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ivolga24 [154]
3 years ago
13

A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0

cm from the center of the ring.
Physics
1 answer:
Grace [21]3 years ago
8 0

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2    where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

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Explain how the earth is round and not flat
frosja888 [35]
It is a conglomeration of rock that collected together over time, and it rotating in a spinning ball that is orbiting a sun that is in the middle of our solar system. 

You should look at a picture of the earth from a satellite.<span />
5 0
3 years ago
Read 2 more answers
Tarzan and Jane. Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you
creativ13 [48]

Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

4 0
3 years ago
A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a
saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

  • Boyle's Law,   (P₁V₁ =P₂V₂)
  • Charles' Law  (V₁/T₁ =V₂/T₂)
  • Gay-Lussac's Law.  (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) \frac{5}{278} =\frac{P_2}{313}  \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm

7 0
3 years ago
You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine
DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

F_x = F cos \theta

with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

F_f = \mu mg

where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N

8 0
3 years ago
i will give brainliest Which of the following structures are present in plant cells but absent in animal cells? Choose 1 answer:
Sonja [21]

Answer:

A. cell walls

Explanation:

Plants have cell walls but animals dont.

3 0
3 years ago
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