Ask question

Ask question

All categories

2 years ago

9

A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h

ow much did the machinist increase the feed rate

2 answers:

LekaFEV [45]2 years ago

6 0

Explanation:

Given that,

Current feed rate of a machine is 5.921 IPM

Final feed rate of the machine is 6.088 IPM

To find,

Increase in feed rate

Solution,

We need to find the increase in feed rate of the machine. It is calculated simply by subtracting final feed rate to the initial feed rate as :

Increase in feed rate = 6.088 IPM - 5.921 IPM

Increase in feed rate = 0.167 IPM

or

Increase in feed rate = 16.7 %

So, the increase in feed rate of the machine is 0.167 IPM or 16.7 %

lukranit [14]2 years ago

4 0

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

You might be interested in

Which statement correctly compares and contrasts the information represented by the chemical formula and model of a compound?

Alina [70]

Models show how the atoms in a compound are connected.

6 0

2 years ago

Read 2 more answers

You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

(C) $- 45^{\circ}$ :

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

5 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

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

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

2 years ago

Ball 1 travels with a momentum of 48.0 kg-m/s east and strikes Ball 2, which is initially at rest.. Ball 1 separates at an angle

vladimir1956 [14]

Answer:

Momentum of 2nd ball is

$P = 31.6 kg m/s$

direction is given as

$\theta = -37.66 degree$

Explanation:

As we know that there is no external force on the system of balls so momentum before and after collision will be conserved

So we have

$P_i = 48 \hat i + 0$

now after collision momentum of two balls is must be same as initial

so we have

$P_i = P_f$

$48\hat i = (30 cos40 \hat i + 30 sin40\hat j) + (P_{2x}\hat i + P_{2y}\hat j)$

so we have

$48 = 23 + P_{2x}$

$P_{2x} = 25 kg m/s$

for other component we have

$0 = 19.3 + P_{2y}$

$P_{2y} = -19.3 kg m/s$

Momentum of 2nd ball is given as

$P = \sqrt{P_2x}^2 + P_{2y}^2}$

$P = 31.6 kg m/s$

direction is given as

$tan\theta = \frac{P_{2y}}{P_{2x}}$

$tan\theta = \frac{-19.3}{25}$

$\theta = -37.66 degree$

5 0

2 years ago

You are moving at a speed 2/3 c toward randy when randy shines a light toward you. at what speed do you see the light approachin

yarga [219]

I see the light moving exactly at speed equal to c.

In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.

3 0

2 years ago