All categories

3 years ago

Heat from the surface is transferred underground by _______.

Physics

2 answers:

gulaghasi [49]3 years ago

7 0

Answer: Option (D) is the correct answer.

Explanation:

When we place two objects of different temperature adjacent to each other then there will occur transfer of heat from hotter object to colder object till their temperature becomes equal.

This process is known as conduction.

Hence, when heat is being transferred from surface to the underground then it is due to the process of conduction.

Convection is defined as the process in which a fluid (liquid or gas) when heated then more dense (cooler) substance sinks ate the bottom whereas less dense (hotter) substance will rises at the top.

Radiation is defined as a process in which an electromagnetic energy will travel through space in the form of waves or particles.

Condensation is a process in which vapor state of a substance changes into liquid state.

Thus, we can conclude that heat from the surface is transferred underground by conduction.

Ostrovityanka [42]3 years ago

5 0

A, convection, is your answer

You might be interested in

Question 10 of 34

labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

#SPJ1

3 0

2 years ago

PLEASE EXPLAIN AND YOU WILL GET BRAINLIST Ms. R is curious if the type of gasoline she uses in her car affects how many miles sh

inessss [21]

Option 3 is the most reasonable

I hope this helped <3

Please give brainliest :)

8 0

3 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

7 0

3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

The drawing shows a skateboarder moving at 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the

Anna11 [10]

Let M = mass of the skier,
v2 = his speed at the end of the track.
By conservation of energy,
1/2 Mv^2 = 1/2 Mv2^2 + Mgh
Dividing by M,
1/2 v^2 = 1/2 v2^2 + gh
Multiplying by 2,
v^2 = v2^2 + 2gh
Or v2^2 = v^2 - 2gh
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46
Or v2^2 = 23.04 - 9.016
Or v2^2 = 14.024 m^2/s^2-----------------------------(1)
In projectile motion, launch speed = v2
and launch angle theta = 48 deg
Maximum height
H = v2^2 sin^2(theta)/(2g)
Substituting theta = 48 deg and value of v2^2 from (1),
H = 14.024 * sin^2(48 deg)/(2 * 9.8)
Or H = 14.024 * 0.7431^2/19.6
Or H = 14.024 * 0.5523/19.6
Or H = 0.395 m = 0.4 m after rounding off
Ans: 0.4 m

The answer in this question is 0.4 m

4 0

3 years ago