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rosijanka [135]
3 years ago
15

How many ways can you see that heat can transfer from the inside to the outside of your home on a cold day? List at least three

and describe in detail how heat moves through the different kinds of materials (glass, brick, wood, insulation, etc.)—by conduction, convection or radiation. Which process do you suppose dominates? Explain your reasoning.
Physics
1 answer:
Yanka [14]3 years ago
4 0

Answer: Heat transfers through brick walls and glass through conduction. In conduction heat is transferred by vibration of molecules. most energetic molecules vibrate and pass on the energy to less energetic molecules. Then they vibrate and further pass on the  energy. In this way heat is transferred out of the home. Heat also transfers through the leakage of warm air from doors and windows. This occurs through convection. In convection energy is transferred through bulk movement of liquid and air molecules. Heat also transfers from insulation. in insulation there is no material in between the layers. So heat transfer through insulation occurs through radiations that occurs by X-Rays, Ultravoilet rays etc.

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What is a material that reduces the flow of heat by conduction, convection, and radiation?
Oksi-84 [34.3K]
The answer is a insulator
4 0
3 years ago
Read 2 more answers
A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l
MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F = -kx

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

F=mg

Inserting this into Hooke's Law

mg=-kx

computing it for x,

-x=mg/k

This is the model which will tell the length of the spring against change in the mass located in the pan.

3 0
2 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
2 years ago
the brightness of stars as they appear from earth is measured by _______ magnitude. a. light b. apparent c. relative d. absolute
xxTIMURxx [149]
How bright a star appears from Earth is the star's apparent magnitude.
8 0
2 years ago
If an impulse of 400 Ns acts on an object for 15s, what is the force of the object?
Ksju [112]

Answer:

J for impulse

t for time

F for force

formula is J=F×t

Explanation:

putting values in eqs after rearranging

we need to find force so

F=J ÷t

F=400÷15

=26.67

=27(rounded off)

27N is the Force applied.

7 0
3 years ago
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