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sashaice [31]
2 years ago
7

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

t of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I
Physics
1 answer:
nikklg [1K]2 years ago
8 0

Answer:

4.77\ \text{A}

Explanation:

F = Magnetic force = 4.11 N

I_n = Net current

I_2 = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

\theta = Angle between current and magnetic field = 65^{\circ}

l = Length of wires = 2.64 m

I = Current in the other wire

Magnetic force is given by

F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}

Net current is given by

I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}

The current I is 4.77\ \text{A}.

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A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium
Kaylis [27]

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

         w² = \frac{k}{m}

to find the constant (k) of the spring, we use Hooke's law with the initial data

         F = - kx

where the force is the weight of the body that is hanging

        F = W = m g

we substitute

        m g = - k x

        k = - \frac{m g}{x}

we calculate

        k = - \frac{9.8 m}{- 2.6 \ 10^{-2}}

        k = 3.769 10² m

we substitute in the first equation

       w² = \frac{ 3.769 \ 10^2 \ m }{m}

       w = 19.415 rad / s

angular velocity and frequency are related

       w = 2πf

        f = \frac{w}{2\pi }

        f = 19.415 / 2pi

        f = 3.09 Hz

7 0
2 years ago
BRAINLYIST ASAPPP write and claim evidence and reasoning statement that explains why mercury and Venus do not have any moons
umka2103 [35]

answer:They are too close to the sun!

Explanation:Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.Same with Venus!

5 0
2 years ago
A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?
ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

6 0
3 years ago
Match the type of heat transfer with its description
VikaD [51]

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

5 0
3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
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