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Mkey [24]
2 years ago
5

A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the norm

al to the loop, and it increases from 0 to 0.90 T in a time of 0.66 s. The wire has a resistance per unit length of 2.9 x 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire
Physics
1 answer:
bazaltina [42]2 years ago
6 0

Given Information:  

Radius of circular loop = r = 9.1 cm = 0.091 m  

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻²  Ω/m

Number of turns = N = 1

Required Information:  

Electrical energy dissipated = E = ?  

Answer:  

Electrical energy dissipated = 50.09x10⁻³ Joules  

Step-by-step explanation:  

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.

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The spring scale will read 559 Newton's or 125.7 pounds.
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3 years ago
Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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2 years ago
A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. Suppose at f
Airida [17]

Answer:

c. The steady-state value of the current depends on the resistance of the resistor.

Explanation:

Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and  switch to the negative terminal of the battery.

But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.

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What does the person in front of the ambulance experience?
kow [346]

A higher frequency. I just got this answer on usa test prep.

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