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2 years ago

13

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you.

2 answers:

Ivan2 years ago

5 0

<span>Right-hand rule reveals a connection between the current and the magnetic field lines in the magnetic field. It is also called Ampere`s right hand screw rule. It determines the direction of the magnetic field surrounding a current carrying wire. The thunb points direction of current and fingers point the direction of magnetic lines of force. Answer: C ) a counterclockwise rotation.</span>

Nostrana [21]2 years ago

4 0

B. A clockwise direction.

Good luck!

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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

2 years ago

Hi, Solve for λ<br> E=hc/λ

Paul [167]

Answer:

λ=hc/E

Explanation:

E=hc/λ

Eλ=hc

λ=hc/E

4 0

2 years ago

A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto

lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift we have

$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}$

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

$\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}$

Solving for $D_{2}$ we get

$D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm$

5 0

2 years ago

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Sidana [21]

Answer:

acceleration of the rocket is given as

$a = 7.45 m/s^2$

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

$v_i = 0$

$v_f = 447 m/s$

$t = 1 min = 60 s$

so we have

$a = \frac{v_f - v_i}{t}$

$a = \frac{447 - 0}{60}$

$a = 7.45 m/s^2$

7 0

2 years ago

Elena L [17]

The answer yr looking for would b true!

6 0

2 years ago

Read 2 more answers