All categories

2 years ago

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.

You might be interested in

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