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2 years ago

14

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

1 answer:

k0ka [10]2 years ago

7 0

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

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How can you describe the motion of an object in a race?

Hitman42 [59]

You can describe the motion of an object by its position, speed, direction, and acceleration

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2 years ago

A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration

Luda [366]

Answer:

$a \approx \: 0.001 \: m {s}^{ - 2}$

Explanation:

Given:

$initial \: velocity \: (u) = 0 \\ \\ Final \: Velocity \: (v) = 72 km /h \\ \\ Time \: (t) = 5 \: hours \\ \\ Acceleration \: (a) =? \\ \\ \because \: a = \frac{v - u}{t} \\ \\ \therefore \: a = \frac{72 - 0}{5} \\ \\ a = \frac{72}{5} \\ \\ a = 14.5 \: km {h}^{ - 2} \\ \\ a = \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2} \\ \\ a = 0.00111882716 \\ \\ a \approx \: 0.001 \: m {s}^{ - 2}$

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2 years ago

An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

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2 years ago

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

natima [27]

The upward force exerted on the board by the support is 530.8 N.

<h3>Upward force exerted on the board by the support</h3>

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

Learn more about upward force here: brainly.com/question/6080367

#SPJ1

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1 year ago

Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y? In 2.00 y? In 2.

Alex17521 [72]

2.0y i think this is guess but if it right then thats good

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2 years ago