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cestrela7 [59]
3 years ago
8

A coil with 150 turns and a cross-sectional area of 1.00 m2 experiences a magnetic field whose strength increases by 0.65T in 1.

80 s. The plane of the coil is perpendicular to the plane of the applied magnetic field. What is the induced emf in the coil
Physics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

54.17volts

Explanation:

Induced emf in a coil placed in a magnetic field can be expressed as E = N¶/t where

N is the number of turns = 150turns

¶ is the magnetic flux = magnetic field strength (B) × area(A)

¶ = BA

B = 0.65T

A = 1.0m²

t is the time =1.8s

Substituting this value in the formula

E = NBA/t

E = 150×0.65×1.0/1.8

E = 54.17Volts

The induced emf in the coil is 54.17Volts

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

4 0
2 years ago
Select all that apply
borishaifa [10]
a,b,c  is your answer light and sound are not considered matter and heat is energy created from matter and electricity is particles moving basically therefore electricity is matter hope this helps
5 0
3 years ago
A skateboarder who travels 60 meters in 30 seconds has a speed of
natulia [17]

2m/s

Explanation:

Given parameters:

Distance traveled = 60m

time taken = 30seconds

Unknown:

Speed of skateboarder = ?

Solution:

Speed is the rate of change of distance with time taken. It is a scalar quantity that only revers magnitude;

    Speed = \frac{distance}{time}

    Speed = \frac{60}{30} = 2m/s

learn more:

Speed brainly.com/question/1548911

#learnwithBrainly

4 0
3 years ago
A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th
PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

   \omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}

   \omega = 1.864\times 10^{-7}\ rad/s

b) tangential speed

   v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}

       v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

      a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}

          a = 3.62 x 10⁻³ m/s²

8 0
3 years ago
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