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kherson [118]
2 years ago
13

As an audio CD plays, the frequency at which the disk spins changes.

Physics
1 answer:
Yuliya22 [10]2 years ago
3 0

Answer:

The velocity is 2.5997 m/s.  

Explanation:

The given condition :-

1) N = 210 rpm

speed ( v ) = 1.3 m/s.

We know that ,

v = r ω

where r = radius of disk.

ω = angular velocity.

But ω = \frac{2*\pi*N }{60}

v = \frac{r* 2*\pi*210 }{60}

1.3 = \frac{r*2*\pi*210 }{60}

r = 0.059144 m

2) N = 420 rpm

v = \frac{r* 2*\pi*N }{60} = \frac{0.059144*2*3.14*420}{60}    

v = 2.5997 m/s.  

So we get the velocity v = 2.5997 m/s.  

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Answer:

Explanation:

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gram / centimeter³

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So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

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2 years ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
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Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

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Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
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