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Maurinko [17]
2 years ago
15

A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output

power of the charger in unit of watt?
Physics
1 answer:
Arte-miy333 [17]2 years ago
3 0

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

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3 years ago
What is the gravitational potential energy of a 65.7 kg person standing on the roof of a 135 meter building?
horrorfan [7]
For the purpose we will use the following equation for potential energy:
U = m * g * h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth).
When we plug values into the equation, we get following:
U= 65.7kg * 9.8 N/kg *135m = 86921.1 J = 86.92 kJ

8 0
3 years ago
a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0
3 years ago
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