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IgorLugansk [536]

3 years ago

9

A magnetic pickup is used to detect the shaft speed of an experimental high speed compressor. a gear with 12 teeth is used to ex

cite the sensor. if the shaft is turning at 18,000 rpm, how many positive pulses will be created per second?

2 answers:

Harlamova29_29 [7]3 years ago

8 0

The answer is 39 seconds.

Hope this helps!! ;))
Have a grat day!! <3

worty [1.4K]3 years ago

5 0

The answer would be 39 seconds

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Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i

maks197457 [2]

Answer:

1 greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s) Y (m)

1 -4.9

2 -19.6

3 -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

3 0

3 years ago

B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

prohojiy [21]

I Lolo my name is keshav and

5 0

3 years ago

You run away from a plane mirror at 1.80 m/s. At what speed does your image move away from you?

Vlad [161]

Answer:

3.60m/s

.................

4 0

3 years ago

Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir

yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) $T=2\pi \sqrt{\frac{r^{3}}{GM}}$

$M=\frac{4\pi ^{2}r^{3}}{GT^{2}}$

$M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}$

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

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

$v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}$

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

$E = \frac{GM}{r^{2}}$

$E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}$

E = 0.28 N/kg

6 0

3 years ago

When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0

3 years ago