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2 years ago

Most pedestrian fatalities occur

Engineering

1 answer:

Makovka662 [10]2 years ago

6 0

Answer: Outside an intersections

Explanation:

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Elan Coil [88]

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4 0

2 years ago

Read 2 more answers

The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The s

REY [17]

Answer:

0.34232

Explanation:

See attachment

3 0

2 years ago

Read 2 more answers

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

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$

Latent heat for ice H=336 KJ/kg

Specific heat for ice $C_p=4.187\ KJ/kg.K$

We know that sensible heat given as

$Q=mC_p\Delta T$

Heat for -15F to 32 F:

$Q=mC_p\Delta T$

$Q=9\times 2.03(32+15) KJ$

Q=858.69 KJ

Heat for 32 Fto 200 F:

$Q=mC_p\Delta T$

$Q=9\times 4.187(200-32) KJ$

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0

3 years ago

The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. Ifh=3ft, determin

KATRIN_1 [288]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0

3 years ago