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neonofarm [45]
2 years ago
10

A teacher lifted a 10 N box and placed it down. Did she do any work? Explain your answer using math

Physics
1 answer:
Stells [14]2 years ago
8 0

Answer:

i think it is 45 kilograms not really sure?

Explanation:it might be wrong but i think it is right

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What are conductors and insulators? Give at least five example of each ​
Leona [35]

Answer: conductors are substances that allow heat or electricity to pass through . It deals with only the flow of electrons.

Eg. water, copper wire, iron rod, some ceramic materials, metallic nail.

Insulators are materials that do not allow heat or electricity to pass through.

Eg. Book, plastic, rubber, glass, paper

Explanation:

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2 years ago
after an investigation kuri determines that her hypothesis was wrong what is the best thing for kuri to do next​
aksik [14]
: You could make slight changes to the process or consider wheather the experiment was carried out correctly
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3 years ago
Read 2 more answers
Hey solution that contains all of the salute it can normally hold at a given temperature is
ankoles [38]

Answer:

A saturated solution

Explanation:

A saturated solution is one that contains the most amount of solute that can be dissolved in it at a given temperature

An example of a saturated solution is carbonated water, which readily gives off bubbles of carbon dioxide gas from areas within the solution to the region above the top surface of the gas in liquid solution

A saturation solution of salt in water can be created by continuing to dissolve salt in a given amount of water until it can no longer dissolve any more salt. However, heating the saturated salt solution, increases the amount of salt that can be dissolved.

Therefore, a solution that contains all of the solute it can normally hold at a given temperature is <u>a saturated solution</u>

8 0
2 years ago
An object rolls East at a steady speed of 12 m/s for 3 seconds. What distance did the object travel
horrorfan [7]
Answer: 36 meters.


Equation to find distance:
Speed x time
8 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
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