During destructive interference, two waves moving through the same medium will

determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the

Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0

3 years ago

An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol

iogann1982 [59]

Answer:

(a) B = 2.85 × $10^{-6}$ Tesla

(b) I = I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = $\frac{mv}{qB}$

⇒ B = $\frac{mv}{qr}$

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = $\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }$

= $\frac{9.11*10^{-27} }{3.2*10^{-21} }$

B = 2.85 × $10^{-6}$ Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒ I = B ÷ μN/L

where B is the magnetic field, μ is the permeability of free space = 4.0 × $10^{-7}$ Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

= $\frac{2.85*10^{-6} }{4*10^{-7} *25}$

I = 0.285 A

5 0

3 years ago

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is

Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_{f}$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_{f}$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_{f}$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

5 0

3 years ago

A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the

Sonja [21]

2.72 N

Explanation:

Step 1:

From the basic formula in electrostatics

F = E * q

where F = Force due to charges

E = Electric field strength

q = Charge

Step 2:

From the given question

q= $8.5*10^{-6} C$

E = $3.2 * 10^{5} N/C$

F = $8.5 * 10^{-6} * 3.2 * 10^{5} = 2.72$ N

8 0

4 years ago

if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?

Katena32 [7]

A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

Learn more about harmonics here brainly.com/question/17315536

#SPJ10

7 0

2 years ago