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DaniilM [7]
3 years ago
9

In the process of melting, what best describes what happens to a solid?

Physics
2 answers:
lions [1.4K]3 years ago
6 0

The correct answer is A.

In the process of melting, a solid will gain kinetic energy and become a liquid.

Hope this helps,

Davinia.

Eduardwww [97]3 years ago
5 0
<span>a. A solid will gain kinetic energy and become a liquid.</span>
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An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
dsp73

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

3 0
3 years ago
Read 2 more answers
How might the climate be affected if you were people drove automobiles and more people road bikes or use public transportation l
Hoochie [10]
The level of greenhouse gases in our atmosphere would decrease, due to less automobiles.
8 0
3 years ago
50200 J of heat are removed from
Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation Q=mcΔT to find m

Plug in all variables -50200=x\cdot 2000\cdot -8

Answer: 3.1375

4 0
4 years ago
An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad
Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0
4 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
3 years ago
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