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Ad libitum [116K]
4 years ago
10

If the Force exerted by the intern is doubled and the distance is halved, does the done by the intern increase, decrease, or rem

ain the same?
Physics
1 answer:
Jlenok [28]4 years ago
3 0

Remain the same

Explanation:

If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.

 Work done is the force applied to move a body through a distance.

Work done = F x d

where F is the applied force

            d is the distance moved

Now;

if:

    f = 2f

    d = \frac{1}{2}d

Input the parameter:

 Work done = fxd = 2f x \frac{1}{2}d  = fd

The work done will still remain the same

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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Determine the speed of sound in air at 300 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3
amm1812

Answer:

c_3_0_0_K=347.19m/s

M=0.864

Explanation:

The speed of sound in the air increases 0.6 m / s for every 1 ° C increase in temperature.  An approximate speed can be calculated using the following empirical formula:

c=331.5+0.6\vartheta

Where:

\vartheta=T-273.15K\\\\

A more exact equation, usually referred to as adiabatic velocity of sound, is given by the following formula:

c=\sqrt{k*R*T}

Where:

R= Gas\hspace{3}constant\hspace{3}of\hspace{3}air=0.287kJ/kg*K=287J/kg*K\\k=Specific\hspace{3}heat\hspace{3}ratio=1.4\\T=Temperature=300K

Hence:

c=\sqrt{(287)*(1.4)*(300)} =347.1887095\approx347.19m/s

Now, the Mach number at which an aircraft is flying can be calculated by:

M=\frac{u}{c}

Where:

u= Velocity\hspace{3}of\hspace{3}the\hspace{3}moving\hspace{3}aircraft\\c= Speed\hspace{3}of\hspace{3}sound\hspace{3}at\hspace{3}the\hspace{3}given \hspace{3}altitude

Therefore:

M=\frac{300}{347.19} =0.8640833984\approx0.864

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Ross and Chandler are carrying a 150 kg crate up a flight of stairs. The crate is 2.0 m long and 1.5 m high, and its center of g
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Answer:

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Explanation:

The reaction at each lifter due weight and overturning effect (R₁, R₂) is given as:

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g = 9.8 m/s²

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