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Elenna [48]
3 years ago
5

What would be some benefits of living on a planet with less surface gravity

Physics
2 answers:
Ket [755]3 years ago
4 0
This is a tricky one.

-- IF I evolved on that planet, then my body is perfectly matched to the
conditions and the environment over there, and I am approximately just
as happy and care-free la de da over there as I really am on Earth,
and I can't imagine living anywhere else.

-- IF I evolved on Earth, and I move this body to a planet with less
surface gravity, then there are all kinds of complications.

. . . Even if the atmosphere over there has the same gases in the
same ratios as Earth's atmosphere, the lower gravity means that the
atmosphere has lower weight.  In turn, that means that the surface
pressure is lower, and I have to work harder to take in enough oxygen to
keep my Earth body functioning.  Breathing alone could be a full-time job.

. . . The acceleration of gravity is less than 9.8 m/s² over there.
This means that when I drop something, it falls slower than I'm used to,
and I can usually grab it before it hits the floor.
When I lift something with the normal force of my Earth arm, it jumps up
faster than I expect.  Until I get used to things, I'll probably overshoot,
lift things too high and too fast, maybe spill the coffee etc.

. . . With my Earth muscles, almost everything takes less force. 
It's a lot easier to walk around.  I bob up and down more than usual,
and my steps are longer.  When I run, it's really funny.  Every time
I take a step, I sail several feet into the air, and come down farther
than I ever did, but it all happens in slow motion.

All of this is fun while it lasts, but it doesn't last long.  My Earth body
gets adjusted to the new planet.  Before long, I actually lose muscle mass
AND bone, so that if I were to go back to Earth, I would be so weak that
I'd have to be carried around for a while.

We know this from what happens to the astronauts who spend 6 months
or a year in zero G on the International Space Station.  Even when they
do resistance exercises for an hour a day, they come back with less
muscle strength and less bone mass.
zepelin [54]3 years ago
3 0
You could jump high!
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Please I really need help with this one
bixtya [17]
Cutting pepper and onions
7 0
3 years ago
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0
3 years ago
What is the gravitational force between two trucks, each with a masof 2.0x10^4 k
eduard

Answer:

6.67\cdot 10^{-3} N

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the two masses

r is the separation between the two masses

In this problem, we have

m_1 = m_2 = 2.0 \cdot 10^4 kg the mass of each truck

r = 2.0 m is their separation

Substituting,

F=(6.67\cdot 10^{-11}) \frac{(2.0\cdot 10^4 kg)^2}{(2.0 m)^2}=6.67\cdot 10^{-3} N

7 0
4 years ago
Instructions:Select the correct answer. A meteor is approaching Earth. Which statement about its motion is true? The meteor trav
Lynna [10]
<span> The meteor accelerates after it is brought into earths force field 

</span>
3 0
3 years ago
Read 2 more answers
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
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