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3 years ago

How do wavelength and wave period relate to a wave's speed

Physics

1 answer:

BigorU [14]3 years ago

4 0

The quotient of (wavelength) divided by (wave period) is equal to (wave speed).

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A thin rod rotates at a constant angular speed. Consider the tangential speed of each point on the rod for the case when the axi

ASHA 777 [7]

Answer:

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

Explanation:

They indicate that the angular velocity is constant, that is

w = dθ / dt

Where θ is the radius swept angle and t the time taken.

The tangential velocity is linear or

v = dx / dt

Where x is the distance traveled in time (t)

In the definition of radians

θ = s / R

Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal

θ = x / R

We substitute in the speed equation

v = d (θ R) / dt

The radius is a constant for each point

v = R dθ / dt

v = R w

With this expression we see that for each point at different radius the tangential velocity is different

6 0

2 years ago

A satellite is always being pulled by gravity.<br> a. True<br> b. False

tangare [24]

This is true. Gravity is constantly pulling on anything and everything (even light!), no matter how far away it is from another object.

3 0

2 years ago

A proton is a subatomic particle that carries a ___________ charge

Dovator [93]

The answer would be a positive charge

4 0

3 years ago

The distance from home plate to the pitchers mound is 18.5 meters. For a pitcher capable of throwing at 3.85m/s(86mi/hr), how mu

Taya2010 [7]

Given that:

Distance , s = 18.5 m

Velocity , v = 3.85 m/s

Time , t =?

Since,

Velocity = distance/time

Time= distance/velocity

time= 18.5/ 3.85

time= 4.8 s

So the time elapse between the release of the ball and the ball passing home plate is 4.8 seconds.

7 0

3 years ago

sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he

podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0

3 years ago