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Soloha48 [4]
3 years ago
6

A light bulb has a resistance of 5 ohms and a maximum current of 10 A. How much voltage can be applied before the bulb will brea

k?​
Physics
1 answer:
Degger [83]3 years ago
8 0

The maximum voltage which can be applied before the bulb will break is 50 Volts.

<u>Given the following data:</u>

  • Resistance of light bulb = 5 Ohms
  • Current = 10 Amps.

To find how much voltage can be applied before the bulb will break, we would apply Ohm's law:

Ohm's law states that at constant temperature, the current flowing through an electrical circuit is directly proportional to the voltage applied across the two terminals of the circuit and it's inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

<u>Where;</u>

  • V is voltage measured in voltage.
  • I is current measured in amperes.
  • R is resistance measured in ohms.

Substituting the given parameters into the formula, we have;

Voltage = 10 × 5

<em>Voltage </em><em>=</em><em> 50 Volts.</em>

Therefore, the maximum voltage which can be applied before the bulb will break is 50 Volts.

Read more: brainly.com/question/23754122

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Answer:

2 m/s²

Explanation:

If changes speed by 2 meters per second each second means:

2 m/s²

Because it changes constantly it veloctity.

Remember the aceleration changes the velocity.

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•Every action has an equal and opposite reaction (the object is putting force on the target, and the target is putting an equal amount of force back)
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3 years ago
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

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