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strojnjashka [21]

4 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]4 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:

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

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At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

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3 years ago

Particles of m1 and m2 (m2>m1) are connected by a line in extensible string passing over a smooth fixed pulley. Initially, bo

Tresset [83]

Answer:

The velocity with which the mass will hit the floor is $v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}$

Explanation:

If the tension in the string is $T$ , for $m_1$ we have

$T- m_1g =m_1a$ ,

and for the mass $m_2$

$T -m_2g = -m_2a$

From these equations we solve for $a$ and get:

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

The kinematic equation

$v_f^2 = v_0^2+2ax$

gives the final velocity $v_f$ of a particle, when its initial velocity was $v_0$ , and has traveled a distance $x$ while undergoing acceleration $a$ .

In our case

$v_0 = 0$ (the initial velocity of the particles is zero)

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

which gives us

$v_f^2 = 2ax$

$v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g$

$\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }$

which is the velocity with which the mass $m_2$ will hit the floor.

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3 years ago

Which shows one example of physical change and the one example of a chemical change?

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Answer:

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Explanation:

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3 years ago

A student must design an experiment to determine the gravitational mass of an object. Which of the following experiments could t

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Answer: a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Explanation:

The options are:

a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

b. Place the object on a surface of negligible friction and pull the object horizontally across the surface with a spring scale at a non constant speed such that a motion detector can measure how the objects speed as a function of time changes.

c. Place the object on a surface that provides friction between the object and the surface. Use a surface such that the coefficient of friction between the object and the surface is known. Pull the object horizontally across the surface with a spring scale at a nonconstant speed such that a motion detector can measure how the objects speed as a function of time changes.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Gravitational mass simply has to do with how the body responds to the force of gravity. From the options given, the correct options are A and D.

For option A, by balancing the torque, the mass can be calculated. Since the known mass and the distance has been given here, the unknown mass can be calculated.

For option D, here the gravitational force can be balanced by the spring force and hence the mass can be calculated.

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