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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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A circular loop of wire of area 10 cm^2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is

S_A_V [24]

The magnetic dipole moment of the current loop is 0.025 Am².

The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.

<h3>What is magnetic dipole moment?</h3>

The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.

Mathematically, magnetic dipole moment is given as;

μ = NIA

where;

N is number of turns of the loop
A is the area of the loop
I is the current flowing in the loop

μ = (1) x (25 A) x (0.001 m²)

μ = 0.025 Am²

The magnetic torque on the loop is calculated as follows;

τ = μB

where;

B is magnetic field strength

B = √(0.002² + 0.006² + 0.008²)

B = 0.01 T

τ = μB

τ = 0.025 Am² x 0.01 T

τ = 2.5 x 10⁻⁴ Nm

Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.

Learn more about magnetic dipole moment here: brainly.com/question/13068184

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3 years ago

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3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated

I am Lyosha [343]

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

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3 years ago