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2 years ago

What is convection? what is radiation ? what conduction?

1 answer:

3 0

--ⓒ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).

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Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

Name and draw the circuit symbols for the

pishuonlain [190]

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Explanation:

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3 years ago

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DanielleElmas [232]

Answer:

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3 years ago

A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e Newton's first law of motion.

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

8 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago