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Dovator [93]
4 years ago
9

Why can protons not be rubbed off one material and transferred to the other?

Physics
1 answer:
Pani-rosa [81]4 years ago
7 0
I think it is that protons are stuck with the neutrons in a strong force that is in the nucleus
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The voltage leaving the secondary coil of a transformer is equal to 240 V. The secondary coil is wrapped 8 times around the core
Lina20 [59]

The initial voltage at the primary= 750 V

Explanation:

for a transformer

Es/ Ep= Ns/ Np

Es= secondary voltage= 240 V

Ep= primary voltage

Ns= number of turns in the secondary=8

Np= Number of turns in the primary=25

so 240/Ep=8/25

Ep=750 V

7 0
3 years ago
Read 2 more answers
What are hot clouds of gaseous elements and compounds that act as nurseries for stars
Aleks04 [339]
Interstellar gas clouds are common in many galaxy, like the Orion nebulae which many young stars are being born. A typical nebula is many light years in diameter and contains enough material mass to make several thousand stars the size of our sun. The majority of the gas in nebulae consist of molecules of hydrogen and helium-but most nebulae also contain atoms of other elements. All known element in our periodic table is also being made inside this crucible of this immense hot gas. The source of the organic molecules is still a mystery. Irregularities in the density of the gas causes a net gravitational force that pull the gas molecules close together.
3 0
3 years ago
What is the direction of the tangential speed when a car is driving around a circular curve?
pantera1 [17]
You're not going to believe this coincidence:. The "tangential" speed is the speed in the direction of the "tangent" to the circle. That's straight ahead, in the direction the car is moving. Since the car is driving in a circle, the tangential direction keeps changing.
8 0
4 years ago
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Which of these statements suggests that the orbital period of Venus is shorter than that of Saturn?
Snowcat [4.5K]

Answer:

C) Venus is closer to the Sun than Saturn is.

Explanation:

This is the right answer.

5 0
3 years ago
Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

v_{f}=1721.1m/s

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
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