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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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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
Calculate the force exerted by a mental ball having a mass of 70kg moving with speed of 20m/s&gt;2
lord [1]

Answer:

F = 1400 N

Explanation:

It is given that,

Mass of the ball, m = 70 kg

It is moving with an acceleration of 20 m/s². We need to find the force exerted by the ball.

Force is given by the product of mass and acceleration. So,

F = ma

F=70\ kg\times \ 20m/s^2\\\\F=1400\ N

So, the force of 1400 N is exerted by a metal ball.

8 0
3 years ago
What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?
sergey [27]

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

6 0
1 year ago
Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona
erica [24]

Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

Mass of the person, m = 60 kg

The acceleration due to gravity on the surface of gravity is 3.7 m/s²

We need to find the weight of a person on the surface of Mercury.

Weight of an object is given by :

W = mg

So,

W = 60 kg × 3.7 m/s²

W = 222 N

Hence, the person will weigh 2222 N on the surface of Moon.

4 0
3 years ago
What are the conditions under which electricity is conducted ​
dolphi86 [110]

There must be a conducting wire and electromotive force or free electrons

3 0
3 years ago
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