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Maslowich
2 years ago
15

A man can jump 1.5 m on earth. Calculate the approximate height

Physics
1 answer:
nataly862011 [7]2 years ago
7 0

Answer:

Mass = volume x density. Volume(2) = 1/3, density is 1/4, so

Mass = 1/3 * 1/4 = 1/12 that of Earth.

Surface gravity = m /r^2 = 1/12 / (1/4)^2 = 1/12 / 1/16 = 16/12 = 4/3 Earth’s

So… surface gravity would by 13 m/s^2 vs. 9.8m/s^2

Someone able to jump up 1.5m vs. 9.8m/s^2 is imparting 5.4 m/s of velocity on themselves. Imparting 5.4m/s of velocity on yourself against 13m/s^2 gravity, you would reach 0 velocity at:

1.12m in height.

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Plsssss I need te answer quick​
jok3333 [9.3K]

Explanation:

reflection ... . .......

3 0
3 years ago
Read 2 more answers
What’s the acceleration if the average velocity is 3.5 and the time is 8.7
Monica [59]
Vf = 0 + 3.5•8.7
= 30.45 m/s
6 0
3 years ago
Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma
Sveta_85 [38]

Answer:

9\cdot 10^9 N

Explanation:

The magnitude of the electrical force between the two point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 =q_2 = +3.0 C is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N

3 0
3 years ago
Plz help >:
svlad2 [7]

Answer:

10m

Explanation:

The object distance and image distance is the same from the mirror. so the image is 5m behind the mirror.

5+5=10

5 0
2 years ago
Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of t
Kipish [7]

Answer:

speed of golf ball is 1.15 × 10^{-30} m/s

and % of uncertainty in speed =  2.07 × 10^{-30} %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = 10^{-3} m

to find out

speed of the golf ball and  % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ \frac{h}{4\pi }    ......1

Δx × ΔPx  ≥ \frac{h}{4\pi }

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × 10^{-34} Js

so put here all these value in equation 1

10^{-3} × 0.0459 × ΔVx =  \frac{6.626*10^{-34}}{4\pi }

ΔVx = 1.15 × 10^{-30} m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed =  1.15 × 10^{-30}  / 55.5

% of uncertainty in speed =  2.07 × 10^{-30} %

3 0
3 years ago
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