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Lady_Fox [76]
3 years ago
12

How long does it take a cup of fresh water to freeze, boil and evaporate compared to fresh water

Physics
1 answer:
eduard3 years ago
7 0
They exhibit identical behaviors with regard to changes of phase.
If there is any difference, it is too small to be measured, and none
has been observed.
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Dont knwo the answers for question 4,5,6
algol [13]
4. bc it is softer--made of a different material-- has a weaker reaction force
5. the sand would not leave footprints and would rather act as a whole
6. there is no gravity in outer space-- she can just float back to the spaceship no problem!
3 0
3 years ago
A stone is thrown in the upwards direction at the velocity of 4 m/s. It attains a certain height and then it falls back. During
Sati [7]

Answer: 0.8 m

Explanation:

Velocity of throw = 4m/s

Maximum Height attained(h) =?

Downward acceleration experienced = 10m/s^2

Using the relation:

v^2 = u^2 + 2aS

v = final Velocity = 0 (at maximum height)

u = Initial Velocity = 4

a = g downward acceleration = - 10

0 = 4^2 + 2(-10)(S)

0 = 16 - 20S

20S = 16

S = 16 / 20

S = 0.8m

Maximum Height attained = 0.8m

3 0
3 years ago
A cubical furnace with external dimensions of 0:6 m is constructed from reclay brick. If the wall thickness is 40 mm, the inner
Alja [10]

Answer:

Explanation:

surface area of furnace = 6 x .6²

=2.16 m²

Rate of heat loss = k A ( θ₂ -θ₁ ) / t where k is coefficient of thermal conductivity , A is surface involved , θ₂ -θ₁  is temperature difference and t is thickness .

For fire clay thermal conductivity

= .05 W / m . K

.05 x 2.16 x ( 630 - 70 ) / .04

= 1512 W

= 1.5  kW .

8 0
3 years ago
When they are far apart, the momentum of a proton is 3.2 ✕ 10−21, 0, 0 kg · m/s as it approaches another proton that is initiall
Rina8888 [55]
<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction. (3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s After the interaction (2.6x10^-21, 1.40x10^-21, 0) kg*m/s and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be 3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of (6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>
5 0
3 years ago
50 POINTS! pls answer
fredd [130]

Thanks for points...

But can you type this question.. As it becomes hard to answer if it's in the picture..

4 0
3 years ago
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