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2 years ago

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A 12-V DC automobile head lamp is to be used on a fishing boat with a 24-V power system. The head lamp is rated at 50 W. A resis

tor is to be connected in series with the lamp to permit it to operate on 24 V. What should be the resistance and power rating of the resistor?

1 answer:

spin [16.1K]2 years ago

8 0

Answer:

The resistance is $R = 2.88 \ \Omega$

Explanation:

From the question we are told that

The voltage rating of the headlamp is $V_1 = 12 \ V$

The voltage of the power system is $p = 24 \ V$

The power rating of the headlamp is $P = 50 W$

Generally the power which the resistor dissipates is mathematically represented as

$P = V_L * I$

=> $50 = 12 * I$

=> $I = 4.1667 \ A$

Generally the resistance is

$R = \frac{V_1 }{I}$

$R = \frac{12 }{4.1667}$

$R = 2.88 \ \Omega$

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4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Basile [38]

The strength of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 is the number of turns in the coil

$\Delta \Phi$ is the change in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ is the time interval

$\epsilon = 0.166 V$

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

$\Delta \Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area of the coil

$\theta=0^{\circ}$ is the angle between the normal to the coil and the field

The area of the coil can be written as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ is its radius

Substituting everything into eq.(1) and solving for B, we find:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

Learn more about magnetic fields:

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#LearnwithBrainly

8 0

2 years ago

Read 2 more answers

An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

Aneli [31]

We are given: Final velocity ( $v_f$ )=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

$v_f = v_i + at.$

Let us plug values of final velocity, and time in above equation.

$20=v_i+a(2.51)$

$20=v_i+2.51a$

Subtracting 2.51a from both sides, we get

$20-2.51a=v_i$ -----------equation(1)

Using another equation of motion

$v_f-v_i=2as$

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.

5 0

2 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

2 years ago

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ioda

Answer:

$Q_{week} = 458884.6\, BTU$

Explanation:

The weekly water consumption of Kimonoski is:

$m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{bath.week} = 948.205\,lbm$

$m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{others, week} = 1346.218\,lbm$

$m_{week} = m_{bath,week} + m_{others, week}$

$m_{week} = 2294.423\,lbm$

The total energy required per week for hot water is:

$Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T$

$Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)$

$Q_{week} = 458884.6\, BTU$

3 0

1 year ago

The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma

Molodets [167]

Answer:

the order of importance must be b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

b e a f c

5 0

2 years ago