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2 years ago

(b) The output current of the generator is 2.0A. The output potential difference (p.d.) of the

Physics

1 answer:

Luden [163]2 years ago

8 0

Answer:

P = 22 watts

Explanation:

Given that,

The output power of the generator = 2 a

The output potential difference = 11 V

We need to find the output power of the generator. The formula for the output power is given by :

$P=V\times I\\\\P=11\times 2\\\\P=22\ W$

So, the output power of the generator is equal to 22 Watts.

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Perform the following mathematical operation and report answer to the correct number of significant figures 143.6 divided by 21.

Svetllana [295]

The answer to the correct number of significant figures is 6.774.

<h3>What is quotient?</h3>

When a number(big) divided smaller number, the answer obtained greater than zero is called a quotient.

Divide 143.6 ÷ 21.2

143.6/21.2 = 1436/212

=6.77358

The quotient is rounded to three significant figures after decimal

143.6 ÷ 21.2 = 6.774

Thus, the answer to the correct number of significant figures is 6.774

Learn more about quotient

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A space shuttle travels around the Earth at a constant speed of 28000 kilometers per hour. If it takes 90 minutes to complete on

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3. Which of the following magnetic scenarios will repel each

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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

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1 year ago