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valina [46]
3 years ago
11

Positive charge is distributed uniformly throughout a non-conducting sphere. The highest electric potential occurs: A. Far from

the sphere B. Just outside the surface C. At the surface D. Halfway between the center and surface E. At the center
Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:E

Explanation:

At the center

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Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
3 years ago
Energy transformations when you cook sausages on a campfire burning wood
Rina8888 [55]

<span>Actually in this case heat energy is being transferred. Heat energy or thermal energy is transferred from the burning of wood to the sausages for it to be cooked. The sausage is being heated by the fire and is absorbing the heat or thermal energy.</span>

8 0
3 years ago
Need help quick!!!!
BARSIC [14]
The answer is B. Unbalanced force
3 0
3 years ago
Read 2 more answers
A light ray hits a plane surface at 20 degrees. What is the angle between the incident and reflected rays.
Eduardwww [97]

Answer:

140°

Explanation:

The law of reflection states that the angle of redlection equals to the angle of incidence.

When light rays hit surface at 20°, they also leave the surface at the same angle

Since the whole surface has 180° then subtracting these two angles from total angle gives the the angle between the incident and reflected rays.

180°-20°-20°=140°

The angle of incidence and reflection are equal hence 140/2=70°

The question needed the angle between the incident and reflected rays which is already calculated as 140°

7 0
3 years ago
How much momentum will a dumb-bell of mass 10 kg transfer
frosja888 [35]

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

Vf = ±4m/s

Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

m = 10kg

v = 4m/s (from previous calculation)

Plug in the values and solve for p:

p = 10(4)

p = 40kg×m/s

6 0
3 years ago
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