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3 years ago

The _______ is the time required for one complete wave oscillation to occur.

Physics

2 answers:

valentinak56 [21]3 years ago

7 0

The period is the time required.....

aleksandrvk [35]3 years ago

6 0

Period is the answer

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A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A

Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

where

$d_i$ is the input distance

$d_o$ is the output distance

For the pulley system in this problem, $d_i = 3.90 m$ and $d_o = 0.975 m$ , so the IMA is

$IMA=\frac{3.90 m}{0.975 m}=4$

B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

$MA=\frac{F_o}{F_i}$

where $F_o$ is the output force and $F_i$ is the input force. For the pulley system in this problem, $F_i = 375 N$ and $F_o = 1345 N$ , so the MA is

$MA=\frac{1345 N}{375 N}=3.59$

C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

$\eta = \frac{MA}{AMA} \cdot 100$

Therefore, in this case,

$\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%$

3 0

3 years ago

PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)

Luden [163]

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

3 0

3 years ago

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.