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inysia [295]
3 years ago
10

a 15 ohm resistor and a 25 ohm resistor are connected in parallel to a 60 volt battery. what is the power used by each device

Physics
1 answer:
andreev551 [17]3 years ago
8 0

25 ohm resistor

E = I * R                    General formula

  • E = 60 volts    Givens
  • R = 25 ohms

60 = I * 25                Divide by 25

60/25 = I*25/25      

2.4 = I

P = E * I

E = 60 volts

I = 2.4 amps

P = 60*2.4

P = 144 watts.                 Answer: 25 ohm Resistor

==============  

15 Ohm Resistor

E = I R

  • E = 60 volts
  • R = 15 ohms

I = ??

60 = I * 15                Substitute

60/15= I * 15/15       Divide by 15

I = 4

P = E * I

  • E = 60
  • I = 4

P = 60 * 4

P = 240 Watts.               Answer: 15 ohm resistor                      

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Voltmeters are used to measure voltage power (volt)

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How does newtons 1st law apply to a project where you drop and egg?
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Put a parashoot on it so it touches the ground lightly and carefully
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4 years ago
Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through diamond. The wavelength o
vladimir1956 [14]

Answer:

speed =  1.24 ×  10⁸m/s

frequency = 4.74 × 10¹⁴Hz

wavelength = 262nm

Explanation:

the speed of the helium-neon light in zircon is given by,

v = c / n

c = 3 × 10⁸m/s is the speed of light in vacuum (and in air)

n = 2.419 is the refractive index of diamond

v = 3 × 10⁸ / 2.419

= 1.24 ×  10⁸m/s

(b) Frequency

The wavelength of the light in air is:

λ₀ = 632.8 × 10⁻⁹

The frequency of the light does not depend on the medium, so it is equal in air and in diamond. Therefore, we can calculate the frequency by using the speed of light in air and the wavelength in air:

f₀ = c / λ₀

= 3 × 10⁸ / 632.8 × 10⁻⁹

= 4.74 × 10¹⁴Hz

and the frequency of the light indiamond is the same:

f¹ = f₀ =  4.74 × 10¹⁴Hz

(c) Wavelength

To calculate the wavelength of the light in daimond, we can use the relationship between speed of light in diamond and frequency:

λ¹ = v / f¹

= 1.24 ×  10⁸ / 4.74 × 10¹⁴

= 2.62 × 10⁻⁷m

= 262nm

3 0
3 years ago
How much work (MJ) must a car produce to drive 125 miles if an average force of 306 N must be maintained to overcome friction?
baherus [9]

Answer:

The right solution is "61.557 MJ". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 306 N

Drive,

D = 125 miles,

i.e.,

  = 201168

meters

As we know,

The work done will be:

= F\times S

On substituting the given values, we get

= 306\times 201168

= 61557408 \ J

On converting it in "MJ", we get

= 61.557\times 10^6 \ J

= 61.557 \ MJ

6 0
3 years ago
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