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Charra [1.4K]
3 years ago
7

A Cannon makes a noise, and it's echo is heard 4.5 seconds later. if sound travels at 343 m/s:

Physics
1 answer:
kramer3 years ago
7 0

a).  (343 m/s) x (4.5 sec)  =  1,543.5 meters

b).  The surface it reflected from is  (1543.5 / 2) = 771.75 meters away.

     The sound had to go there AND come back before you heard the echo.
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Difference between the fertility rate and mortality rate of developed and least development countries
dybincka [34]

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5 0
3 years ago
A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?
Advocard [28]

Answer:

v=697.2km/h

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

V=\frac{d}{t}

The distance is clearly 1743 km and the time is:

t=2h+30min*\frac{1h}{60min} =2.5h

Thus, the velocity turns out:

v=\frac{1743km}{2.5h}\\ \\v=697.2km/h

Which is a typical velocity for a plane to allow it be stable when flying.

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5 0
3 years ago
Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside
jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0
1 year ago
At exactly 3:14PM, the velocity of a dog running in a park points toward a group of flowers. Which of the following best describ
dmitriy555 [2]

Answer:

Options A, B, and C are all possible.

Explanation:

We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?

If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.

If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.

If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers  nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.

4 0
3 years ago
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