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3 years ago

The current through a 10 ohm resistor is 1.2 amperes.What is the potential difference across the resistor?

Physics

1 answer:

Natalija [7]3 years ago

6 0

Answer: The potential difference across the resistor is 12 volts.

Explanation:

To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.

Mathematically,

$V\propto I\\V=IR$

Where,

V = potential difference = ?V

I = Current flowing = 1.2 A

R = Resistor = $10\Omega$

Putting values in above equation, we get:

$V=1.2\times 10=12V$

Hence, the potential difference across the resistor is 12 volts

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Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d

ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

$f(x) = 8x^3-3x^2-5x^8$

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

$f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8$

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

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4 years ago

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Answer:

Explanation:

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3 years ago

94 points if you hurry The phase of matter with the least kinetic energy:

OlgaM077 [116]

Answer:

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3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

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4 years ago

What does the revolution of the moon do to it?

Degger [83]

Answer:

The Moon revolves or moves around the Earth in a path called its orbit and rotates, or spins, in space.

Explanation: The Moon's movements cause the phases of the Moon and the Earth's ocean tides.

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3 years ago