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Ne4ueva [31]
3 years ago
13

What is the correct water depth for an echo travel time of 6 seconds? (for the purposes of this exercise, assume pressure and te

mperature effects on sound wave velocity are negligible.)?
Physics
1 answer:
saul85 [17]3 years ago
7 0
In physics, there are empirical values for common important parameters. For example, the speed of sound is equal to 340 meters per second. Unlike the speed of light, the speed of sound is dependent on temperature and pressure. But in standard room conditions, the speed is 340 m/s. Using this value of speed, we can find the depth given the time. You should also note that the distance the echo travels is exactly the same distance that object travelled. Therefore,

Distance = 340 m/s * 6 seconds
Distance = 2,040 meters deep
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What are some of the challenges of scuba diving?
EastWind [94]
Some of the challenges are the unpredictable fish and the risk of scratching againest coral or drowning for not focusing on your oxygen tank.
7 0
3 years ago
Objects falling through air experience a type Of friction called ____
DaniilM [7]

Answer: That's air resistance.

Explanation: Well, air resistance is an upward force exerted on falling objects.

( I hope this helped <3 )

6 0
3 years ago
Read 2 more answers
A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0
3 years ago
Helpp me pleassee....
lilavasa [31]

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz

Now fundamental wavelength is :

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m

So, the fundamental wavelength of the vibrating string is 1.7 m.

4 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
3 years ago
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