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3 years ago

What is the reading of this Dial Caliper?

Engineering

1 answer:

lesya [120]3 years ago

7 0

Answer:

Explanation:

bdgdsfggsfg

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5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

marta [7]

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

8 0

3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago

Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t

HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid F

$F=\rho AV^2$

So force exerted in x-direction

$F_x=\rho AV^2$

$F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So force exerted in y-direction

$F_y=\rho AV^2$

$F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2$

F=0.763 N

So the resultant force R

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{0.763^2+0.763^2}$

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0

3 years ago

A ________ can be installed in a cast-iron block to repair a worn or cracked cylinder. Question 24 options:

VLD [36.1K]

Answer:

Cylinder sleeve

Explanation:

7 0

3 years ago

A(n)___ branch circuit supplies two or more receptacles or outlets for lighting and appliances

Juli2301 [7.4K]

Answer: General purpose branch circuit

Explanation:

General purpose branch circuit are the type of circuits that are used mainly to supply light to two or more receptacle outlets for small appliances. This circuits are about 120v can be used either in residential, commercial and industrial buildings.

6 0

3 years ago