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natita [175]
2 years ago
7

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?
Physics
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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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?
Kay [80]
There was no net force on the stuffed toy, because the kids might have the same strength,  The same force is on both sides of it.  T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move.  </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving. 
3 0
2 years ago
Read 2 more answers
Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the
Ratling [72]
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
5 0
1 year ago
Read 2 more answers
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
1 year 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
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