Silicon chips are used primarily in ?

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:

i)

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

So

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

A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit

Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

$W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf$

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0

3 years ago

You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall

choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are $\rightarrow 120ft5in+2(10ft)$

The two long are $\rightarrow 122ft1in=122.08ft$

The two shors are $\rightarrow 86ft4.5in = 86.375ft$

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

$Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3$

That in cubic yards is equal to $180.15 (1cy=27ft^3)$

Hence, we need order 5% plus that represent with the quantity

$Q_{ordered}=1.05*180.15=189.15cy$

8 0

2 years ago

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = $\frac{A}{A-C}$

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = $\frac{A}{A-C}$

= $\frac{1034}{1034-675 \cdot 6}$

Apparent specific gravity = 2.88

b) Bulk specific gravity $G_{B}^{O D}=\frac{A}{B-C}$

$G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}$

= 2.76

c) Bulk specific gravity (SSD):

$G_{B}^{S S D}=\frac{B}{B-C}$

$=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}$

$G_{B}^{S S D}$ = 2.80

d) Absorption% :

$=\frac{B-A}{A} \times 100 \%$

$=\frac{1048 \cdot 9-1034}{1034} \times 100$

Absorption = 1.44 %

e) Bulk Volume :

$v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}$

$=\frac{1048 \cdot 9-675 \cdot 6}{1}$

= $373.3 cc$

5 0

2 years ago

What type of engineer works to create a practical and safe energy source?

Fittoniya [83]

Why did you put this on here when you know the answer lol

4 0

3 years ago