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11Alexandr11 [23.1K]
3 years ago
7

A battery of voltage V delivers power P to a resistor of resistance R connected to it. By what factor will the power delivered t

o the resistor change if the following changes are made? Give your answers as a factor of the original power P.(a) The resistance is changed to 2.90R.(b) The voltage of the battery is now 2.90V, but the resistance is R.(c) The resistance is 2.90R and voltage is 2.90V.(d) The resistance is 2.90R and the voltage is V/2.90.
Physics
1 answer:
Anettt [7]3 years ago
4 0

Answer:

Explanation:

Power P = V² / R

a ) The resistance is changed to 2.90R

Power will become 1 / 2.9 times .

b )The voltage of the battery is now 2.90V, but the resistance is R

P = (2.9V)² / R

= 8.41 x V² / R

So power becomes 8.41 times

c )The resistance is 2.90R and voltage is 2.90V

Power P = (2.9V)² / 2.9 R

= 2.9 V²/R

So power becomes 2.9 times

d ) The resistance is 2.90R and the voltage is V/2.90

Power P = ( V/2.90)² x 1 / 2.90R

1 / ( 2.9 )³ x V² / R

= 1 / 24.389 x V² / R

So power becomes  1 / 24.389 times .

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
3 years ago
What is the answer to this question number 2?
anzhelika [568]

Answer:

1⁺ ion

Explanation:

Metals in the first group on the periodic table will prefer to form 1⁺ ion. This is because the 1 valence electron in their orbital.

Most metals are electropositive and would prefer to lose electrons than to gain it.

Like all metals, the group 1 elements called the alkali metals would prefer to lose and electron.

On losing an electron the number of protons is then greater than the number of electrons. This leaves a net positive charge.

3 0
2 years ago
The diagram illustrates the movement of sound waves between an observer and a race car. As the race car drives away from the obs
Andrei [34K]
I believe the answer is d
6 0
3 years ago
2. A body is thrown vertically upward with a speed of 100 m/s.The time taken to be
Pachacha [2.7K]

Answer:

b. 20 sec

Explanation:

y = y₀ + v₀ t + ½ g t²

0 = 0 + (100) t + ½ (-10) t²

0 = 100t − 5t²

0 = t (100 − 5t)

t = 0, t = 20

The body lands after 20 seconds.

4 0
2 years ago
You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.
Vikki [24]

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

5 0
3 years ago
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