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Temka [501]
3 years ago
14

Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4

C. Find the electric force exerted on one sphere by the other.
Physics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

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OlgaM077 [116]

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

3 0
3 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0
3 years ago
A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
3 years ago
An ideal gas is in a sealed rigid container. the average kinetic energy of the gas molecules depends most on
finlep [7]
It depends most on the temperature of the gas.
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if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?​
quester [9]

Answer:

See explanation

Explanation:

The acceleration due to gravity on an object is independent of the mass of the object. This is so because, the acceleration due to gravity depends only on the radius of the earth and the mass of the earth.

As a result of this, all objects are accelerated to the same extent and should reach the ground at the same time when released from a height as long as other forces other than gravity are not at work.

5 0
3 years ago
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