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ch4aika [34]
3 years ago
15

a light bulb has a resistance of 360 . what is the current in the bulb when it has a potential difference of 120 v across it? 0.

33 a 3 a 480 a 43,200 a
Physics
2 answers:
xenn [34]3 years ago
7 0
Presume we are looking for the current:

V = IR

120 = I*360

120/360 = I

1/3 = I

I = 1/3 = 0.333..

Current ≈ 0.33 Ampere.
tekilochka [14]3 years ago
6 0

Answer:

i = 0.33 A

Explanation:

As per ohm's law we know that voltage across any ohmic resistor is product of resistance and current through it

So it is given as

V = i R

now we know that

V = 120 Volts

R = 360 ohm

so here we have

120 = 360 \times i

i = \frac{1}{3} A

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when you tell someone over and over to stop bothering you and they dont so i think you should tell a teacher

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3 years ago
1) how does reflection differ from diffraction?
Evgesh-ka [11]

Answer: (1) The correct answer is A.

(2) The correct answer is D.

Explanation:

(1)

Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.

Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.

Therefore, the correct answer is A.

(2)

Amplitude is the maximum displacement in the medium from the rest position.

The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.

Therefore, the correct option is D.

5 0
3 years ago
Read 2 more answers
What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0
2 years ago
A 20.5 kg ball moving at 38.5 m/s on a horizontal, frictionless surface runs into a light spring of force
tekilochka [14]

Answer:

7.68 m

Explanation:

Kinetic energy in ball = elastic energy in spring

KE = EE

½ mv² = ½ kx²

mv² = kx²

x = v √(m / k)

x = (38.5 m/s) √(20.5 kg / 515 N/m)

x = 7.68 m

5 0
3 years ago
A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

8 0
4 years ago
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