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2 years ago

During crystallisation the crystals separate out from the hot ________solution of a substance on cooling

Physics

1 answer:

alekssr [168]2 years ago

6 0

Answer:

The process of separation or deposition of crystals from a hot saturated solution on gentle cooling of the solution is called 'crystallisation'.

Explanation:

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M= 1000 g g= 10 m/s2 h= 10 K.E at GROUND = 400 j M.E = ....... ........J the answer is 400 so plz explain how

Simora [160]

Answer:

M = 500 J

Explanation:

Given that,

Mass, M = 1000 g = 1 kg

Height, h = 10 m

Potential energy is given by :

P = mgh

P = 1×10×10

P = 100 J

The kinetic energy at ground = 400 J

Mechanical energy = sum of kinetic and potential energy

So,

M = 100 + 400

M = 500 J

So, the mechanical energy of the system is 500 J.

4 0

3 years ago

PLEASE PLEASE HELP! Referring to the diagram above, predict what will happen when the switch is closed. Explain your answer.

IgorC [24]

the batteries would heat up due to the over load of power not going into any thing and the screw driver is giving it a boost of energy

8 0

3 years ago

0.0010 kg pellet is fired at a speed of 52.2 m/s at a motionless 3.3 kg piece of balsa wood. When the pellet

Paraphin [41]

The formula for momentum is p = m*v

The conservation of momentum suggests:

m*vi = m*vf (initial mass times initial velocity = final mass times final velocity or initial momentum = final momentum)

(0.0010)(52.2) = (0.0010 + 3.3)vf

vf = (0.0010)(52.2)/(0.0010 + 3.3) = 0.0522/3.301 ≈ 0.01581 m/s

To the nearest thousandth ≈ .016 m/s

7 0

3 years ago

How much force is needed to accelerate a 1.5 kg physics book to an acceleration of 6<br> m/s^2?

aleksley [76]

Answer:

Force = 8.0 k g m / s

Explanation:

Force = mass x acceleration

Mass = 4.0 k g Acceleration = 2.0 m / s 2

Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2

3 0

2 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

2 years ago