 0
Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m
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an astronaut on earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. if she were instead in
nadya68 [22]

The gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

<h3>What is gravitational force?</h3>

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

\rm F = \dfrac{Gm_1m_2}{r^2}

Where G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

Weight of ice cube = weight of soft drink displaced

mg = ρVg

m = ρV

g does not affect astronauts on earth in a lunar module.

Thus, the gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.

Learn more about the gravitational force here:

brainly.com/question/12528243

#SPJ1

3 0
1 year ago
A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel
Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

3 0
3 years ago
How do we use light in space?
bulgar [2K]

Answer:

They use LED lights.

Explanation:

Hope this helps

-A Helping Friend (mark brainliest pls)

5 0
3 years ago
A circular loop of radius 0.0400 m is
Alenkinab [10]

Let's see

Find enclosed area of loop

\\ \rm\dashrightarrow \pi r^2

\\ \rm\dashrightarrow 3.14(0.04)^2

\\ \rm\dashrightarrow 0.005m^2

Now

\\ \rm\dashrightarrow \phi=BAcos\theta

  • We need B

\\ \rm\dashrightarrow 9.59\times 10^{-7}=B(0.005)cos75

\\ \rm\dashrightarrow 9.59\times 10^{-7}=0.0046B

\\ \rm\dashrightarrow B=2084.78\times 10^{-7}

\\ \rm\dashrightarrow B=2.08\times 10^{-4}T

\\ \rm\dashrightarrow B=0.208mT

4 0
2 years ago
An 8μC charge is moving through a 10T magnetic field at a speed of 2.5⋅107 m/s perpendicular to the direction of the field. What
tangare [24]

check photo for solution

6 0
3 years ago
Read 2 more answers
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