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Sergio [31]
3 years ago
15

What happens to the potential energy of a stationary charge when it begins to move freely from one point to another under the in

fluence of an electric field?
It increases.
It decreases.
It remains the same.
It becomes zero.
Physics
2 answers:
Ulleksa [173]3 years ago
8 0
It decreases and the decrease in the energy level appear as a kinetic energy of the charge
german3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

Potential energy is the energy obtained by an object due to its position. Whereas kinetic energy is the energy obtained by an object due to its motion.

For example, when a charge moves from one point to another then it means the charge has come to motion.

Thus, we can conclude that potential energy of a stationary charge when it begins to move freely from one point to another under the influence of an electric field decreases.

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PLS ANSWER FAST WILL GIVE BRAINLEST!!!!
amid [387]

Answer:

F = 500 N

Explanation:

F = m × a

F = 1,000 × 0.5

F = 500 kg m/s²

F = 500 N

7 0
3 years ago
. An iron cube (ρ = 7860 kg/m3) having sides of 1 cm is submerged in open fresh water (ρ = 1000kg/m3) with its top surface at a
Korolek [52]

Answer:

\Delta P = 98.07\,Pa

Explanation:

The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:

\Delta P = \rho_{w}\cdot g \cdot \Delta h

\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.01\,m)

\Delta P = 98.07\,Pa

5 0
3 years ago
How much force is needed to accelerate a 2500 kg car at a rate of 3.5 m/s^2?
Sati [7]

F = ma  

F = applied force in newtons = to be determined  

m = mass of the car = 2,500 kg  

a = acceleration of the car = 3.5 m/s²  

F = (2,500 kg)(3.5 m/s²)  

F =8750

4 0
3 years ago
A train travels 83 kilometers in 2 hours, and then 79 kilometers in 2 hours. What is its average speed?
snow_tiger [21]
If you could please give me a already given speed I could estimate it. since there is no speed shown you wouldn't be able to estimate the speed of the moving train.
7 0
3 years ago
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= \frac{241.75}{69.1*9.8} *l * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0
3 years ago
Read 2 more answers
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