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BlackZzzverrR [31]
2 years ago
12

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

electric field due to this sphere at a point 5.0 cm outside the sphere's surface? (k=1/4πϵ0=8.99×109 N · m2/C2) A metal sphere of radius cm carries a charge of μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point cm outside the sphere's surface? ( N · m2/C2) 4.0×109 N/C 4.0×107 N/C 8.0×107 N/C 4.2×106 N/C 8.0×109 N/C
Physics
1 answer:
frozen [14]2 years ago
3 0

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

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Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
2 years ago
A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph
Nonamiya [84]

Answer:

The electric field is  E = 7.5 *10^{6} \ N/C

Explanation:

From the question we are told that

    The radius of the metal sphere is  R = 2.0 \ cm  =  0.02 \ m

     The excess charge which the metal sphere carries is  q =  3.0 \mu C  =  3.0*10^{-6} \ C

      The distance of the position being to the center is D = 6.0 \ cm  = 0.06 \ m

       The coulomb constant is   k =9*10^{9} \  N \cdot m^2 /C^2

Generally the electric field is mathematically represented as  

        E = \frac{k *  q}{D^2}

substituting values

        E = \frac{9*10^{9} *  30.*10^{-6}}{(0.06)^2}

      E = 7.5 *10^{6} \ N/C

5 0
1 year ago
If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
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D) All of the above.
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4 0
2 years ago
Read 2 more answers
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
1 year ago
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
gregori [183]

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

7 0
2 years ago
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