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2 years ago

When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance?

Physics

1 answer:

grandymaker [24]2 years ago

3 0

Answer:

Explanation:

Power is expressed as P = V²/R

R = resistance

V = supplied voltage

Given P = 600W and V = 30V

R = V²/P

R = 30²/600

R = 900/600

R = 1.5ohms

magnitude of its resistance is 1.5ohms

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Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward

KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal =30 degree

Density $\rho \ of\ water= 1000 kg/m3$

Drag force FD is given as

$F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A$

$= 0.5\times 1000\times 4.32\times 1.2\times 1.3$

Drag force = 14422.2 N acting opposite to the motion

As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force

TCos30 = F_D

$T = \frac{F_D}{cos30}$

$T =\frac{ 14422.2}{cos 30}$

T = 16653.32 N

7 0

2 years ago

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0

2 years ago

The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the

Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

$m = IS$

Where,

I = Current

S = Area

$m = IA \rightarrow m= I( \pi r^2)$

Replacing with our values we have that,

$8*10^{22} = I \pi(\frac{3000*10^3}{2})^2$

Re-arrange to find I,

$I = 1.1317*10^{10} A$

Part B) To find the Current density we need to find the cross sectional area of the Wire:

$A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2$

Finally the current density is simply J

$J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2$

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

$A = \pi (10-3)^2 \\A = 49\pi$

Therefore the current density would be

$J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2$

Comparing the two values we can see that the 2mm wire has a higher current density.

4 0

2 years ago

What’s the equation that links with total input, efficiency, energy and useful output energy transfer

weqwewe [10]

Efficiency. The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. Efficiencies can be written as decimals like 0.33 or percentages 33%. To convert a efficiency expressed as a decimal to a percentage you need to multiply by 100.

8 0

2 years ago

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

natima [27]

Answer:

The attractive force between them is $1.296 \times 10^{18}$ N