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Levart [38]
2 years ago
7

Pls answer ASAP pls and thnk you my friends

Physics
1 answer:
rodikova [14]2 years ago
6 0

Answer:

Since this is old, hope you don't mind if I scoop up these points

Explanation:

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Does uranium use steam?
Neko [114]

Answer:

Yes.

Explanation:

Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.

4 0
2 years ago
A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the
solong [7]

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

4 0
3 years ago
How much time does it take 9,560 joules of work with 860 watts of power
tia_tia [17]

Explanation:

use equation power=watts/time

to find time rearrange to make time = power/wats

so you have your equation substitute the numbers

so 9560J/860W is 11 minutes

5 0
3 years ago
Read 2 more answers
Any one tell me about the earth rotation it rotatining or not with any proof? ​
Dafna11 [192]
The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.
6 0
3 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
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