# A SCUBA diver produces a spherical air bubble with bulk modulus = 1.2 x 105 Pa and radius = 2.9 mm while exploring a shipwreck below the surface of the ocean. The air bubble rises to the surface of the ocean and expands to a radius = 3.8 mm. What is the difference in pressure between the surface and at the SCUBA diver's depth, ( )?

ΔP = 0.497 10⁵ Pa

Explanation:

All materials are elastic, so they deform for the pressures applied, in bulk module is the deformation spreads for a given pressure, it is defined by

B = - P /(ΔV/V)

Where the negative sign is for the module to be positive, P is the pressure and ΔV/V is the unit volume reduction

Let's apply that equation to our case, let's look at the volumes of the bubbles

In the water

r₁ = 2.9 mm = 2.9 10⁻³ m

V = 4/3 π R3

V₁ = 4/3 π (2.9 10⁻³)³

V₁ = 102.2 10⁻⁹ m³

On the surface

r₂ = 3.8mm = 3.8 10⁻³m

V₂ = 4/3 π (3.8 10⁻³)³

V₂ = 229.8 10⁻⁹ m³

Δv = V₁-V₂

ΔV = (102.2 - 229.8) 10⁻⁹ m³

ΔV = -127.6 10⁻⁹  m³

Let's calculate the pressure

B = - P / (ΔV / V)

P = - B ΔV / V

P = - 1.2 10⁵ (-127.6 10⁻⁹/102.2 10⁻⁹)

P = 1,498 10⁵ Pa

The pressure difference between this depth and the surface is the difference with the atmospheric pressure (1.01 105 Pa)

ΔP = P2 - Patm

ΔP = (1,498 -1.01) 10⁵

ΔP = 0.497 10⁵ Pa

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I would say A is the correct answer.

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If a scientist unknowingly breaks the law, he is guilty of _____.

Explanation:

If a scientist unknowingly breaks the law, he is guilty of _Noncompliance and such a scientist is said to be noncompliant scientist

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noncompliance

Explanation:

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Bill does 1500 j of work on a leaver, which is a simple machine. The lever does 1350 j of work on a stone. What is the efficiency of the leaver?

c) 90% is the efficiency of the leaver.

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Efficiency = (1350/1500)×100

= 90%

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A second class lever is the handiest lever which can promise that the attempt arm will usually be more extra than the load arm. This arrangement results in a larger attempt arm-to-load arm ratio, making the second magnificence lever the maximum robotically high-quality.

The question is incomplete. Please read below to find the missing content.

Bill does 1500 J of work on a lever, which is a simple machine. The lever does 1350 J of work on a stone.

What is the efficiency of the lever?

a)0.9%

b)1.1%

c)90%

d)111%

#SPJ2

C. 90%

The efficiency of the lever is 90%.

A block with a weight of 3.0 N is at rest on a horizontal surface.A 1.0 N upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface?

Explanation:

Given

weight of block

A force is of is applied on the block

As 1 N is less than weight of block so block exert  Force less than its weight

So Block exert a force of 2 N downward on floor

An airplane pilot fell 370 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot's mass was 80 kg and his terminal velocity was 50 m/s, estimate: (a) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; and (c) the work done on him by air resistance as he fell.

A) The work done by the snow in bringing him to rest is; 100000 J

B) The average force exerted on him by the snow to stop him is; F = 90909.09 N

C) Work done on him by the air resistance is;

W_s = 190362 J

We are given;

Mass of pilot; m = 80 kg

Terminal velocity; v = 50 m/s

A) Let us first estimate the change in kinetic energy which is workdone by the snow in bringing him to rest;

W = ∆K = Final kinetic energy - initial kinetic energy

∆K = ½mv² - ½mu²

u is initial velocity = 0 m/s

Thus;

∆K = ½(80 × 50²)

∆K = 100000 J

B) The average force exerted on him by the snow to bring him to stop is;

F = W/∆x

We are given ∆x = 1.1 m

Thus;

F = 100000/1.1

F = 90909.09 N

C) Final velocity without drag is;

v = √(2gx)

We are given x = 370 m. Thus;

v = √(2 × 9.8 × 370)

v_f = 85.2 m/s

Terminal velocity is velocity without drag. Thus, the work done on him by the air resistance as he fell is;

W_s = ½m((v_f)² - (v²))

W_s = ½ × 80 × (85.2² - 50²)

W_s = 190362 J

a) 100 kJ

b) 90.909 × 10³ N

c) 190.08 kJ

Explanation:

Given:

Initial velocity, v₁ = 50 m/s

Final velocity, v₂ = 0

Mass of the pilot, m = 80 kg

Distance covered in snow, d = 1.1 m

Now,

Work done (W) = change in kinetic energy

or

W =

in substituting the values, we get

W =

or

W = 100000 J = 100 kJ

b) Average force exerted by snow

Now, Force = (change in momentum/time taken)

or

F = ΔM/Δt

now,

ΔM = m(v₁ - v₂)

Δt = distance/ (average speed) = 1.1/[(50 + 0)/2] = 0.044 s

F = 80 × (50-0)/0.044

F = 90.909 × 10³ N

c) Work done by air resistance

W = change in energy by the air resistance

now, the change is energy is as:

from potential energy at the height, H = 370 m to the potential energy acquired just before touching the snow

thus,

W = mgH - (1/2)mv₁²

on substituting the values, we get

W = 80 × 9.8 × 370 - (1/2) × 80 × 50² = 190080 J = 190.08 kJ

What must be 'n' of a sphere surrounded by water so that the parallel rays that affect one of its faces converge on the second vertex of the sphere?

The refractive index of the sphere is 2.66

Solution:

The refractive index, and since the sphere is surrounded by water.

Therefore, according to the question, the parallel rays that affect one of the faces of the sphere converges on the second vortex:

Thus the image distance from the pole  of surface 1, v' = 2R

where

R = Radius of the sphere

Now, using the eqn:

Since, v is taken as infinite

n = 2.66

What is the effect on the current in a simple circuit when both the voltage and the resistance are divided by 2?