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Gala2k [10]
3 years ago
7

A particle of mass m = 4.5 kg has velocity of v = 7 i m/s, when it is at the origin (0,0). Determine the z-component of the angu

lar momentum of the particle about each of the following reference points. The coordinates of the reference points have units of (meters, meters). Assume the positive z-axis is directed out of the screen.
Physics
1 answer:
Dima020 [189]3 years ago
4 0

Answer:

Explanation:

Momentum P = Mass x Velocity

M = 4.5kg

V = 7m/s

4.5kg x 7m/s

= 31.5xkgm/s

L=IW( Angular momentum) at stationary origin (0,0)

I = 1/2 x Mr^2

L = 1/2 x 4.5x 31.5

L = 70.8kgm/s

At stationary point, (0,0) No coordinate exist

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What were Rosalind Franklin, James Watson, and Francis Crick’s discovery, and what did it lead to? In three to five sentences, e
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8 0
2 years ago
_________ is the angle of incidence is equal to the angle of reflection
ExtremeBDS [4]
Law of reflection is the angle of incidence = angle of reflection.
8 0
3 years ago
Examples of forces in motion that you can see?
lidiya [134]

Answer:

tensional force

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6 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
3 years ago
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