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1 year ago

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

con rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from the point A?

1 answer:

Mice21 [21]1 year ago

6 0

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

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A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, $L_{a} = 1 m$

Frequency, $\vartheta = 1.5 MHz = 1.5\times 10^{6} Hz$

Power transmitted, $P_{t} = 4 W$

Now,

For a monopole antenna:

$\lambda_{a} = \frac{c}{\vartheta}$

where

$\lambda_{a}$ = wavelength transmitted by the antenna

c = speed of light in vacuum

$\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m$

Now,

Since, the value of $\lambda_{a}$ >> $L_{a}$ thus the monopole is a Hertian monopole.

The resistance is calculated as:

$R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}$

$R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega$

$P_{radiated} = P_{t}$

$P_{radiated} = \frac{R}{I^{2}}$

Now, the current I is given by:

$I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A$

5 0

2 years ago

A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform

kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

$B=\mu_{0} ni$

Put the value into the formula

$B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}$

$B_{c}=0.0000100\ T$

$B_{c}=10.02\times10^{-6}\ T$

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance $d=\dfrac{5}{2}$

$d=2.5\ cm$

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

$B_{w}=\dfrac{\mu_{0}I}{2\pi d}$

Put the value into the formula

$B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}$

$B_{w}=4.0\times10^{-6}\ T$

This field is perpendicular to the wire.

The magnitude of magnetic field is

$B=\sqrt{B_{c}^2+B_{w}^2}$

$B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}$

$B=0.00001078\ T$

$B=10.78\times10^{-6}\ T$

$B=10.78\ \mu\ T$

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

6 0

1 year ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

horrorfan [7]

Answer:

The value is $t_1 = 9 \ s$

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

$v = u + at$

Here u = 0 \ m/s

a = 13.5 $m/s^2$

So

$v = 0 + 13.5 * 3.10$

=> $v = 41.85 \ m/s$

The is distance it covers at this time is

$s = u * t + \frac{1}{2} a * t^2$

=> $s = + \frac{1}{2} * 13.5 * 3.10^2$

=> $s =64.87$

Now when sled stops its the final velocity is $v_f = 0 m/s$ while the initial velocity will be the velocity after its acceleration i.e $v = 41.85 \ m/s$

So

$v_f = v + a_1t_1$

Here $a_1 = - 4.65$ , the negative sign shows that it is deceleration

So

$0 = 41.85 - 4.65 * t_1$

=> $t_1 = 9 \ s$

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1 year ago

Two basketball teams are resting during halftime. While they are resting, a truck driver asks them for help pushing his broken d

viva [34]

Answer:

It is a superordinate goal because both teams could have helped with the task.

Explanation:

If both teams pushed then they could have made it happened

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2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago