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2 years ago

14

Thermal energy transfers from a cup of tea at 350 K to the hand holding it. A)True B)False

1 answer:

Alexandra [31]2 years ago

6 0

Answer:

true

Explanation:

it is an energy transfer

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A sample of gas has an initial volume of 4.5 L at a pressure of 754 mmHg . Part A If the volume of the gas is increased to 8.5 L

Firdavs [7]

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

6 0

3 years ago

f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

2 years ago

A technician services the carburetor, and then, performs a complete governor system adjustment. The governor system on the engin

madreJ [45]

Answer: Either of the two

Explanation:

Either of the two - the governed idle spring or the normal primary governor spring can be adjusted first.

The main function of the carburetor is to mix air with fuel.

5 0

2 years ago

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

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2 years ago

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A printed circuit board (PCB) is supported by a chassis that is attached to a vibrating motor. The board is 1.6 mm thick, 200 mm

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2 years ago