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2 years ago

Define Impulse And MomentumPlsss Answer Now I Need Now

Physics

1 answer:

andrew-mc [135]2 years ago

8 0

Answer:

Impulse-The effect of force acting over time to change the momentum of an object.

momentum-The product of mass of a particle and its velocity.

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A sinusoidal sound wave moves through a medium and is described by the displacement wave function

ZanzabumX [31]

By applying the wave equation we know that the maximum speed of the element's oscillatory motion is 1716 micrometer / s.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

k = 2π / λ

ω = 2πf

where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

v = dy / dt

v = d(2.00cos (15.7x - 858t)) / dt

v = -858 x (-2.00sin(15.7x - 858t))

v = 1716 sin(15.7x - 858t) micrometer/s

maximum velocity can be reached when (sinθ = 1), hence

v = 1716 sin(15.7x - 858t)

v = 1716 x 1

v = 1716 micrometer / s

For more on wave equation on: brainly.com/question/25699025

#SPJ

4 0

2 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago

A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g

Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0

3 years ago

Read 2 more answers

If the Sun were the size of a small exercise ball (about 0.5 meters (m) in

suter [353]

Answer:

the size of a pea

Explanation:

4 0

3 years ago

Can someone plz help me DUE TOMORROW

aksik [14]

Hey the answer to the question is
m = 0.40

3 0

3 years ago

Define Impulse And MomentumPlsss Answer Now I Need Now​

Define Impulse And MomentumPlsss Answer Now I Need Now