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3 years ago

When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored

Physics

1 answer:

adoni [48]3 years ago

4 0

The energy stored in a capacitor is

E = (1/2) · (capacitance) · (voltage)²

E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

<em>E = 4.32 x 10⁻⁴ Joule</em>

(That's 0.000432 of a Joule)

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What is the average power output of an athlete who can life 9.0 * 10^2 kg 2.5 m in 2.0 s?

eduard

Answer:

Average power output of athlete = 11025 Watts

Explanation:

Work done is defined as the product of force applied and the displacement perpendicular to the force.

Work = $Force\times Displacement$

Power is defined as work done per unit time.

Power = $\frac{Work done}{Time}$

Here the person lifts 900 kg.

Height = 2.5 m

Time interval = 2 seconds

Force = $weight \times gravity$

= $900 \times 9.8$

= 8820 N

Work done = $Force\times Displacement$

= $8820 \times 2.5$

= 22050 J

Power = $\frac {22050}{2}$

= 11025 Watt

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3 years ago

What is the part of the cell that stores food?

Alenkasestr [34]

Part of the cell that stores food is called vacuole

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3 years ago

Before you conduct your experiment, you need to form a hypothesis. A hypothesis is a prediction of what you think will happen in

Stolb23 [73]

Example: A apple rotting.

If I put my apple in a fridge, then it would not rot as fast because it is in a cooled area. (example)

Hope it helps! Brainiest Answer would be amazing!

5 0

3 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is:

$7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}$

$p_{B,2} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is:

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}$

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0

3 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

8 0

3 years ago