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Mrac [35]
3 years ago
10

A hollow metal sphere is electrically neutral (no excess charge). A small amount of negative charge is suddenly placed at one po

int P on this metal sphere. If we check on this excess negative charge a few seconds later we will find one of the following possibilities:
a. All of the excess charge remains right around P.
b. The excess charge has distributed itself evenly over the outside surface of the sphere.
c. The excess charge is evenly distributed over the inside and outside surface.
d. Most of the charge is still at point P, but some will have spread over the sphere.
e. There will be no excess charge left.
What is the The original magnitude of the force the acharge was magnitude of the force on now?
Physics
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:

b. The excess charge has distributed itself evenly over the outside surface of the sphere.

Explanation:

When a neutral hollow metal sphere is touched externally with a charge, the charge distributes itself evenly on the surface of the sphere. One thing to note here is that for a hollow sphere, the charges stays on the surface alone. This phenomenon is why the inside of airplanes are free from electric charge when struck by a lightning when they travel through a lightning storm.

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A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will
gtnhenbr [62]
1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t
y(t)=h- \frac{1}{2}gt^2
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2, and initial height h=1.85 m.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0:
0=h- \frac{1}{2}gt^2
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }=  0.61 s

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one: 
v_x=18.1 m/s
On the y-axis, the velocity is given by
v_y(t)=gt
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s
And the speed of the ball is the magnitude of the resultant of the two components:
v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s
8 0
3 years ago
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My workout schedule is day on - day off

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4 0
4 years ago
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What is the same for all types of electromagnetic radiation?
Mariulka [41]
<span>Radio Waves, Microwaves, infra-Red, Visible spectrum, Ultraviolet radiation, x-rays, Gamma Rays. Then again I could be wrong.</span>
5 0
3 years ago
A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distan
OleMash [197]

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Speed of the car, v = 100 km/h = 27.77 m/s

The coefficient of kinetic friction of the tires, \mu_k=0.5

Let f is the net force acting on the body due to frictional force, such that,

-f=ma

a=\dfrac{-f}{m}

a=\dfrac{-\mu _k mg}{m}

a=\mu_k g

a=-0.5\times 9.8

a=-4.9\ m/s^2

We know that the acceleration of the car in calculus is given by :

v.dv=a.dx, x is the stopping distance

\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}

\dfrac{v^2}{2}|_v^0=ax

0-(27.77)^2=-2\times 4.9x

On solving the above equation, we get, x = 78.69 meters

So, the stopping distance for the car is 78.69 meters. Hence, this is the required solution.

6 0
3 years ago
Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.
Serhud [2]

Answer:

Correct answer:  E total = 2,800 J

Explanation:

Given:

m = 4 kg   the mass of the object

V = 20 m/s  the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H   we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

4 0
4 years ago
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