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Mazyrski [523]
3 years ago
15

Define amplitude and frequency as they relate to a wave?

Physics
1 answer:
Solnce55 [7]3 years ago
3 0
<span>frequency definition. In physics, the number of crests of a wave that move past a given point in a given unit of time. The most common unit offrequency is the hertz (Hz), corresponding to one crest per second. Thefrequency of a wave can be calculated by dividing the speed of the wave by the wavelength.</span><span>amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.</span>
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A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1
lidiya [134]

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as  

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and  

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people  

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum  

2000 ×  1 = (2000 + 3000 ) ω

solve we get now  

ω = 0.4 rad/s  

5 0
3 years ago
Which of the following examples best describe using an inclined plane
Alinara [238K]
The best possible Answer is c
8 0
3 years ago
What is the amount of the force accelerate a 20kg object at a rate of 5m/s
Anni [7]

Answer:

100 N is the answer of the question

6 0
2 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0
3 years ago
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