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OverLord2011 [107]
2 years ago
12

The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

ttery.
Physics
2 answers:
posledela2 years ago
4 0

The power dissipated by a resistance ' R '
powered by a battery ' V ' is                            
                                                 P = V² / R .

Before we play with actual numbers, look at that formula, and
notice that the power is proportional to the square of the voltage. 
So if the voltage suddenly drops by half, then the power that's
dissipated by the resistance drops to 1/2² = 1/4 of where it started out.

Now that we have the pure Science worked out, we're ready to do
the Engineering, and apply our findings to some actual technology.

We have a light bulb that's designed to dissipate energy at the rate
of 24W when you connect it to a source of 12V .  But all we have
in the lab today is a 6V battery.  What power will our bulb dissipate
if we connect it to the 6V that we have, instead of the 12V that it's
designed for ?

We're all standing around scratching our beards, when the new hot-shot
post-doc lab assistant reminds us of the formula up in the first paragraph,
and shows us that since we're running it at 1/2 of the design voltage, we
expect it to dissipate 1/4 of design power = 6 watts.

It'll give us a bit of light so we can still walk around the lab without bumping
into things, but it won't get hot enough to warm our lunches.  We'll have to
do that over a Bunsen burner or in the microwave oven today, and one of
you junior members of the lab staff will have to stop at Lady-o'-Rack and
pick up either a new battery or a new bulb on your way in to work tomorrow.
tresset_1 [31]2 years ago
3 0
It would be 12W because: 6v is half of 12v so half of 24w would be 12w
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Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

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3 years ago
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Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the
scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

how \frac{KQ^{2} }{r^{2} } = 0.42

Then

F = \frac{0.42*3}{8}

F = 0.1575 N

3 0
3 years ago
A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper
alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

4 0
2 years ago
Ann (mass 50 kg) is standing at the left end of a 15-m-long, 500 kg cart that has frictionless wheels and rolls on a frictionles
ivanzaharov [21]

Answer:

13.5 m

Explanation:

M = Mass of cart = 500 kg

m = Ann's mass = 50 kg

v_m = Velocity of Ann relative to cart = 5 m/s

v_M = Velocity of Cart relative to Ann

As the linear momentum of the system is conserved

Mv_M+mv_m=0\\\Rightarrow v_M=-\frac{mv_m}{M}\\\Rightarrow v_M=-\frac{50\times 5}{500}\\\Rightarrow v_M=-0.5\ m/s

Time taken to reach the right end by Ann

Time=\frac{Distance}{Speed}\\\Rightarrow Time=\frac{15}{5}=3\ s

Distance the cart will move in the 3 seconds

Distance=Speed\times Time\\\Rightarrow Distance=-0.5\times 3=-1.5\ m

The negative sign indicates opposite direction

Movement of Ann will be the sum of the distances

15+(-1.5)=13.5\ m

The net movement of Ann is 13.5 m

5 0
2 years ago
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