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rjkz [21]
3 years ago
9

A solid spherical conductor has a radius of 12 cm. The electric field at 24 from the center of the sphere has a magnitude of 640

N/C. What is the charge density (in C/m2) on the sphere
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

Explanation:

Given:

Radius of sphere (r) = 12 cm = 0.12 m

Distance from the electric field R = 24 cm = 0.24 m

Magnitude (E) = 640 N/C

Find:

Charge density on the sphere

Computation:

Charge on the sphere (q) = (1/K)ER²            (K = 9 × 10⁹)

Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²

Charge on the sphere (q) = 4 × 10⁻⁹ C

Charge density on the sphere = q / [4πr²]

Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]

Charge density on the sphere = [4 × 10⁻⁹] / [0.18]

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

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A 140 g baseball is moving horizontally to the right at 35 mis when it is hit by the bat. the ball fl ies off to the le ft at 55
Anit [1.1K]

Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

When it hits the bat, it flies left with a velocity of 55 m/s at an angle of 25°. To reiterate:

mass = 0.140kg, Initial: 35m/s, Final: 55m/s at an angle of 25°

For this problem, we have to use the impulse equation, but before that lets break the velocity into components (It will be apparent towards the end):

The initial velocity is moving only in the horizontal direction, so:

v_{0x} = 35 m/s

The final velocity has an x and a y component:

v_{fx} = 55cos(25) = 49.84692829m/s\\v_{fy} = 55sin(25) = 23.2440044m/s

Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

J = dp= m*dv

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s

Finally, substitute into the equation

J = m*dv\\J = 0.140kg * 87.97320604m/s\\J = 12.31624885kg*m/s\\J = 12.32 kg*m/s

J = 12.32kg*m/s

8 0
3 years ago
What is the magnitude of the force a 25 charge exerts on a 3 mc charge 35cm away?
lara31 [8.8K]
The electrostatic force between two charges is given by
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges

In our problem, 
q_1 = 25 \mu C=25 \cdot 10^{-6} C
q_2 = 3 mC = 3 \cdot 10^{-3} C
r=35 cm=0.35 m
Therefore the electrostatic force is
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(25 \cdot 10^{-6}C)(3 \cdot 10^{-3}C)}{(0.35 m)^2}= 5510 N
4 0
4 years ago
How can one object affect the motion of another without touching it?​
stealth61 [152]

By non-contact forces (e.g. gravitational force and electric force)

Explanation:

In order for an object to exert a force on another object, the two object can also be not touching each other. In fact, there exist some non-contact forces in nature.

Concerning macroscopic objects, the two main non-contact forces acting between objects are:

- The gravitational force: this force is exerted between every object that has mass. It is always attractive, and its magnitude is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

- The electric force: this force is exerted between objects that have electric charge. It can be either attractive or repulsive, and its magnitude is given by

F=k\frac{q_1 q_2}{r^2}

where:

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

q_1, q_2 are the two charges

r is the separation between the two charges

Learn more aboit gravitational and electric force:

brainly.com/question/1724648

brainly.com/question/12785992

brainly.com/question/8960054

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#LearnwithBrainly

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When a puddle dries up what are the particles really doing
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The particles are either being absorbed or evaporating
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jeka57 [31]
We classify the elements as follows:

<span>Cd2+ = Diamagnetic
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