Which of the following types of light can humans see with their eyes?

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a

In-s [12.5K]

Answer:

$T = 1.108\,s$

Explanation:

The period of a physical pendulum is:

$T = \sqrt{\frac{I_{O}}{m\cdot g \cdot L} }$

$T=2\cdot \pi \sqrt{\frac{\frac{1}{3}\cdot m \cdot L^{2} }{m\cdot g\cdot L} }$

$T=2\cdot \pi \sqrt{\frac{L }{3\cdot g} }$

The length of the leg is approximately the height of the person:

$L = 0.915\,m$

The period is:

$T = 2\cdot \pi \sqrt{\frac{0.915\,m}{3\cdot (9.807\,\frac{m}{s^{2}} )} }$

$T = 1.108\,s$

4 0

3 years ago

1. My grass is dying, and I believe it's because it is not getting enough water. Sol

trasher [3.6K]

Answer: I actually need the same answer

Explanation:

5 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

Using the graph, predict how many paper clips a 7.5 V battery would pick up for both the 25-coil electromagnet and the 50-coil e

Svet_ta [14]

NOTE: The strength of an electromagnet is directly proportional to not only the current but the number of windings. Doubling the number of windings doubles the strength of the magnet.

5 0

2 years ago

A rocket ship is accelerating at 200 m/s2, its mas is 135,000,000 kg. What is the force generated by this acceleration?

Rina8888 [55]

Acceleration does NOT "generate" force. Acceleration NEEDS force to make it happen. Without force ... provided by something else ... acceleration can't happen.

The force NEEDED to accelerate a mass with a certain acceleration is

Force needed = (mass) times (acceleration)

For the rocket ship in the question,

Force = (135,000,000 kg) times (200 m/s²)

Force = (135,000,000 x 200) kg-m/s²

<em>Force = 27 Giga-Newtons </em>(27,000,000,000 Newtons)

The gas-generator cycle F-1 rocket engine, developed in the US by Rocketdyne in the late 1950s, was used in the Saturn V rocket, the main launch vehicle of NASA's Apollo moon lander program . Five F-1 engines were used in the first stage of each Saturn V.

==> The thrust of each F-1 engine at full throttle is 7,770 kilo-Newtons.

It would take <em>3,475 </em>of these F-1 rocket engines, running full-throttle, to provide the force calculated in the answer to this question. If you didn't have 3,475 F-1 rocket engines, then you couldn't accelerate 135,000,000 kg at 200 m/s².

(And by the way ... the mass of each F-1 engine is 8,400 kg. So 3,475 engines alone account for 22% of the mass you're trying to accelerate. And don't even get me started about the mass of the FUEL you'd need to carry.)

5 0

3 years ago