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sergey [27]
3 years ago
12

Part A Which of the following statements is/are true? Check all that apply.

Physics
1 answer:
viva [34]3 years ago
4 0

Answer:

A) True

C)True

D)True

Explanation:

As we know that power is the rate at which work is done.

We know that mechanical work w given as

 w = ∫F.ds

F=Force

ds = Elemental displacement

So power P = F.V

V=Velocity

F=Force

The unit of power in SI unit is Watt or joule per second (J/s).But on the other hand the unit of power in English system is horsepower.

The is also can be defined as the rate at which energy is transformed.

Following option are corrects

A) True

C)True

D)True

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If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?
Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy mgh, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature {\Delta}T then will be

{\Delta}T=\frac{{\Delta}Q}{mc} .

Where m is the mass of water in the pool, c is the specific heat capacity of water, and {\Delta}Q is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0
3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
In which of the following conditions would there be the most resistance?
Natasha2012 [34]

Answer:

A) A warm wire

Explanation:

A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.

8 0
3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
Which of the following best describes what muscles do.
kow [346]

I think b support body weight

8 0
3 years ago
Read 2 more answers
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