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lord [1]
3 years ago
11

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
________ × 10–4 J
What is the electric potential energy of the point charge at 1.3 m?

_________ × 10–3 J
Physics
2 answers:
tatyana61 [14]3 years ago
8 0

E = electric field = 1 x 10³ N/C

q = charge = 4 x 10⁻⁶ C

d = distance = 0.10 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (0.1)

U = 4 x 10⁻⁴ J


when distance = d = 1.3 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (1.3)

U = 5.2 x 10⁻³ J

schepotkina [342]3 years ago
7 0
4 for the first one and 5.2 for the second one.
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Help me please! i will give brainliest i need it done i know its easy but i have a lot of work to do
cricket20 [7]

Answer:

Down below :)

Explanation:

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5. Evaporation - F

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5 0
2 years ago
Read 2 more answers
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
2 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a
USPshnik [31]
Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²

Answer: 105.93 kg-m²

6 0
2 years ago
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
Which steps are important when designing and conducting a scientific experiment?
Pepsi [2]
<span>Make an Observation. Scientists are naturally curious about the world. ... Form a Question. After making an interesting observation, a scientific mind itches to find out more about it. ... Form a Hypothesis. ... Conduct an Experiment. ... <span>Analyze the Data and Draw a Conclusion.</span></span>
8 0
2 years ago
Read 2 more answers
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