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3 years ago

If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?

1 answer:

8 0

Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing 1/nth1/nth of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

With this assumption, the energy loss becomes

ΔE = 1/2 mv2 − 1/2 m (v−v/n) 2

and the number of planks NN becomes

N = 1/2mv2 /ΔE = n2/ 2n−1

Otherwise, if you assume that the bullet loses 1/nth1/nth of its velocity per plank, then the answer is N=∞

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What is the period and frequency of the second hand on a clock?

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Answer:

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2 years ago

Four identical charges, Q, occupy the corners of a square with sides of length a. A fifth charge, q, can be placed at any desire

Nat2105 [25]

Answer:

q = - ( 2*sqrt(2) + 1 )*Q / 4

Explanation:

Given:

- Side of each square L = a

- Charge q is placed among 4 other identical charges Q.

Find:

- Find the location and magnitude of the fifth charge q such that net Electric Force at its position is zero.

Solution:

- Compute distance r from charge Q to q i.e center of all four charges Q:

r = 0.5*sqrt ( a^2 + a^2 )

r = a / 2sqrt(2)

- Compute the individual Electrostatic forces @ point A:

F_b = F_d = k*Q^2 / a^2

F_c = k*Q^2 / (a*sqrt(2))^2 = k*Q^2 / 2*a^2

F = k*Q*q / (a / 2sqrt(2))^2 = 2*k*Q^2 / a^2

- Use Electrostatic Equilibrium conditions:

F_b + F_c*cos(45) = F*cos(45)

F_d + F_c*sin(45) = F*sin(45)

- Plug in the values and equate:

{ (Q/a)^2 + (Q^2 / 2*a^2*sqrt(2)) = sqrt(2)*Q*q / a^2

- Canceling all k's and a^2:

Q * ( 1 + (1 /2*sqrt(2)) ) = sqrt(2)*q

q = - ( 2*sqrt(2) + 1 )*Q / 4

- Note: In attachment Q's and q's are interchange but the solution here provided is according to the question at hand.

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3 years ago

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3 years ago

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Sveta_85 [38]

B is the correct answer

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3 years ago

Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here : 750 J electrical energy i

Charra [1.4K]

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So here we can see that 350 J of energy is lost against many other type of frictional and resistive loses.

So here we can say that out of 750 J of energy only 400 J is used to run the fan and rest amount of energy is lost against friction.

also we can say that efficiency of this fan will be

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3 years ago