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Vesnalui [34]
3 years ago
10

Select all that apply.

Physics
2 answers:
umka2103 [35]3 years ago
8 0
The true statements are

"Its temperature will rise until reaching 0°C."
And
"Its temperature will remain at 0°C until all the ice melts."

So, basically once the ice has reached 0°C we can conclude that the ice has probably fully melted. However the temperature has to stay there if we want the ice to melt.

We must remember that the melting point of ice is 0°C, and that Ice is completely melted at approximately 0°C so we must have those two statements true
klemol [59]3 years ago
4 0
Temperature will rise until reaching 0°C.
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A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten
djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

<u>Given</u>

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.  

<u>Required </u>

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

(c) r = 40.0 cm  

<u>Solution</u>

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4\pi∈_o                       (1)

r is the distance where the potential is measured and the term 1/4\pi∈_o equals  9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m  

V= 1*q/4\pi∈_o*r

 =60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by  

V = 1*q/4\pi∈_o*R                (2)    

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m  

V = 1*q/4\pi∈_o*R

    = 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)  

V= 125 V

7 0
3 years ago
Scientific investigations
snow_tiger [21]
I don’t know what you are trying to ask complete my but here are the steps to a scientific investigation i hope this helps you

6 0
3 years ago
You are writing a science report and want to find accurate trustworthy information. Which would be the best resource?
fgiga [73]
<span>an encyclopedia

Happy studying!</span>
8 0
4 years ago
Read 2 more answers
How can a cyclist's acceleration change even if its velocity remains constant?
aleksley [76]

It can't.  Acceleration IS a change in velocity.


5 0
4 years ago
A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm
dezoksy [38]

Answer:

7.65 m

Explanation:

P_1 = Initial pressure = 0.03 atm

P_2 = Final pressure = 1 atm

r_1 = Inital radius = 21 m

V_1 = Intial volume of gas = \frac{4}{3}\pi r_1^3

V_2 = Final volume of gas = \frac{4}{3}\pi r_2^3

T_1 = Initial temperature = 200 K

T_2 = Final temperature = 323 K

From ideal gas law we have

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m

The radius at liftoff is 7.65 m

7 0
3 years ago
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