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2 years ago

Damage reports should contain the information needed for what?

Engineering

1 answer:

SIZIF [17.4K]2 years ago

3 0

Answer:

<h2>it's damage kindly be careful</h2>

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A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang

AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

We know that angular speed given as

$\omega =\dfrac{d\theta}{dt}$

We know that for one revolution

θ=2π

Given that time t= 2 hr

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

$V=\sqrt{\dfrac{GM}{R}}$

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

$V=\sqrt{\dfrac{GM}{R}}$

$V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}$

V=7059.44 m/s

iii)

We know that centripetal fore given as

$F=\dfrac{mV^2}{R}$

Here given that m= 200 kg

R= 8000 km

so now by putting the values

$F=\dfrac{mV^2}{R}$

$F=\dfrac{200\times 7059.44^2}{8000\times 10^3}$

F=1245.8 N

3 0

2 years ago

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

guajiro [1.7K]

Answer:

Technician B only

Explanation:

hope this helps :)

5 0

1 year ago

A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret

andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

Explanation:

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = $\frac{8000 X 10^-^3}{0.7}$

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = $\frac{0.002}{5}$

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

$E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi$

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0

2 years ago

What is differences Between hard shoulder & soft shoulder in civil Engineerin?

r-ruslan [8.4K]

Answer:

The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'

Mark as brilliant........... 

5 0

3 years ago

Which of the following is a common use for commas?

andreyandreev [35.5K]

Answer:

connecting two independent clauses

4 0

2 years ago

Read 2 more answers