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Anton [14]
3 years ago
7

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

stance, measured from the center of the sphere, does the electric field reach a value equal to half its maximum value?
Physics
1 answer:
BlackZzzverrR [31]3 years ago
4 0

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo
tiny-mole [99]
The answer is left and right. 
4 0
3 years ago
Read 2 more answers
A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the
ehidna [41]

Answer: 170.67 N

Explanation:

Given

Mass of skier is m=82.9\ kg

Height of the inclination is h=20\ m

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N

Thus, the horizontal friction force is 170.67 N.

7 0
3 years ago
Why does current flow through a circuit?
Ghella [55]

Answer:

Explanation:

cause it do

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3 years ago
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Una locomotiva traina un solo vagone, che ha la sua stessa massa, con la forza di
gregori [183]

Answer:

c)    F = 16000 N

Explanation:

For this exercise we use Newton's second law

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they tell us that adding the other wagons the acceleration of the locomotive must be maintained

 

      F = m a

by adding the other four wagons

mass = 4 no

therefore to maintain the force you must also raise the same factor

         Fe = 4Fo

         Fe = 4 4000

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3 0
3 years ago
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