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BARSIC [14]
3 years ago
13

A 10.0-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

re and rail is 0.300 Ω . Pulling the wire at a steady speed of 4.00 m/s causes 3.90 W of power to be dissipated in the circuit.
Physics
1 answer:
Triss [41]3 years ago
8 0

Question:

What are the magnitudes of the  pulling force and the magnetic field

Answer:

The magnetic field strength is 2.7 T

The pulling force F_p is 0.975 N

Explanation:

Here we have;

Length of wire, l = 10.0 cm = 0.1 m

Resistance of wire, R = 0.300 Ω

Speed of wire, v = 4.00 m/s

Power dissipated = 3.90 W

Based on the given data, we apply the relation;

I =\frac{Blv}{R}..............(1)

Where:

B = Magnetic field strength

I = Current

Since P = I²R, we have;

3.9 = I²·0.300Ω

∴ I² = 3.9/0.300 = 13

From which I = √13

Substituting the value of I in equation (1) above, we have;

\sqrt{13}=  \frac{B \times 10.0 \times 4.00}{0.300}

Therefore;

B = \frac{\sqrt{13} \times 0.300}{0.1 \times 4.00} = 2.70416 \ T\approx 2.7 \ T

The magnitude of the pulling force is given by the following relation;

F_p = F_m = IlB = l\sqrt{\frac{P}{R} }  \frac{\sqrt{PR} }{lv} =\frac{P}{v} = \frac{3.9 \ W}{4.00 \ m/s} = 0.975 \ N

The pulling force F_p = 0.975 N.

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0
3 years ago
A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
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Answer:

3.59 m/s

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We are given that

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Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

Initial momentum=Final momentum=(m+m')V

627.6=(85+90)V=175V

V=\frac{627.6}{175}=3.59 m/s

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