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3 years ago

The pinion gear A rolls on the gear racks B and C.

Physics

1 answer:

FinnZ [79.3K]3 years ago

5 0

A it makes more sense

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What is the net force on a water bottle?

Bad White [126]

Answer:

if it is not moving it is zero, like all things that aren't in motion

Explanation:

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3 years ago

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Use the table to answer the question.

QveST [7]

Answer:

Mechanical Waves

Explanation:

I took the practice

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2 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.

Drupady [299]

A I hope it helpsss youuu ;:))))

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3 years ago

A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz

Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, $\bar f$ is given by

$\bar f=f\frac {V}{V-V_s}$ where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and $V_s$ is the speed of car which is unknown, \bar f is given as 440 Hz

$440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s$

8 0

3 years ago