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2 years ago

7

Hell please thanks!!!!!!’

1 answer:

Slav-nsk [51]2 years ago

5 0

Answer:

liquid phase

Explanation:

it is liquid phase because molecules are not that tightly packed as solid and not that far away from each other as in gas phase.

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The catapults on the uss george h.w. bush can launch aircraft from rest to a speed of 150 mph over a distance of 270 feet. find

vlabodo [156]

Hope this helps, have a great day ahead!

8 0

2 years ago

Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry

USPshnik [31]

Answer is given below

Explanation:

This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle
So, most of the weight of full loaded wheelbarrow's load goes on that's wheel and due to friction force between wheel and surface it can easy to move
He uses force to rotate the wheel, much more than the force applied to the rim of the wheel on the axis of rotation or torque

3 0

3 years ago

The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the gr

WINSTONCH [101]

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

6 0

2 years ago

Your environment can play a role in whether or not a specific gene is expressed

Hatshy [7]

I don't think so. How are you supposed to write the answer?

4 0

3 years ago

Read 2 more answers

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

2 years ago