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Serhud [2]
4 years ago
10

If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the two

increases?
Physics
1 answer:
Mrac [35]4 years ago
5 0
The Car Will Cause Some Kind Of Wind Wave And Hit The Person's Body But Its Not Harmful And You Will Hear The Loud Noise Passing By

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Stars within the halo move in random orbits that extend above and below the disk of the galaxy, while stars in the disk move in
Blizzard [7]
<span>Answer: Spherical Distribution Feedback: Correct The stars in the halo component have highly-inclined random orbits that orbit the center of our Galaxy. The stars within the halo would therefore make up a spherical distribution of stars surrounding the center of the Galaxy. In comparison, the disk stars move in elliptical orbits, which are nearly circular and are confined to the disk of the Galaxy. Disk stars therefore have very small inclinations and do not move above or below the plane of the Galactic disk.</span>
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3 years ago
Help ASAP ITS DUE IN 30 MIN
Katena32 [7]

can't read it, need larger picture

5 0
4 years ago
A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra
gavmur [86]
I believe it would be 144 m!
7 0
3 years ago
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Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0
4 years ago
The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch
AnnZ [28]

Answer:

Explanation:

Given an LC circuit

Frequency of oscillation

f = 299 kHz = 299,000 Hz

AT t = 0 , the plate A has maximum positive charge

A. At t > 0, the plate again positive charge, the required time is

t =

t = 1 / f

t = 1 / 299,000

t = 0.00000334448 seconds

t = 3.34 × 10^-6 seconds

t = 3.34 μs

it will be maximum after integral cycle t' = 3.34•n μs

Where n = 1,2,3,4....

B. After every odd multiples of n, other plate will be maximum positive charge, at time equals

t" = ½(2n—1)•t

t'' = ½(2n—1) 3.34 μs

t" = (2n —1) 1.67 μs

where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

t''' = ½(2n—1) 1.67μs

t"' = (2n —1) 0.836 μs

where n = 1,2,3...

6 0
3 years ago
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