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lys-0071 [83]
3 years ago
8

Plzzz help with the question below!

Physics
1 answer:
KatRina [158]3 years ago
7 0
The answer is place.
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At one point in space the Electric potential is measured to be 119 V at a distance of 1 meters away it is measured to be 43 V. F
Anni [7]

Answer:

76 V/m

Explanation:

V_{i} = Electric potential at initial location = 119 V

V_{f} = Electric potential at final location = 43 V

d = distance between initial and final location = 1 m

E = magnitude of electric field

Magnitude of electric field is given as

E = \frac{- (V_{f} - V_{i})}{d}

E = \frac{- (43 - 119)}{1}

E = 76 V/m

8 0
3 years ago
If a load of snow contains 3 tons it will weigh how many pounds
stiv31 [10]

Answer: 6,000 pounds ( lbs)

1 ton equals 2,000 pounds 3 X 2,000lbs = 6,000lbs

7 0
2 years ago
An electromagnet is created using a battery, an insulated copper wire and an iron nail. The wire is wrapped around the nail 30 t
Veseljchak [2.6K]

Answer:

Explanation:

If one assume that each turn is like a strand of electromagnet, which can then be added up. Therefore, increase in the number of turns will yield to increase in the magnetic strength. Also if the current increases, then there will be increase in the magnetic field strength.

From Ohm's law

V = IR

I = V/R

That is a direct increase in voltage will lead to increase in current.

Increase the voltage of the battery and increases the number of turns of the coil. Will suit the situation

4 0
2 years ago
The smallest possible part of an element that can still be identified as the element is called
andreyandreev [35.5K]

Answer:

BLM

Explanation:

ACAB

4 0
2 years ago
A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan
In-s [12.5K]

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

\text{Initial Energy}=\text{Final Energy}

\frac{1}{2} mv^2=\frac{1}{2} kx^2

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

mv^2 = kx^2

v^2 = \frac{kx^2}{m}

v = \sqrt{\frac{kx^2}{m}}

Our values are,

m = 5.22*10^4kg

k = 4.58*10^5N/m

x = 32cm = 0.32m

Replacing our values we have,

v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}

v = 2.733*10^5m/s

Therefore the velocity is 2.733*10^5m/s

8 0
3 years ago
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