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3 years ago

Describe briefly difference between conduction, convection, and radiation.

Physics

1 answer:

Fed [463]3 years ago

3 0

The three types of heat transfer are conduction convection and radiation. Explanation:

Heat transfer happens only when there is said to be a temperature difference. (i.e) heat transfer occurs with a higher temperature region to a lower temperature region.
Heat is transferred usually via some medium in the case of convection and conduction whereas in the radiation no medium is required ( i.e it take place in a vacuum ).
Conduction is a type of heat transfer that is mainly a character for the conducting and semiconducting solid material.
Convection is the heat transfer method that usually takes place in the liquids and gases.
Radiation is the heat transfer process in which heat is transferred via infrared or electromagnetic waves.

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I E I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I E I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

2 years ago

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0

2 years ago

Which of the given will facilitate a normal Diels–Alder reaction?

babymother [125]

Answer:

Explanation:

- The rate of the Diels-Alder is orders of magnitude faster if there is an electron-withdrawing group on the dienophile. For example, replacing a hydrogen on ethene with the electron-withdrawing group CN results in about a 10^5 increase in the reaction rate.

- Other common electron withdrawing functional groups that will accelerate the Diels Alder reaction of dienophiles include aldehydes, ketones, and esters.

- In short, any functional group conjugated with the pi bond which can act as a pi acceptor will accelerate a Diels-Alder reaction with a typical diene.

- See attachment for graphical explanation.

7 0

3 years ago

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0

3 years ago

Marisol drives north at 64.3 km/h. How far does Marisol travel after 5.8 h?

Nady [450]

Answer:

372.94 km

d= st

= (64.3 km/h)(5.8h)

=372.94 km

3 0

3 years ago

Read 2 more answers