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mezya [45]
2 years ago
5

A 0.137 kg mass on a string

Physics
1 answer:
I am Lyosha [343]2 years ago
7 0

brainly.com/question/26500837

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Lactic acid is produced during low-moderate exercise <br> True <br> False
Marizza181 [45]
The answer to your question is true.
3 0
3 years ago
Read 2 more answers
How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its
LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0
3 years ago
Please help me
melamori03 [73]
<h2>Question:</h2>

An automobile is driving uphill. Which form of energy is not involved in this process?

<h2>Choosing:</h2>

electromagnetic

potential

kinetic

chemical

<h2>Answer:</h2>

<u>Electromagnetic</u><u> </u>

<h3><u>#READINGHELPSWITHLEARNING</u><u> </u></h3><h3><u>#CARRYONLEARNING</u><u> </u></h3><h3><u>#STUDYWELL</u><u> </u></h3>
7 0
3 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0
3 years ago
Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster
astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0
3 years ago
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