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3 years ago

5

can someone please please please help me find the spring force constant thank you thank you thank you thank you

1 answer:

Levart [38]3 years ago

3 0

Answer:

spring force frictonal

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The law of conservation of matter states that matter can be neither created nor destroyed. Your friend shows you the following c

MrRa [10]

Yes, our friend is right, because there is no contradiction to the law of conservation of mass in the above equation. It just the mass of the product is equal to the mass of reactants.. and that is shown in the equation you have presented earlier

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3 years ago

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jasenka [17]

Answer:

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3 years ago

PLEASEEE HELP, thank you :)

telo118 [61]

Answer:

The answer is B.

Explanation:

Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :

Total resistance = 20Ω + 40Ω + 60Ω

= 120Ω

Next, we have to find out that 1Ω is equal to how many voltage by dividing :

120Ω = 60V

1Ω = 60V ÷ 120

1Ω = 0.5V

Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :

1Ω = 0.5V

20Ω = 0.5V × 20

20Ω = 10V

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3 years ago

Can someone help me please

iogann1982 [59]

Examine the water molecules in the animation. After the wave has passed which of the following is true?

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3 years ago

Read 2 more answers

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

3 years ago