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3 years ago

What real world measurable property of waves is the A value in the equation relating frequency and wavelength?

Physics

1 answer:

navik [9.2K]3 years ago

3 0

Answer:

Hello There!!

Explanation:

The frequency of a wave is the number of times per second that the wave cycles. Frequency is measured in Hertz or cycles per second.The period and frequency are closely related to each other. The period equals 1 over the frequency and the frequency is equal to one over the period.The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

hope it helps,have a great day!!

~Pinky~

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A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

$R = u² sin2θ/g$

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

$u² = \frac{Rg}{sin2\theta}$

$u^2 = \frac{9.1 x 9.80}{sin26}$

$u^2 = 113.17$

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

$v^2 = (v_osin\theta)^2 -2gh$

$h =\frac{(v_o^2sin\theta)^2}{2g}$

$h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}$

h = 1.10 m

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3 years ago

Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul

Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

$v_1=\sqrt{2gh}$

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

$J=m(v_2-v_1)$

$J=2m\sqrt{2gh}$

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

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