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3 years ago

Where are a hurricanes fastest winds and heaviest rain found?

Physics

1 answer:

PSYCHO15rus [73]3 years ago

8 0

The fastest winds and heaviest rain is found in the eye of the hurricane.

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When do we land on Mars ?

lorasvet [3.4K]

Answer:

I think like 2024 or 25

Explanation:

Elon musk will probably go

5 0

3 years ago

Read 2 more answers

Which statement about acids is true? A. Weak acids ionize completely. B. Acids turn red litmus paper blue. C. Acids conduct elec

Vinil7 [7]

Id say d because it releases hydrogen and on the other hand a base receives it
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6 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago

What type of energy does friction produce?

VMariaS [17]

Answer:

The correct option is A = adiant negy.

6 0

3 years ago

May

GalinKa [24]

Answer:

a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s

Explanation:

all these exercises are about kinematics

a) The body is released from rest,

y = y₀ + v₀ t - ½ g t²

in this case when reaching the ground y = 0 and its initial velocity is vo = 0

0 = y₀ + 0 - ½ g t²

t² = 2 y₀ / g

t² = 2 4 /9.81

t² = 0.815

t = √0.815

t = 0.90 s

b) It is thrown upwards at v₀ = 4 m / s

y = y₀ + v₀ t - ½ g t²

in this case the initial and final height is the same

y = y₀ = 0

0 = v₀ t -1/2 g t²

t = 2 v₀ / g

t = 2 4 /9.81

t = 0.815 s

c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + 0 - ½ g t²

t² = 2 i / g

t² = 2 4 / 9.81

t² = 0.815

t = 0.90 s

d) the horizontal distance traveled is

x = v₀ₓ t

x = 4 0.90

x = 3.6 m

e) We can calculate the time to fall from I = 2 m

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + 0 - ½ g t²

t² = 2 y₀i / g

t² = 2 2 /9.81

t² = 0.4077

t = 0.639 s

Therefore, when making measurements, you should find readings around this value.

8 0

3 years ago