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andrezito [222]
2 years ago
7

The acceleration due to gravity on the earths surface is?​

Physics
2 answers:
Gennadij [26K]2 years ago
8 0

Acceleration due to gravity depends directly on the mass of Earth and inversely to the square of radius of Earth.

marta [7]2 years ago
8 0

Answer:

9.8 m/s2

Explanation:

Acceleration due to gravity is the acceleration gained by an object due to the gravitational force. Its SI unit is m/s 2. It has both magnitude and direction, hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2.

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In science, the metric units are most often the units used for experiments. These units are not often used in America otherwise.
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Assume your computer has a power rating of 50.0 W, and you use your computer for 7.0 hours a day for a normal school day. If the
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HTTPS:/www.word doc.com
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2 years ago
A 10-μF capacitor in an LC circuit made entirely of superconducting materials ( R = 0 Ω ) is charged to 100 μC. Then a supercond
satela [25.4K]

Answer:

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

8 0
2 years ago
At the bottom of a large cylindrical tank filled with fresh water the gauge pressure is 11.6 psi. What is the height (in feet) o
Art [367]

Answer:

8.19m

Explanation:

Parameters given:

Pressure, P= 11.6 psi = 79979.185 Pa

Gauge pressure is given as:

P = h*d*g

=> h = P/(d*g)

Where

h = height of tank

d = density

g = acceleration due to gravity

Density of water = 997 kg/m³

Therefore, the height of the tank is:

h = 79979.185/(997 * 9.8)

h = 8.19m

3 0
2 years ago
The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the
wel

Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0
3 years ago
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