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1 year ago

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

Physics

1 answer:

Yuliya22 [10]1 year 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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A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

Liula [17]

The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
   Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D = (1/2) (g) (t²)

Multiply each side by 2 :         2 D =            g    t²

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

   and smaller 'g' ==> longer 't' .--

They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.

Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.

5 0

2 years ago

Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

Debora [2.8K]

Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

$H = \vec d$

work done is given as

$W = \vec F . \vec d$

Potential energy is given as

$PE_g = mgh$

force due to gravity is given as

$\vec F_g = mg$

now here if we plug in the value of distance and force in the formula of work done then we will have

$W = (mg)(h)$

so here we got

$W = PE_g$

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

3 0

1 year ago

Read 2 more answers

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

1 year ago

En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0

1 year ago