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3 years ago

What is a caliper used for?

Physics

2 answers:

pav-90 [236]3 years ago

7 0

a caliper is used to make a precise measurement of length within tubing or on nuts/bolts, or anything really

uysha [10]3 years ago

5 0

<h3><u>Answer; </u></h3>

To make a precise measurement of length

<h3><u>Explanation;</u></h3>

<em><u>Calipers are used to make a precise measurement of length such as diameters of circular solids, such as the diameter of a pipe.</u></em>

<em><u>All calipers may be used to take the same measurements, that is the inside, outside, depth and the step measurements.Those calipers such as vernier caliper that can be used to measure the external size of an object have a pair of jaws. </u></em>
Vernier caliper is a type of caliper that is used to precisely measure outside dimensions such as length, width, or diameter.

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How do you say scuba dive in spanish

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An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

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3 years ago

Give an example of how the apparent motion of an object depends on the observers motion

QveST [7]

In a moving car the outside looks to be moving. however if viewed from the outside, the car appears to be moving. so motion is relative to the person observing.

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3 years ago

Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a

Nonamiya [84]

Answer:

$1486.5\frac{Btu}{s}$

Explanation:

The inlet specific volume of air is given by:

$v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \ \ \ \ \ \...i$

The mass flow rates is expressed as:

$\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}$

The energy balance for the system can the be expresses in the rate form as:

$E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}$

Hence, the mass flow rate of the air is 1486.5Btu/s

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3 years ago

The change in internal energy during one complete cycle of a heat engine A. equals the net heat flow into the engine. B. equals

Stels [109]

Answer:

B. equals zero

Explanation:

Given data

one complete cycle = heat flow

solution

we have given that when heat engine complete 1 cycle change in energy = net heat flow

that is always equal to zero

from first law of thermodynamics that

ΔU = Q + W

we know ΔU is the change internal energy in system and Q is net heat transfer in system and W is net work done in system

therefore change of internal energy during one cycle

ΔU = Ufinal - Uinitial

ΔU = Uinitial - Uinitial = 0

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3 years ago