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1 year ago

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Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

moves 6 m downward before coming to rest and assuming uniformly accelerated motion, determine (a) the angular acceleration of the drum, (b) the time required for the cylinder to come to rest.

1 answer:

Xelga [282]1 year ago

7 0

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

4 0

1 year ago

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean ( $\mu$ ) plus or minus one standard deviation ( $\sigma$ ), $P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%$

at the range of mean plus or minus two standard deviation, $P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%$

at the range of mean plus or minus three standard deviation, $P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%$

So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ( $P(X \leq \mu+2\sigma)$ )

So we know that the 95.5% is between $\mu - 2\sigma = 0.3 -2*0.1 = 0.28$ and $\mu + 2\sigma = 0.3 +2*0.1 = 0.32$ , hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So $P(X \leq \mu+2\sigma) \approx 2.25%$

8 0

1 year ago

ANSWER BELOW QUESTIONS:

Bogdan [553]

Answer:

why should we do , do by your own , no sense

Explanation:

3 0

1 year ago

Read 2 more answers

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

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2 years ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0

1 year ago