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Juliette [100K]
3 years ago
7

What is the strength of the electric field in a region where the electric potential is constant?

Physics
1 answer:
Roman55 [17]3 years ago
4 0

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>

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Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic
marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in  cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = 1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 0.063 m³/s

flow rate in  liters per minute

1 gallon = 3.78541 L

 Q = 1000\times \dfrac{3.78541\ m^3}{1\ gallon}

 Q = 3785.41 m³/min

flow rate in cubic feet per second

 1 gallon = 0.133681 ft³

 1 min = 60 s

Q = 1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 2.23 ft³/s

4 0
2 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
While hovering motionless 3.0 meters an asteroid, a small
AnnZ [28]
I have no idea what you are trying to ask sorry
4 0
3 years ago
If the Sun were the size of a small exercise ball (about 0.5 meters (m) in
suter [353]

Answer:

the size of a pea

Explanation:

4 0
2 years ago
A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step
Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

7 0
3 years ago
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