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Ghella [55]
3 years ago
7

Noise from highways is a large issue, particularly for those who live close to the highway. Engineers have come up with several

solutions to the problem. One solution can be a high vertical wall. Second, planting vegetation, like trees. Third, using a large buffer zone or open space.
Explain how trees, a high wall or buffer zones can reduce the noise from a highway that is heard at a nearby home.
Think…
· How would the noise be for someone in a home?
· How would the noise be for someone driving on the highway?
· How does the DOPPLER EFFECT fit in to this scenario?
You must write a response with a MINIMUM of 5 sentences to receive full credit.
WILL MARK AS BRAINILEST
Physics
1 answer:
kifflom [539]3 years ago
8 0
Noise from highways are a significant issue. particularly for those who live close to the highway
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Select the correct answer. An airplane is flying at a constant speed in a positive direction. It slows down when it approaches t
Triss [41]

An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going to land. this is an example of negative acceleration (D).

7 0
3 years ago
Consider this situation: A baseball player dives head-first
siniylev [52]
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.

I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
7 0
3 years ago
Is the matching correct?
iragen [17]

Yes thats right :)

have a great day!!!


7 0
3 years ago
A man-made satellite orbits the earth in a circular orbit that has a radius of 32000 km. The mass of the Earth is 5.97e 24 kg. W
Gwar [14]

Answer:

= 3521m/s

The tangential speed is approximately 3500 m/s.

Explanation:

F = m * v² ÷ r

Fg = (G * M * m) ÷ r²

(m v²) / r = (G * M * m) / r²

v² = (G * M) / r

v = √( G * M ÷ r)

G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴

r = 32000km = 32 * 10⁶ meters  

G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶

v = √1.24 * 10⁷  

v = 3521.36m/s

The tangential speed is approximately 3500 m/s.

8 0
4 years ago
Read 2 more answers
A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o
Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

s=ut+\frac{1}{2}at^2

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

v_x =\frac{d}{t}

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

v_x = \frac{286}{5.05}=56.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0
3 years ago
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