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3 years ago

2 Physics Questions WILL GIVE BRAINLIEST!!

Physics

1 answer:

In-s [12.5K]3 years ago

5 0

Answer:

a) The force of friction is 1.5 N.

b) The new force is 4/9 times as large as the original force.

Explanation:

It is given that the book is moving on the table.

∴ The friction acting on the block should be kinetic.

∴ Coeffecient of kinetic friction , k = 0.3

The force of friction = kN , where 'N' is the normal force acting on the surface where friction is acting.

In this case , N = mg , where 'm' is the mass of the book and 'g' is the acceleration due to gravity.

∴ Force of friction , f = kmg = 0.3×0.5×10 = 1.5N

b) We know that ,

Gravitational force , F ∝ m , where 'm' is the mass of the body.

F ∝ $x^{2}$ where 'x' is the distance between the two bodies.

Since mass of both the bodies is doubled , the force will be 4 times as large compared to the original force F from the first proportionality equation.

However , as the distance is 3 times as large , the force will be 9 times as small as compared to the original force F

∴ The new force is 4/9 times as large as the old force

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The coordinates of a bird flying in the xy plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2. Calculate

yuradex [85]

The derivative of the function space as a function of time is equal to a function of speed as a function of time.

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3 years ago

Read 2 more answers

B) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected

levacccp [35]

Answer:

i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

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