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Kobotan [32]
3 years ago
7

20 POINTS!

Physics
1 answer:
CaHeK987 [17]3 years ago
6 0
Potential energy is the answer
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The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10
Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0
2 years ago
How do kinetic and potential energy transfer to one throughout a roller coaster ride?
mojhsa [17]

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

6 0
2 years ago
John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Sedbober [7]

Answer:

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

\omega =\sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

x=A\cos(\omega t)

Finally, the equation of the motion of the system is:

x=(0.088m)\cos(\omega t)

or

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

7 0
3 years ago
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