All categories

3 years ago

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

1 answer:

8 0

GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!

You might be interested in

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

6 0

2 years ago

A taxi starts from Monument Circle and travels 5 kilometers to the east for 5 minutes. Then it travels 10 kilometers to the sout

Morgarella [4.7K]

Answer: The taxi is moving with reference to A) Monument Circle. For each leg of the trip, the taxi's A) Average speed stays the same, but it's B) Average velocity changes.

Explanation: Brainliest Please!!!!

7 0

3 years ago

A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume

Assoli18 [71]

Answer:

Hence the pressure is $3\times 10^5 Pa$

Explanation:

Given data

Q=1500 J system gains heat

ΔV=- 0.010 m^3 there is a decrease in volume

ΔU= 4500 J internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is $3\times 10^5 Pa$

3 0

3 years ago

A 5 kg rock falls from an 8 m high cliff. What is the speed of the rock when it hits the ground?

Levart [38]

Answer:

3k mph

Explanation:

don't take my answer it is wrong

3 0

2 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

5 0

3 years ago