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sweet-ann [11.9K]
3 years ago
6

Prove that the rate of heat production in each of the two resistors connected in parallel is inversely proportional to the resis

tances
Physics
1 answer:
Viktor [21]3 years ago
4 0
I believe that the answer to the question provided above is that with increase in resistance provided with constant current, Power dissipated will be lessen since power loss is high. Low power dissipation has low heat production.
Hope my answer would be a great help for you.  
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Convert -90.0°F to -5.0°c to kelvin
Ratling [72]

-90.0 °F  =  -67 and 7/9 °C

-90.0 °F also = +205.372 K

-5.0 °C  =  +23 °F

-5.0 °C also  =  +268.15 K

4 0
3 years ago
Diffraction is observable when the opening is ______ than the wavelength.
masya89 [10]
Diffraction is observable when the is smaller than the wavelength 
5 0
3 years ago
Read 2 more answers
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0
3 years ago
What are the cahnges that a force can bring out on a body ? <br><br><br>Give examples
djyliett [7]
Hi Pupil Here is your answer ::




➡➡➡➡➡➡➡➡➡➡➡➡➡



1 The shape of the Body

Example : The shape of the ball lying on a floor can be changed by pressing it.


2 Direction of the Body

Example : The direction of motion of moving ball can be changed by hitting it with a bat.


3 The speed of the Body

Example : A ball at rest can be set in motion if force is applied only


4. Size of the Body

Example : The length of a spring tied and on one end can be increased by pulling it.




⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅




Hope this helps .......
5 0
3 years ago
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PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru
LuckyWell [14K]
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
4 0
3 years ago
Read 2 more answers
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