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2 years ago

If the current is cut in half and the resistance is increased by a factor of three, how will the power be affected?

1 answer:

8 0

One formula for the power dissipated in an electric circuit is

   Power = (current)² x (resistance) .

I think that's the one we should use, because it has exactly
all of the quantities that we know or need to find.

OK.
   Original power = (original current)² x (original resistance)
and
   New power = (new current)² x (new resistance)

   = (0.5 x old current)² x (3 x old resistance)

   = (0.25) (old current)² x (3) (old resistance)

   = (0.75) (original current)² (original resistance)

   = (0.75) (original power) .

Summary: New power = (0.75) (original power)

That looks just like
choice B:
   The power will decrease to 75 percent of its previous value.

You might be interested in

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

FrozenT [24]

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = $M_{0} V$

where $M_{0}$ is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = $M_{0}/ \sqrt{1 - \beta ^{2} }$

where

$\beta = V/c$

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> $M_{0}V/ \sqrt{1 - \beta ^{2} }$

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75 $M_{0} V$ = $M_{0}V/ \sqrt{1 - \beta ^{2} }$

the equation reduces to

1.75 = $1/ \sqrt{1 - \beta ^{2} }$

square both sides of the equation, we have

3.0625 = 1/ $(1 - \beta ^{2} )$

3.0625 - 3.0625 $\beta ^{2}$ = 1

2.0625 = 3.0625 $\beta ^{2}$

$\beta ^{2}$ = 0.67

β = 0.819

substitute for $\beta = V/c$

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

6 0

3 years ago

Which sentence accurately uses the homophones “they’re,” “there,” or “their”?

Inessa05 [86]

Answer:

They're going to come home as soon as the movie is over.

4 0

2 years ago

Read 2 more answers

**Fill in the blanks**

Nataly_w [17]

Answer:

Word for the first blank: gravity

Word for the second blank: matter

Explanation:

The only way debris from the impact with Earth can be held close to Earth is due to a force. The only force that could be acting from Earth is "the force of gravity".

The gravitational pull of this new object being formed, increases proportional to its mass as more and more "matter" accumulates. And the accretion process is now on its way.

4 0

3 years ago

Read 2 more answers

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$