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kolbaska11 [484]
3 years ago
9

In a uniform electric field what factors does the electrical potential energy depend on

Physics
1 answer:
NARA [144]3 years ago
6 0
The electrical potential energy of a particle in a uniform electric field depends on 1) the charge of the particle 2) the distance from the source of the field 3) the magnitude of the electric field

In fact, the electrical potential energy is defined as
U=q V (1)
where q is the charge and V is the voltage. Since the electric field is uniform, the voltage increase linearly with the distance from the source:
V=Ed (2)
where E is the electric field strength and d is the distance. Putting (2) into (1), we find
U=qEd
therefore, we see that the potential energy of the particle depends on the charge q, the distance d from the source and the electric field strength E (which is constant at every location, because the field is uniform)
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An underwater diver sees the sun at an apparent angle of 45.00 from the vertical. How far is the sun above the horizon? [n in wa
pantera1 [17]

Answer:

19.872 degrees

Explanation:

Mathematically;

Using Snell’s law

n1 sin A = n2 sinB

Where ;

n1 = refractive index in air = 1

n2 is refractive index in water = 1.33

A = ?

B = 45

Substituting the values in the equation;

1 sin A = 1.33 sin45

Sin A = 1.33 sin 45

A = arc sin (1.33 sin 45)

A = 70.12

Thus, the actual direction of the Sun with respect to the horizon = 90-A = 19.872 degrees

6 0
4 years ago
The sugars produced by photosynthesis are stored and later may be consumed by an organism such as a deer, mouse, or human. The c
Andre45 [30]

Answer: This isn't a question but the statement is quite accurate to a well thought out science lesson.

3 0
2 years ago
When Javier touches a hot mug of tea, energy is transferred from the surface of the mug to his hand. What kind of energy is this
Tpy6a [65]

Answer:

Thermal energy

Explanation:

Thermal energy refers to the energy due to the random motion of the particles inside a substance/objects. The faster the particles moves in an object, the more thermal energy it can transfer. This is also directly related to the concept of temperature of an object: the higher the temperature of an object, the faster its particles move, the more thermal energy the object can transfer.

According to the basic laws of thermodynamics, thermal energy (also called heat) is always transferred spontaneously from an object at higher temperature to an object at lower temperature. In this problem, the hot mug is warmer than Javier's hand, so thermal energy is transferred from the hot mug to Javier's hand.

6 0
3 years ago
Read 2 more answers
The absolute temperature of a sample of monatomic ideal gas is doubled at constant volume. What effect
Advocard [28]
<h3><u>Answer;</u></h3>

Doubles and Remains the same

<h3><u>Explanation;</u></h3>
  • The effect of doubling the absolute temperature of a sample of a monoatomic ideal gas is that,the pressure doubles and density of the sample of gas remains the same.
  • <em><u>According to ideal gas equation; PV = nRT; Where P is pressure and V is the Volume, n is the number of moles, R is the ideal gas constant and T is the absolute temperature.</u></em>
  • <em><u>Therefore, when the temperature of the mono atomic ideal gas is doubled, the pressure of the gas will also doubles.</u></em>
  • However, in a closed chamber mass of the ideal gas is invariant, since density depends only on the mass and volume therefore the density of the ideal is gas will remain the same.
5 0
3 years ago
A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j
fenix001 [56]

Answer:

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

\vec F_e = q\vec E

\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3

\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j

Now force on moving charge due to magnetic field is given as

\vec F_b = q(\vec v \times \vec B)

\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )

\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i

\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k

Now net force due to both

F = F_e + F_b

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

7 0
3 years ago
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