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Margarita [4]
3 years ago
10

Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

ow long will it take for him to hear his own voice echo off of the wall?
Physics
1 answer:
Klio2033 [76]3 years ago
7 0

Answer:

2.62seconds

Explanation:

Echo occurs when sound reflects and causing a delay before getting to the listener.

Let x be the distance from between the source and the reflector

t be the time taken for the echo to occur

v be the velocity of sound in air.

Using the relationship 2x = vt

t = 2x/v

Given x = 450metres

v = 343m/s

t = 2(450)/343

t = 900/343

t = 2.62seconds

It will take 2.62 seconds for him to hear his own voice echo off of the wall.

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Sam stands on a 20 m high cliff and throws a 45 g rock with an initial velocity of 5 m/s [forward] to the water below. Use the c
storchak [24]

Answer:

v = 12.52 [m/s]

Explanation:

To solve this problem we must use the energy conservation theorem. Which tells us that potential energy is transformed into kinetic energy or vice versa. This is more clearly as the potential energy decreases the kinetic energy increases.

Ep = Ek

where:

Ep = potential energy [J] (units of joules]

Ek = kinetic energy [J]

Ep = m*g*h

where:

m = mass of the rock = 45 [g] = 0.045 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = (20 - 12) = 8 [m]

Ek = 0.5*m*v²

where:

v = velocity [m/s]

The reference level of potential energy is taken as the ground level, at this level the potential energy is zero, i.e. all potential energy has been transformed into kinetic energy. In such a way that when the Rock has fallen 12 [m] it is located 8 [m] from the ground level.

m*g*h = 0.5*m*v²

v² = (g*h)/0.5

v = √(9.81*8)/0.5

v = 12.52 [m/s]

5 0
3 years ago
A 735 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two
antiseptic1488 [7]

Answer:

Fg=1.02*10^-4 N

Explanation:

Fg=Gm1m2/r^2

8 0
3 years ago
A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?
AnnyKZ [126]

To answer this question, it helps enormously if you know
the formula for momentum:

           Momentum = (mass) x (speed) .

Looking at the formula, you can see that momentum is directly
proportional to speed.  So if speed doubles, so does momentum.

If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.

6 0
3 years ago
Read 2 more answers
Please help me with both of them
8_murik_8 [283]

Answer:

101011010101010100101000101010100010011010100010100000101041204105210241012021012021021012022221222122122345788981633333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333311111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111114565656565+4+652147

21212121512

546213171549895465621324547998995656565656565656565722426579898541321447985331321

Explanation:

4 0
4 years ago
When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

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

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
4 years ago
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