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Vesna [10]
3 years ago
15

Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would

one headlight and the starter consume if connected in series to a 12.0 V battery
Physics
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

 P_{s } = 29.64 W

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At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0
1 year ago
All objects emit ______ radiation?<br> A-electromagnetic<br> B-kinetic<br> D-solar
natulia [17]
A, electromagnetic radiation
6 0
3 years ago
Some help me on this please
satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0
3 years ago
Please. Physics is so difficult.
Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

4 0
1 year ago
Read 2 more answers
what is the name of the area around a charged object where the object can exert a force on other charged objects?
shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

4 0
2 years ago
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