Ask question

Ask question

All categories

2 years ago

5

A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc

ated 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed.
(a) What is the angular acceleration (in rad/s^2) of the centrifuge as it spins up?
(b) How many revolutions does the centrifuge make as it goes from rest to its final angular speed?

1 answer:

Dafna11 [192]2 years ago

6 0

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s, u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest), = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

You might be interested in

A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards

8_murik_8 [283]

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

3 0

2 years ago

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0

2 years ago

Read 2 more answers

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

DENIUS [597]

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

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

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

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

$y=7.39\ m$

5 0

2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

1 year ago

A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an

Ainat [17]

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

4 0

1 year ago