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Nata [24]
2 years ago
9

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

at a speed of 100 m/s. The plane of the coil is perpendicular to the field lines. If the breakdown voltage of air is 4 000 V/cm on that day, the largest gap you can have between the two wires connected to the ends of the coil and still get a spark is:
Physics
1 answer:
Kay [80]2 years ago
6 0

Explanation:

It is given that,

Area of square coil, A=4\ cm^2=0.0004\ m^2

Side of the square, L = 0.02 m

Number of turns, N = 10000

Uniform magnetic field, B = 1.5 T

Speed, v = 100 m/s

An emf is induced in the coil which is given by :

E=NBLv

E=10000\times 1.5\times 0.02\times 100

E = 30000 V

Breakdown voltage of air, V=4000\ V/cm=400000\ V/m

Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,

Electric field, E'=\dfrac{V}{d}

\dfrac{30000}{d}=400000

d = 0.075 m

Hence, this is the required solution.

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The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1
Andrei [34K]

Answer:

The expression of gravitational field due to mass m_! at a distance l_a

Explanation:

We have given mass is m_1

Distance of the point where we have to find the gravitational field is l_a

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We have to find the gravitational filed

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2 years ago
A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterbac
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6 0
2 years ago
A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally
vazorg [7]

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec  

6 0
2 years ago
A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i
aniked [119]

Answer:

a. N = 2.49W b.  0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ  = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

6 0
2 years ago
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