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Roman55 [17]
3 years ago
6

What output will come for the expression (4.00/(2.0+2.0)) answer​

Engineering
1 answer:
True [87]3 years ago
5 0
The answer for the question that you have asked on this website formally known as brainly is 1
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What resources did Margaret Hutchinson Rousseau use ?
blagie [28]

Answer:

Explanation:

During the Second World War, she oversaw the design of production plants for the strategically important materials of penicillin and synthetic rubber.Her development of deep-tank fermentation of penicillium mold enabled large-scale production of penicillin.She worked on the development of high-octane gasoline for aviation fuel.Her later work included improved distillation column design and plants for the production of ethylene glycol and glacial acetic acid.

3 0
3 years ago
Why is the newtons law of cooling and explain how to derive it/
SIZIF [17.4K]

Answer:

For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. qf = final temperature of object

Explanation:

hope this helps you sorry if it doesn’t help you

5 0
3 years ago
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth
Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

5 0
3 years ago
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

8 0
4 years ago
Al Quiz
lbvjy [14]
5000! Hope this helps! If you are confused feel free to ask again
4 0
3 years ago
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