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Anvisha [2.4K]
3 years ago
13

What is the volume of this bubble when it reaches the surface?

Physics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.

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Which descriptor applies to the circuit?<br><br> A- on<br> B- closed<br> C- incomplete <br> D- open
HACTEHA [7]
Open circuit is the answer
3 0
3 years ago
Read 2 more answers
What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil
lora16 [44]

Answer:

\theta=145

Explanation:

The amplitude of he combined wave is:

B=2Acos(\theta/2)\\

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145

5 0
3 years ago
the density of aluminum is 2700 kg/m3. if transverse waves travel at in an aluminum wire of diameter what is the tension on the
Sunny_sXe [5.5K]

The tension on the wire is 52.02 N.

From the question, we have

Density of aluminum = 2700 kg/m3

Area,

A = πd²/4

A = π x (4.6 x 10⁻³)²/4

A = 1.66 x 10⁻⁵ m²

μ = Mass per unit length of the wire

μ = ρA

μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²

μ = 0.045 kg/m

Tension on the wire = √T/μ

34 = √T/0.045

34² = T/0.045

T = 52.02 N

The tension on the wire is 52.02 N.

Complete question:

The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.

To learn more about tension visit:  brainly.com/question/14336853

#SPJ4

5 0
1 year ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
One of the key disadvantages of buying fitness equipment is __________. A. privacy B. flexibility C. maintenance D. choice Pleas
Alex Ar [27]

Answer:

<h2>C. maintenance </h2>

Explanation:

I personally believe one key disadvantages is the cost of maintaining the equipment unlike the gym where you have to subscribe for the month or the year and forget about anything, owning the gym equipment comes with the extra cost and responsibilities of maintenance to the owners.

5 0
3 years ago
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