Ask question

Ask question

All categories

3 years ago

12

6x10^8 kg ————— 2x10^4 m^3

1 answer:

barxatty [35]3 years ago

4 0

The answer to the question

You might be interested in

Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Lelu [443]

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

5 0

3 years ago

Question one please! This is 8th grade science btw.

yKpoI14uk [10]

Good Luck, NYAYAYAYAYa

8 0

3 years ago

How can tidal force from the moon affect our earth?

VARVARA [1.3K]

Answer:

sorry but I can understand the question

5 0

2 years ago

Read 2 more answers

A 3kg book falls from a 2m tall bookshelf what is the speed <br><br> PICTURE included

mafiozo [28]

Answer:

39.2m/s

Explanation:

The potential energy the book has right before it falls is equal to the kinetic energy in falling.

PE = KE

mgh = (1/2)mv

2gh=v

v=(2)(9.81)(2)

v=39.24m/s

8 0

3 years ago

A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

5 0

3 years ago