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Luba_88 [7]
3 years ago
11

In most cases, what happens to a liquid when it cools?

Physics
1 answer:
Stolb23 [73]3 years ago
5 0
When  a liquid cools is usually turns into a solid.
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What is convection?<br> what is radiation ?<br> what conduction?
stepan [7]

--ⓒᴄᴏɴᴠᴇᴄᴛɪᴏɴ ᴏᴄᴄᴜʀꜱ ᴡʜᴇɴ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ᴀ ʟᴏᴛ ᴏꜰ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ ɪɴ ᴀ ʟɪQᴜɪᴅ ᴏʀ ɢᴀꜱ ᴍᴏᴠᴇ ᴀɴᴅ ᴛᴀᴋᴇ ᴛʜᴇ ᴘʟᴀᴄᴇ ᴏꜰ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ʟᴇꜱꜱ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ. ... ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴇxᴘᴀɴᴅ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ. ᴛʜɪꜱ ɪꜱ ʙᴇᴄᴀᴜꜱᴇ ᴛʜᴇ ᴘᴀʀᴛɪᴄʟᴇꜱ ɪɴ ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴍᴏᴠᴇ ꜰᴀꜱᴛᴇʀ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ ᴛʜᴀɴ ᴛʜᴇʏ ᴅᴏ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ᴄᴏʟᴅ.

--Ⓡʀᴀᴅɪᴀᴛɪᴏɴ ɪꜱ ᴇɴᴇʀɢʏ ᴛʜᴀᴛ ᴄᴏᴍᴇꜱ ꜰʀᴏᴍ ᴀ ꜱᴏᴜʀᴄᴇ ᴀɴᴅ ᴛʀᴀᴠᴇʟꜱ ᴛʜʀᴏᴜɢʜ ꜱᴘᴀᴄᴇ ᴀɴᴅ ᴍᴀʏ ʙᴇ ᴀʙʟᴇ ᴛᴏ ᴘᴇɴᴇᴛʀᴀᴛᴇ ᴠᴀʀɪᴏᴜꜱ ᴍᴀᴛᴇʀɪᴀʟꜱ. ... ᴛʜᴇ ᴋɪɴᴅꜱ ᴏꜰ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀʀᴇ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ (ʟɪᴋᴇ ʟɪɢʜᴛ) ᴀɴᴅ ᴘᴀʀᴛɪᴄᴜʟᴀᴛᴇ (ɪ.ᴇ., ᴍᴀꜱꜱ ɢɪᴠᴇɴ ᴏꜰꜰ ᴡɪᴛʜ ᴛʜᴇ ᴇɴᴇʀɢʏ ᴏꜰ ᴍᴏᴛɪᴏɴ). ɢᴀᴍᴍᴀ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀɴᴅ x ʀᴀʏꜱ ᴀʀᴇ ᴇxᴀᴍᴘʟᴇꜱ ᴏꜰ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ ʀᴀᴅɪᴀᴛɪᴏɴ

--Ⓒᴄᴏɴᴅᴜᴄᴛɪᴏɴ ɪꜱ ᴛʜᴇ ᴡᴀʏ ɪɴ ᴡʜɪᴄʜ ᴇɴᴇʀɢʏ ɪꜱ ᴛʀᴀɴꜱꜰᴇʀʀᴇᴅ (ᴛʜʀᴏᴜɢʜ ʜᴇᴀᴛɪɴɢ ʙʏ ᴄᴏɴᴛᴀᴄᴛ) ꜰʀᴏᴍ ᴀ ʜᴏᴛ ʙᴏᴅʏ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴏɴᴇ (ᴏʀ ꜰʀᴏᴍ ᴛʜᴇ ʜᴏᴛ ᴘᴀʀᴛ ᴏꜰ ᴀɴ ᴏʙᴊᴇᴄᴛ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴘᴀʀᴛ).

3 0
2 years ago
a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he
Naya [18.7K]
Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.
8 0
3 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0
3 years ago
A block lies on a horizontal frictionless surface and
zhenek [66]

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

Fx=\dfrac{1}{2}kx^2

\\\Rightarrow x=\dfrac{2F}{k}

\\\Rightarrow x=\dfrac{2\times 3}{60}

\\\Rightarrow x=0.1\ m

The position of the block is 0.1 from the initial position.

6 0
3 years ago
A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with
Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

              = \dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2

              = 31 J

Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0
3 years ago
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