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Marrrta [24]
3 years ago
10

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

urrent is: A) 2.0 A B) 0.63 A C) 5.9A D) 300 A E) 26000 A
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

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For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?
Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

Explanation:

The motion of a projectile consists of two independent motions:  

- A uniform motion (with constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction  

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

v=u+at

where

u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)

a=g=9.8 m/s^2 downward is the acceleration of gravity

t is the time

So the equation becomes

v=gt

This means that

C. The downward component of the projectile's velocity continually increases

Because every second, it increases by 9.8 m/s in the downward direction.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
Sound is an example of a ______ wave.
tester [92]
Sound is a longitudinal wave.
7 0
3 years ago
Read 2 more answers
A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic
Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current  i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28  × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0
3 years ago
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1515
Talja [164]

Complete  Question

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.

Answer:

The  angle is  

Explanation:

From the question we are told that

   The light emerging from second Polaroid is 1/5 the  unpolarized

Generally the intensity of light emerging from the first Polaroid is mathematically represented as

             I_1 = \frac{I_o}{ 2 }

Generally from the Malus law the intensity of light emerging from the second Polaroid  is mathematically represented

      I_2  =  I_1 cos^2 (\theta )

=>   cos^2 (\theta ) =  \frac{I_2}{I_1 }

=>   cos (\theta) =  \sqrt{ \frac{I_2}{I_1} }

From the question I_2  =  \frac{I_o}{5}

     cos (\theta) =  \sqrt{ \frac{ \frac{ I_o}{5} }{\frac{I_o}{2} } }

     cos (\theta) =  \sqrt{ \frac{2}{5} }

=>    \theta =   cos ^{-1} [\sqrt{\frac{2}{5}}  ]

=>    \theta  =  50.77^o

8 0
3 years ago
According to coulombs law, how is the electric force related to the distance between two charges?
lorasvet [3.4K]

Answer:

D

Explanation:

From the formula of coulombs law F = Kq1q2/square of r, we can say the electric force is indirectly related to square of r

4 0
3 years ago
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