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stellarik [79]
3 years ago
15

A car drives 400 meters in 5 seconds. What is the cars speed in m/s?

Physics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

80/1

Explanation:

to figure out the answer of how many miles/meters per second divide the meters/miles by the second

so 400÷5 would be 80

80 meters per second

hope this helps :))

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A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t
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Answer:

The value of spring constant is 266.01 \frac{N}{m}

Explanation:

Given:

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Height difference of pellet rise h_{f} - h_{o} = 6.03 m

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From energy conservation law,

Spring potential energy is stored into potential energy,

  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

  k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }

  k = 266.01 \frac{N}{m}

Therefore, the value of spring constant is 266.01 \frac{N}{m}

6 0
3 years ago
Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan
vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

S = k_{H} \cdot p (1)

<em>where S: is the solubility or concentration of Ar in water, k_{H}: is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!          

8 0
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According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the  electrostatic force will become weak by one fourth of the original force.

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Energy can not be created or destroyed but can change from one form to another.

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