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Elena L [17]
3 years ago
12

A car accelerates uniformly in a straight line from rest at the rate of 3.7 m/s2

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0

Explanation:

<u>Using equations of motion</u> :

(A) we know that,

u = 0

a = 3.7 m/s^2

S = 63

{v}^{2}  =  {u}^{2}  + 2as

v =  \sqrt{0 + 2 \times 3.7 \times 63}  = 21.6 \: m{s}^{ - 1}

(B)

v = u + at

21.6 = 0 + 3.7 * t

<u>t (time to reach 63 m) = 5.83 seconds</u>

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Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en
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c. 2 m/s

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4 years ago
A force of 10 N causes a spring to extend by 20 mm. Find a) the spring constant of the spring in N/m b) the extension of the spr
suter [353]

(a) The spring constant is 500 N/m.

(b) The extension of the spring when 25 N force is applied is 0.05 m.

(c) The applied force to cause an extension of 5 mm is 2.5 N.

The given parameters:

  • Applied force, F = 10 N
  • Extension of the spring, x = 20 mm

The spring constant is calculated as follows;

F = kx\\\\k = \frac{F}{x} \\\\k = \frac{10}{20 \times 10^{-3}} \\\\k = 500 \ N/m

The extension of the spring when 25 N force is applied is calculated as follows;

F = kx\\\\x = \frac{F}{k} \\\\x = \frac{25}{500} \\\\x = 0.05 \ m

The applied force to cause an extension of 5 mm is calculated as follows;

F = kx\\\\F = 500 \times 5 \times 10^{-3}\\\\F = 2.5 \ N

Learn more about Hook's law here: brainly.com/question/12253978

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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!          

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