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devlian [24]
3 years ago
12

Two positive charges of 6 µC are separated by a distance of 50 cm in air. What is the electric field strength at the midpoint of

the line joining the charges?
Physics
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

The electric field strength at the midpoint of the line joining the charges is zero (0)

Explanation:

Given that the two charges are both positive (same charge) and are equal in magnitude that is 6uC. The electric field strength at the midpoint of the line joining the two charges will be equal and opposite in magnitude, therefore they will cancel each other out and the electric field strength at this point will be equal to zero.

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WILL MARK BRAINLIEST
rodikova [14]

Answer:

152,155 J

Explanation:

115,333 + 36,822 = 152,155J

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3 years ago
Write the formula of Lever, Pulleys, wheel and axle and inclined plane.<br>​
maxonik [38]

Answer:

Lever => d_{e} = d_{r}

Pulley => G = M x n (gravitational acceleration)

Wheel and axle => M.A = Radius of the wheel/radius of the axle = R/r

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3 years ago
Describe a good level of body fat
m_a_m_a [10]

Answer: a good level of body fat can be found using your weight and height as a reference.

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6 0
4 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

C = 4.014\cdot 10^{-11}\,F

C = 40.138\,pF

b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

q = (4.014\times 10^{-11}\,F)\cdot (400\,V)

q = 1.606\times 10^{-8}\,C

q = 16.056\,nC

c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)

U = 3.212\times 10^{-6}\,J

U = 3.212\,\mu J

3 0
3 years ago
Read 2 more answers
A 75 kg ball carrier is running to the right at 6.5 m/s. An 80 kg defender is chasing the ball carrier running at 7.0 m/s. The d
Sophie [7]

Answer:

3,544.375Joules

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion, It is expressed as;

Kinetic energy = 1/2mv²

m is the mass of the body

v is the velocity

For the ball carrier;

KE = 1/2(75)(6.5)²

KE = 3168.75/2

KE = 1584.375Joules

For the defender;

KE = 1/2(80)(7)²

KE = 3920/2

KE = 1960Joules

The kinetic energy of the ball carrier/defender system BEFORE the tackle = KE for the carrier + KE for the defender

kinetic energy of the ball carrier/defender system BEFORE the tackle= 1584.375+1960 = 3,544.375Joules

7 0
3 years ago
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