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3 years ago

By approximately how much does your gravitational potential energy change when you jump

1 answer:

4 0

Potential Energy = height*mass*gravity
the height is factually the height of the center of mass, which will be varied the same as the lowest of your feet. So when you hurdle, the height that you jump can be used to distinguish the alteration in height of your center of mass.In other words, it is inclined on the mass of the human, be contingent on how extraordinary the person can jump. Supposing a man is 80 kg and he can jump such that his center of mass rises 0.3 m in Earth's standard gravitational field. This sums to approximately 240 Joules of increase in GPE.

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PLEASE HELP

GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

3 years ago

At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr

sattari [20]

Answer:bill 5 m/s. Jack:10 m/s

Explanation:

Cuz I took it

7 0

3 years ago

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago

What nuclear weapon is the biggest and who currently owns this bomb

erma4kov [3.2K]

The RDS-220 <span>hydrogen bomb, soviet </span>

5 0

3 years ago

Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0

2 years ago