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Naya [18.7K]
2 years ago
12

A pendulum makes 50 complete swings in 2 min 40 s. What is the time period for 1 complete swing?

Physics
2 answers:
wolverine [178]2 years ago
6 0

Answer:

50 swings = 2 mins 40 secs

=> 50 swings = 160 secs

=> 1 swing = 160 / 50 secs

=> 1 swing -= 3.2 secs

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

krok68 [10]2 years ago
3 0
Please give the other person his brianliest
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This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
Olegator [25]

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

3 0
2 years ago
Please re-explain the following phrases in terms of momentum
zheka24 [161]
A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>
8 0
2 years ago
A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans
sertanlavr [38]

Answer:

0.4 A

Explanation:

From the question,

Electric power = Voltage×current

P = VI.......................... Equation 1

Make I the subject of the equation

I = P/V..................... Equation 2

Given: P = 96 J/s, V = 230 V.

Substitute into equation 2

I = 96/230

I = 0.4 A.

Hence the current is 0.4 A

8 0
2 years ago
A small object carrying a charge of -3.00 nc is acted upon by a downward force of 30.0 nn when placed at a certain point in an e
arlik [135]
The working equation for this one is:

E = F/Q, where E is the strength of the electric field, F is the electric force and Q is the charge. Substituting the corresponding values, the strength of the electric field is equal to

E = -30 nN/-3 nC
E = 10 nN/nC


3 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
2 years ago
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