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3 years ago

How does the map scale help to interpret the map?

2 answers:

6 0

Map scale refers to the relationship (or ratio) between distance on a map and the corresponding distance on the ground. For example, on a 1:100000 scale map, 1cm on the map equals 1km on the ground.

blondinia [14]3 years ago

6 0

Answer:

Explanation:

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Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air

statuscvo [17]

Answer:

The kinetic energy of A is twice the kinetic energy of B

Explanation:

5 0

3 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago

- You have a bin wrench turning a 1/2 13 UNC bolt. You overcome 1200 lbs of resistance when you

andrew11 [14]

is my guess but yeah

6 0

3 years ago

A single-cylinder pump feeds a boiler through a delivery

Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 $m^3/hr$

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius $r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m$

We know that area is given by

$A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2$

We know that discharge is given by $Q=AV$ , here A is area and V is velocity

So $Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec$

So net discharge in 1 hour = $981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour$

8 0

3 years ago

A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.0

harina [27]

Answer:

Load carried by shaft=9.92 ft-lb

Explanation:

Given: Power P=4.4 HP

P=3281.08 W

<u><em>Power: </em></u>Rate of change of work with respect to time is called power.

We know that P= $Torque\times speed$

$\omega=\frac{2\pi N}{60}$ rad/sec

So that P= $\dfrac{2\pi NT}{60}$

So 3281.08= $\dfrac{2\pi \times 2329\times T}{60}$

T=13.45 N-m (1 N-m=0.737 ft-lb)

So T=9.92 ft-lb.

Load carried by shaft=9.92 ft-lb

3 0

2 years ago