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3 years ago

How to find gravitational potential energy?

Physics

1 answer:

Nikitich [7]3 years ago

4 0

Mass x height x gravity

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The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom

BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart

Then 1/5 * 100 deg C = 20 deg C the requested temperature

6 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

A force F of magnitude 2x^3 is applied to stop a particle moving with an initial velocity of v0. The particle travels from x=0 t

3241004551 [841]

Answer:

Explanation:

Given that

F=2x³

Work is given as

The range of x is from x=0 to x=D

W=-∫f(x)dx

Then,

W=-∫2x³dx from x=0 to x=D

W=- 2x⁴/4 from x=0 to x=D

W=-2(D⁴/4-0/4)

W=-D⁴/2

W=1/2D⁴

The correct answer is F

5 0

3 years ago

How many cities could 8 TeV power for a week?

HACTEHA [7]

science hasnt figured it out yet

6 0

3 years ago

Read 2 more answers

The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s

Bumek [7]

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: $m_{(t)} =m_{0} e^{-kt}$ , where $m_{0}$ is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

$m_{(1)} =m_{0} e^{-k(1)}$ , but $m_{(1)}=\frac{99m_{0} }{100}$ , so we have $\frac{99m_{0} }{100}=m_{0}e^{-k}$ , then $k=-ln(\frac{99}{100})=0.01$

Now using k found we must to find $m_{(5)}$ .

$m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg$

6 0

3 years ago