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2 years ago

Question 15 of 25

Physics

1 answer:

vichka [17]2 years ago

7 0

Answer: conduction :it transfers heat between objects that are in direct contact with eachother

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An electric lamp creates light by forcing electrons through the filament of its light bulb. The kinetic energy of the moving ele

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electric potential energy..electrons get the the kinetic energy from the voltage applied across the conductor

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Sound from a source vibrating at frequency 200 Hz travels at the speed of 300 m/s in air.Wavelength of the sound is (in m/s)

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3 years ago

PLEASE HELPP MEEE!!!!!!!!!The goal is to increase the power; therefore, it is necessary to

Brrunno [24]

As we know that power is defined as rate of work done

so we will have

$P = \frac{Work}{time}$

so in order to increase the power as per above formula we know that either we need to increase the work or we need to decrease the time to complete that work

So here the correct answer will be

increase the work being done or decrease the time in which the work is completed.

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3 years ago

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What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given th

elena-14-01-66 [18.8K]

Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

number of turns, N = 940 turns

diameter, d = 24 cm = 0.24 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 5 ms = 5 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

N is the number of turns

A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

V₀ = NAB2π/t

A = πd²/4

V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x 5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V

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3 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

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