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AURORKA [14]
3 years ago
5

Newton’s first law of motion is often called the law of?

Physics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

Newton's first law is often called the law of inertia.

searched it up to verify

Also welcome to brainly

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Write a hypothesis about the effect of the wavelength on the diffraction angle. Use the "if . . . then . . . because . . ." form
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A) Wavelength is the distance between two successful crests or troughs in a transverse wave or the distance between two successful compressions or rarefactions in a longitudinal wave. if the wavelength of a wave is long then the wave will diffracts more compared to a wave with a shorter wavelength because the size of the wavelength is proportional to the angle of diffraction. Such that if the gap is larger than the wavelength then the wave passes through the gap and does not spread much on the other side, while when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread greatly out.

b) Diffraction is a phenomenon that occurs when a wave encounters an obstacles or a slit. It is referred to as the bending of light around corners of an obstacles or aperture into the region of geometrical shadow of the obstacle. It is caused by one wave of light is shifted by the diffracting object which will in turn cause the wave to have interference with itself. (either constructive or destructive)
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3 years ago
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Is a puddle of water an object in motion?
Brilliant_brown [7]
Yes, a puddle of water is an object in motion.
4 0
3 years ago
ball a 0.604 kg moving right at 11.6 m/s makes a head-on collision with ball B at rest. after, ball A moves right at 2.09 m/s, a
jolli1 [7]

Answer:

1.142

Explanation:

8 0
3 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
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