Ask question

Ask question

All categories

strojnjashka [21]

3 years ago

10

Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee

d in units of radians/sec. (b) The linear speed in units of inches/sec of a point on the outer edge of the blade.

1 answer:

notka56 [123]3 years ago

3 0

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

$\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}$

$\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}$

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

$v=\omega r$

with v linear speed, ω angular speed and r the radius of the blades. So:

$v=(24.43\frac{rad}{s})(11 in)$

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

$v=268.73\frac{in}{s}$

You might be interested in

Hi Friends!<br><br>please help me with these questions!<br><br>SUBJECT: Chemistry, Physics,Biology

satela [25.4K]

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

6 0

2 years ago

Explain the characteristics of the Gypsum Hills regionIf we have a sample of silicon (Si) atoms that has 14 protons, 14 electron

amid [387]

Answer:

second one

Explanation:

8 0

2 years ago

100 points!! word bank!

Novosadov [1.4K]

Answer:

Except in the crust, the interior of the Earth cannot be studied by drilling holes to take samples. Instead, scientists map the interior by watching how seismic waves from earthquakes are bent, reflected, sped up, or delayed by the various layers.

Explanation:

6 0

2 years ago

What type of energy is formed when an object encounters friction?

kiruha [24]

Answer:

thermal energy

Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force. The forces acting on the package are gravity, the normal force, the force of friction, and the applied force.

Explanation:

6 0

2 years ago

Read 2 more answers

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

1 year ago