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Colt1911 [192]
3 years ago
10

Which two men developed the steam engine?

Physics
1 answer:
Lina20 [59]3 years ago
4 0

Watt was one of them


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Right answer gets brainlist
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Numberrrr 1 Inertiaaaaa
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2 years ago
Which branch of physics deals with the study of force, energy, and motion?
yawa3891 [41]
The branch of physics that deals with the study of force energy and motion is classic mechanics
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3 years ago
Suppose that instead of a long straight wire, a shortstraight wire was used. The distance from the wire to thepoint that the mag
vichka [17]

Answer:

Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.

Explanation:

The magnetic field, B of long straight wire can be obtained by applying ampere's law

B= \frac{\mu_0 I}{2\pi r}

I is here current, and r's the distance from the wire to the field of measurement.

The magnetic field is obviously directly proportional to the current wire. From this expression.

As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.

4 0
3 years ago
A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy
Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis  

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be F_1 and force on +Q charge y axis due to -Q charge on x-axis be F_2.

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = \sqrt{2}d units

k_e= Coulomb constant

F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N

F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N

Net force on +Q charge on y-axis is:

F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N

F_y=F_1-F_2cos45^o

F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})

F_N=\sqrt{F_x^2+F_y^2}

|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|

The net froce on the +Q charge on y-axis is

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

4 0
3 years ago
Read 2 more answers
State how the sun transfers energy to earth
Nookie1986 [14]

The sun transfers energy to Earth in the form of electromagnetic radiation.
That includes radio waves, heat, light, ultraviolet, and X-ray energy.


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3 years ago
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