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3 years ago

A sled of mass 8 kg slides along the ice. It has an initial speed of 4 m/s but

Physics

1 answer:

marysya [2.9K]3 years ago

6 0

As our story begins, the sled ... whose mass is 8 kg ... is sliding along the ice at a speed of 4 m/s.

The sled's kinetic energy is (1/2 m v²) = (4 kg · 16 m²/s²) = 64 J .

After what seems like only the blink of an eye, the sled is no longer sliding. It is stationary. Motionless. At Rest. Just sitting there !

Its speed has been reduced to zero and ... because kinetic energy is the energy of motion ... the sled's kinetic energy is now also zero. Sixty-four Joules of energy have disappeared !

How can this be ? ! ? We know that energy is conserved. It can never just appear out of nothing, and it can never just disappear into nothing. If energy suddenly appears, it had to come from somewhere, and if energy suddenly disappears, it had to go somewhere. So where did our 64 Joules of kinetic energy go ?

It went into the ice, THAT's where ! We can say that the sled did 64J of work, and melted a thin slick layer of water on the surface of the ice. OR we can say that friction did NEGATIVE 64J of work on the sled, to cancel the 64J that it had originally, sap its kinetic energy, and bring it to rest.

I think <em>choice-B</em> was supposed to say "<em>B. -64J</em>", but somebody typed it sloppily and neglected to proofread it before posting.

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4 years ago

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4 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi

sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

$v_1$ = Velocity of sound = 342 m/s

$v_2$ = Velocity of sound = 346 m/s

$f_1$ = First frequency = 440 Hz

Frequency is given by

$f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz$

Beat frequency is given by

$|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s$

Beat frequency is 5.15348 Hz

Wavelength is given by

$\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m$

Relation between length of the flute and wavelength is

$\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m$

At v = 346 m/s

$\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m$

$L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m$

Difference in length is

$\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm$

It extends to 4.55 mm

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3 years ago

During a baseball game, a batter hits a high .

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3 years ago

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3 years ago