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IRISSAK [1]
3 years ago
5

An electron is placed 5.40 cm due north of a charged sphere and experienced a force of 8.30 x 10^-13 N due north. What is the el

ectric field experienced by the electron at that point?
Physics
1 answer:
ira [324]3 years ago
4 0

Answer:

The electric field is E = 5.1*10^6N/C.

Explanation:

The force F on a charge q in an electric field E is given by

F = qE,

which can be rearranged to give

E = \dfrac{F}{q }

Now, the force on the electron is F = 8.30*10^{-13}N, and its charge is

q = 1.6*10^{-19}C; therefore,

E = \dfrac{8.30*10^{-13}N}{1.6*10^{-19}C}

\boxed{E = 5.1*10^6N/C}

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In a related practical activity, 2 litres of a gas were heated from 35 °C to 75 °C. If the pressure was kept constant, calculate
Pachacha [2.7K]

Answer:

2.26l

Explanation:

From the general gas equation:

P1V1/T1  = P2V2/T2

Since pressure remained constant we can say:

V1/T1  = V2/T2

so to convert to kelvin add 273 to both temperature values then we can say:

1 m^3= 1000 L

2l=0.002m^3

Then;

0.002/308=V/348

V=(0.002/308)348

Final volume=0.002259m^3

=2.6l(1 decimal place)

4 0
2 years ago
Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

#SPJ4

5 0
1 year ago
Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball
vlabodo [156]

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

6 0
3 years ago
After three shuttle accidents in three months, one of which resulted in a critical injury to a driver, Get Around Shuttle driver
alexandr1967 [171]

Answer:

Safety

Explanation:

As we know that safety is very important before doing anything .This is also a important part of the human needs.Every human wants both emotional and physical safety from his job or whatever is doing.

Maslow's given a Hierarchy of Needs and the second tier of this Hierarchy is safety.

So the answer of the given question is safety.

7 0
4 years ago
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo
aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

3 0
3 years ago
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