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3 years ago

How does radiation move?

Physics

1 answer:

Degger [83]3 years ago

8 0

Radiation can move through empty space . to transfer heat: convection cannot.

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If two bodies weighing 10 kg and 20 kg respectively are standing at the

kvv77 [185]

Answer:

potential energy = mgh , where m is mass of body , g is acceleration due to gravity and h is given height

since both of the body are at same height above the ground , so the body having greater mass will have more potential energy

that is 20 kg body will have higher P.E

Explanation:

7 0

3 years ago

Giving brainlist so plz help

Volgvan

Answer:

I DON KNOW

Explanation:

6 0

4 years ago

A mercury thermometer is constructed as

Tamiku [17]

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$

Therefore, the change in volume with a change in temperature of 36°C becomes:

$\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3$

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

$V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm$

8 0

3 years ago

If you push twice as hard against a stationary brick wall, the amount of work you do

Alecsey [184]

Answer:

Zero

Explanation:

The work done on an object is given by:

$W=Fd cos \theta$

where

F is the force applied on the object

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

In this problem, you are pushing again a stationary wall: this means that the walls does not move. As a result, the displacement is zero: d=0. Therefore, the work done is also zero: W=0.

8 0

3 years ago

A force of 100 N acts upward. Resolve this force into 2 components; one that acts 30º north of west and one that acts 60º north

GrogVix [38]

To resolve these forces we have to make use of the sines and cosines.

To resolve this force in 30 degree north of west, the answer will be

100*sin(30)

The answer will be 50N

Now to resolve the force acting 60 degree north of east

100* cos(60)

The answer will be 50N.

This also adds to the total force acting that is 50+50=100N. This is the way forces are resolved according to their specified angles.

5 0

3 years ago