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PIT_PIT [208]
3 years ago
12

The number of times that a wave's amplitude cycles from its starting point, through its highest amplitude and its lowest amplitu

de, and back to its starting point over a fixed period of time, is known as what term below?
Physics
1 answer:
Roman55 [17]3 years ago
5 0

Answer:

Frequency

Explanation:

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A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.
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Lowest frequency of wave
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

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3 years ago
Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon
dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, a=7.78\times 10^{11}\ m

Mass of the sun, M=1.99\times 10^{30}\ kg

(a) Let T is the period of Jupiter's orbit. It is given by :

T^2\propto a^3

T^2=\dfrac{4\pi^2}{GM}a^3

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3

T=3.74\times 10^8\ s

(b) We know that,

1\ year=3.154\times 10^7\ s

or

1\ s=3.171\times 10^{-8}\ year

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T = 11.859 earth years

Hence, this is the required solution.

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4 years ago
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kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho
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32 = 0 \times t +  \frac{1}{2} \times 2.5 \times t^{2}     \\  32 = 0 + 1.25 \times t {}^{2}  \\ 32 = 1.25t {}^{2}   \\   \frac{32}{1.25}  =  \frac{1.25t {}^{2} }{1.25}  \\ t {}^{2}  = 25.6 \\  \sqrt{t {}^{2} }  =  \sqrt{25.6}  \\ t = 5.1seconds \\

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