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Zanzabum
3 years ago
9

Gravitational potential is the energy due to an object's:

Physics
2 answers:
Radda [10]3 years ago
7 0

Answer:

The answer is C. position in a gravity field

butalik [34]3 years ago
4 0

position in a gravitation field (c)

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What is the area of 12 1/2 and 17 1/5
jonny [76]
215

I am assuming that those are the width and length. So to find area all you have to do is multiply the two because lw = a. 12 1/2*17 1/5 = 215

25/2*86/5 = 2150/10 = 215
3 0
3 years ago
What is the strength of the electric field in a region where the electric potential is constant?
Roman55 [17]

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx        ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

<em>Therefore, a constant electric potential means that electric field is zero.</em>

4 0
3 years ago
What is the resistance of resistor R3?
Vsevolod [243]

Answer: 2.0

Explanation: I KNOW !

5 0
3 years ago
A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas
Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

\varepsilon' =\dfrac{S_y}{E}

\varepsilon' =\dfrac{240}{110\times 1000}

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0
3 years ago
Which is another term for the free enterprise system
Naily [24]

Answer:

Another term for free enterprise system would be capitalism

6 0
3 years ago
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