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Fed [463]
3 years ago
11

A car, on a straight road, is stopped at a traffic light. When the light turns to green the car accelerates with a constant acce

leration. It reaches a speed of 19.1 m/s (68.8 km/h) in a distance of 101 m. Calculate the acceleration of the car.
Physics
1 answer:
Fantom [35]3 years ago
3 0

Answer:

Acceleration of the car will be equal to a=1.805m/sec^2

Explanation:

As the car is initially stop so initial velocity of the car u = 0 m /sec

It is given that after moving a distance of 101 m velocity of car is 19.1 m/sec

So final velocity of the car v=19.1 m/sec

We have to find the acceleration of the car

From third equation of motion we know that v^2=u^2+2as

So 19.1^2=0^2+2\times a\times 101

a=1.805m/sec^2

So acceleration of the car will be equal to a=1.805m/sec^2

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Which idea is associated with Copernicus? Select one: a. The orbits of the planets are circles. b. The orbits of the planets are
Schach [20]

Answer:

d. The earth rotates around the sun

Explanation:

  • Nicolas Copernicus is considered the first person to give the theory of heliocentric, or Sun-centered system of our planetary system.
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3 years ago
What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = 1 \times 10-12 W/m^2

Now

Intensity level ( or Loudness)is

L = log10 \frac{I}{10}

L = log10 \frac{10}{1\times 10^{-12}}

L = log10 \times 1013

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Therefore  

L = 13 bel

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1 bel = 10 decibels

So,

The  Sound intensity level is

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5 0
3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

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