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3 years ago

9

I need help with one question on my homework. This is on the Specific Heat Capacity required practical.

1 answer:

Leviafan [203]3 years ago

5 0

Answer:

will mostly accord at the top of the boiling water my kind sir

Explanation:

Evaporation takes place only at the surface of a liquid, whereas boiling may occur throughout the liquid. In boiling, the change of state takes place at any point in the liquid where bubbles form. The bubbles then rise and break at the surface of the liquid.

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Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

kolbaska11 [484]

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 $E = 200000 MPa$

The strain of the bar when subjected to 150 MPa is

$\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075$

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

$\epsilon = \frac{\Delta L}{L}$

$L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm$ or 1.8m

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3 years ago

Which explains the downward force produced when air flows over the winglike spoiler on a race car?

professor190 [17]

Newton's principle
newton's 2nd law of motion where F=ma

3 0

3 years ago

base your answers on the following four graphs, which represent the relationship between speed and time for four different objec

Agata [3.3K]

I would say B. Hope this helps Plz give a thx

6 0

3 years ago

A dog is 10 m from a cat, whose speeds are 6 and 5 m / s, respectively. What time does the dog require to catch the cat?

scoray [572]

Answer:

10 seconds

Explanation:

because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover

4 0

4 years ago

Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×

ikadub [295]

Happy Holidays!

We can use the following equation to solve for the gravitational force:

$\large\boxed{F_g = G\frac{m_1m_2}{r^2}}$

Fg = force due to gravity (N)

G = Gravitational constant

m1,m2 = masses of the objects (kg)

r = distance between the objects (m)

Plug in the given values into the equation:

$F_g = (6.67*10^{-11})\frac{(70)(60)}{(2)^2}} = \boxed{7.0 * 10^{-8}N}$

7 0

3 years ago