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Korvikt [17]
3 years ago
8

A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

ular to the plane of the loop, and is decreasing at a constant rate of 0.190 T s. (a) What emf is induced in this loop? (b) If the loop has a resistance of find the current induced in the loop
Physics
1 answer:
erica [24]3 years ago
5 0

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

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Which part of an atom makes up most its volume
VikaD [51]

Hi! The inner most part of the atom (which is most of the atom) is occupied by electrons, therefore it’d take of most of its volume. Have a nice day!

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4 years ago
Who is the richest artist in the Gambia 2020​
Likurg_2 [28]

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3 years ago
In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri
egoroff_w [7]

Answer:

a) v=99.8584\ m.s^{-1}

b) v'=99.366\ m.s^{-1}

Explanation:

Given:

mass of the arrow, m=0.02\ kg

stiffness constant of the bow, k=330\ N,m^{-1}

distance of pulling back the arrow on the bow from its mean position, \Delta x=0.55\ m

height of the apple targeted, h=5\ m

<u>Force on the arrow due to the stiffness of the bow:</u>

F=k.\Delta x

F=330\times 0.55

F=181.5\ N

Now the acceleration of the arrow upwards:

a=\frac{F}{m}

a=\frac{181.5}{0.02}

a=9075\ m.s^{-2}

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

v^2=u^2+2(a-g).\Delta x

where:

v= velocity with which the arrow leaves the bow

u= initial velocity of the arrow after it left

v^2=0^2+2\times (9075-9.81)\times 0.55

v=99.8584\ m.s^{-1}

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

<u>Using eq. of motion:</u>

v'^2=v^2-2\times g.h

where:

v'= final velocity when the arrow hits the target

v= initial velocity after the arrow has been launched

v'^2=99.8584^2-2\times 9.81\times 5

v'=99.366\ m.s^{-1}

3 0
3 years ago
A man standing on top of a 30 m tall building throws a brick downwards with a velocity of 12 m/s. Determine the speed of the bri
BlackZzzverrR [31]

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building  = 30m

Initial velocity  = 12m/s

Unknown:

Final velocity  = ?

Solution:

We apply one of the kinematics equation to solve this problem:

         v²  = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

          v²   = 12²  + (2 x 9.8 x 30)

          v  = 27.1m/s

7 0
3 years ago
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