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A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kirza4 [7]

Answer:

The average induced emf is 29.06 V

Explanation:

Given;

radius of the circular loop, r = 12.0 cm = 0.12 m

magnetic field strength, B = 1.8 T

time interval = 2.8 ms = 0.0028 s

Area of the circular loop, A = πr² = π (0.12)² = 0.0452 m²

$e.m.f =\frac{BA}{t}$

where;

B is magnetic field

A is area

t is time

substitute the values given in the above equation to calculate the average emf that will be induced in the wire loop during the extraction process.

$e.m.f =\frac{1.8 *0.0452}{0.0028} = 29.06 \ V$

Therefore, the average induced emf is 29.06 V

5 0

3 years ago

Kundan who weighs 200N and bijaya who weighs 500N are playing sea saw.If Bijaya is 1m away from the fulcrum,how far should kunda

Tom [10]

Answer:

Kundan should sit at more that 2.5 meters from the location of the fulcrum on the side opposite to the side where Bijaya sits

Explanation:

The given parameters are;

The weight of Kundan, F₁ = 200 N

The weight of Bijaya, F₂ = 500 N

The distance of Bijaya from the fulcrum, d₂ = 1 m

Therefore, for Kundan to lift Bijaya, Kundan should sit at a location further than the point where both the moment of Bijaya and Kundan about the fulcrum are equal

Moment of a force, M = The magnitude of the force, F × The perpendicular distance to the point at which the force acts, d

At equilibrium, the sum of the clockwise moment = The sum of the anticlockwise moment

Taking the moment of Kundan as being in the anticlockwise direction and the moment of Bijaya as acting in the clockwise direction, we have;

F₁ × d₁ = F₂ × d₂

∴ 200 N × d₁ = 500 N × 1 m

d₁ = 500 N·m/(200 N) = 2.5 m

Therefore, when Kundan sits at the point d₁ = 2.5 m, away from the fulcrum on one side of the sea saw and Bijaya sits at the point 1 m from the fulcrum on the other side of the sea saw, both Kundan and Bijaya will have the same moment

For Kundan to lift Bijaya, therefore, Kundan as to sit at a point further than 2.5 m from the fulcrum on the opposite side of the fulcrum from Bijaya.

8 0

3 years ago

1-D Kinematics, Constant Acceleration After falling a distance of 45.0 m from the top of a building, a box is landing on the top

Mrrafil [7]

Answer:

Explanation:

Given

Object fall from a height of $s=45\ m$

Considering initial velocity to be zero i.e. $u=0$

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2(9.8)\cdot 45$

$v=29.69\ m/s\approx 29.7\ m/s$

(b)Average acceleration

After falling 45 m, object strike the car and comes to rest after covering a distance of 0.5 m

again using

$v'^2-u'^2=2as'$

here final velocity will be zero i.e. $v'=0$

initial velocity $u'=v$

$0-(29.7)^2=2\cdot a\cdot 0.5$

$a=-882.09\ m/s^2$

(c)time taken by it to stop

$v'=u'+a't$

$0=29.7-882.09\cdot t$

$t=0.034\ s$

5 0

3 years ago

Plz helpppp............

horsena [70]

Explanation:

anyone use zoom

code:- 2574030731

pass:- HELLO

Z●●M

7 0

3 years ago

Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared t

Rama09 [41]

Answer:

$a_s=4.8\times 10^{-2}~m^2$

Explanation:

Given:

cross sectional area of the bone, $a=4.8 \times 10^{-4} ~m^2$

factor of up-scaling the dimensions, $s=10$

Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.

The scaled up area is:

$a_s=a\times s^2$

$a_s=[4.8 \times 10^{-4}]\times 10^2$

$a_s=4.8\times 10^{-2}~m^2$

7 0

3 years ago