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A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge

Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π× $10^{-7}$ ×0.024 / 2π(1.20× $10^{-2}$ ) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 × $10^{-7}$ × 1.09861 × 120

current = 52.73 × $10^{-6}$ A

so here current in loops is 52.73 μA

8 0

3 years ago

Two parallel plates of area 0.155 m2<br> are separated by 0.00100 m. What<br> is their capacitance?

KonstantinChe [14]

Answer:

1.37 x 10^-9

Explanation:

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4 0

3 years ago

In which ancient civilization did the sport of cucj originate

Burka [1]

It originated in China.

3 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

If a girl is running along a straight road with a uniform velocity 1.5 m /s find her acceleration

Phoenix [80]

Answer:

As velocity is uniform then acceleration will be zero.

Explanation:

6 0

3 years ago