All categories

3 years ago

Heat is the transfer of thermal energy from one object to another because of a difference in

Physics

2 answers:

evablogger [386]3 years ago

8 0

The Answer Is Temprature. Hope This Helps :)

Gelneren [198K]3 years ago

4 0

Answer:

temperature

Explanation:

For any transfer of heat energy from one system to other it always requires temperature gradient.

So here we can say that when two bodies are at different temperatures then the Heat is transferred from one object at high temperature to other object at lower temperature.

The transfer of energy is of dynamic type in which if two bodies comes to same temperature then heat given by one body to other body is equal in magnitude to the heat given by other body.

So here if the temperature of two objects are equal then in that case the net thermal energy transferred will be zero.

So here correct answer would be TEMPERATURE

You might be interested in

Forces that act on an object but are not equal in size

mezya [45]

Forces that are equal in size but opposite in direction and do not cause a change in an object's movement are called balanced forces.

forces that aren't equal in size and do cause a change in movement (what it seems like you're asking for) are called UNBALANCED FORCES

so answer (in case that wasn't clear, as I'm tired) : unbalanced forces

7 0

3 years ago

How long does GSR last ?

SSSSS [86.1K]

Answer:

i think that the answer might be be C. one day. because it last about 4 to 6 hours

4 0

2 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

IRISSAK [1]

To develop this problem we will apply the concepts related to the Electromagnetic Force. The magnetic force can be defined as the product between the free space constant, the current (of each cable) and the length of these, on the perimeter of the cross section, in this case circular. Mathematically it can be expressed as,

$F= \frac{\mu_0}{2\pi} \frac{I^2L}{d}$

Here,

$\mu_0$ = Permeability free space

I = Current

L = Length

d= Distance between them

Our values are,

$L = 0.67m$

$m = 0.082kg$

$d = 7.4*10^{-3}m$

$I = \text{Current through each of the wires}$

Rearranging the previous equation to find the current,

$\frac{mg}{L} = 2*10^{-7} (\frac{I^2}{d})$

$I = \sqrt{\frac{mgd}{(2*10^{-7})L}}$

$I = \sqrt{\frac{(0.082)(9.8)(7.4*10^{-3})}{(2*10^{-7})(0.67)}}$

$I = \sqrt{44377.9}$

$I = 210.66A$

Therefore the current in the rods is 210.6A

8 0

3 years ago

Read 2 more answers

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

$u = 27.7 m/s$ --> initial velocity of the car

$\mu=0.804$ --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

$F=-\mu mg$

where m is the car's mass and $g=9.81 m/s^2$ is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2$

And the negative sign means it is a deceleration.

2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

$v^2 - u^2 = 2aS$

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

$S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m$

3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

$d_t = ut=(27.7 m/s)(0.543 s)=15.0 m$

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

$d_b=48.6 m$

So the total distance your car covers is

$S'=d_t+d_b=15.0 m +48.6 m=63.6 m$

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

$d=S'-S=63.6 m-48.6 m=15.0 m$