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Lubov Fominskaja [6]
3 years ago
8

Express 9.39 x 109 seconds in terms of days.

Physics
1 answer:
Natali5045456 [20]3 years ago
4 0
<span> In 1 day there are 24 hours x 60 min/hour x 60 seconds/min = 86 400 seconds

So divide 9.39 x 10^9 by 86 400 to determine the # of days there are

= (9.39 x 10^9) ÷ (8.64 x 10^4)
= (9.39 ÷ 8.64) x (10^9 ÷ 10^4)
= 1.0868055556 x 10^5
= 108 681 (nearest whole day) </span>
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Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

               h = vot + 1/2gt²

             71.3 = 0t + 1/2(9.81)t²

             2(71.3) = 9,81t²

              t² = 2(71.3)/9.81

              t² = 14.53

              t = 3.81 s

b)

      vf = 0 + (9.81)(3.81)

      vf = 37.4 m/s

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To obtain the same resistance force, a greater force must be exerted in a machine of lower efficiency than in a machine
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Answer: True.

Explanation:

A resistance force is also known as friction. And the efficiency of a machine is affected by friction.

A machine of lower efficiency has higher magnitude of friction than a machine of higher efficiency.

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If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c
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Answer:

\mu_s \geq 0.27

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

f_s=\frac{mv^{2}}{R}

But we know that:

f_s\leq \mu_s N

And the normal force is given by the sum of the forces in the vertical direction:

N-mg=0 \implies N=mg

Finally, we have:

f_s=\frac{mv^{2}}{R}  \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR}  \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27

So, the minimum value for the coefficient of friction is 0.27.

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