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2 years ago

Power is calculated by multiplying voltage by

Physics

1 answer:

Snowcat [4.5K]2 years ago

4 0

Answer:

Power is calculated by multiplying voltage by current.

Explanation:

If you're calculating the power of an athlete who is lifting weights,

or the power of a windmill that is pumping water from a well, or

the power of two horses turning millstones, then those electrical

units won't help at all.

But if you happen to be calculating the power delivered to an

electrical circuit or dissipated by an electrical device, then you

can use ...

(voltage) times (current)

(voltage)² divided by (resistance)

(current)² times (resistance) .

The choice just depends on which quantities you know

or can easily measure.

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a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr

Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

$f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}}$ where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

$f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz$

5 0

2 years ago

A child pulls a sled up a snow-covered hill. The child does 504 J of work on the sled. If the child walks 15 m

Juliette [100K]

Answer: F = 33.6 N

Explanation: work = force · distance or W = F·s

Force F = W/s = 504 J/15 m

7 0

2 years ago

Identify the two pieces of information you need to know the velocity of an object

Georgia [21]

Manuel with and height?

8 0

2 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

2 years ago

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The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

8 0

2 years ago

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