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2 years ago

While on an ice-fishing trip, you find yourself holding a 10 kg fish but stuck in the middle of

Physics

1 answer:

Anna007 [38]2 years ago

3 0

The forward force you exert on the fish and your backward action will allow you to reach the shore.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

Fa = -Fb

Let's assume the fish is held in the hook, this will give you the opportunity to throw the fish forward while still holding it.

When the the fish is thrown forward, you will move backwards with an equal force based on Newton's third law. Your backward momentum towards the shore will help to maintain equal linear momentum between you and the fish.

Thus, this forward force of the fish and your backward action will allow you to reach the shore.

Learn more Newton's third law of motion here: brainly.com/question/25998091

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When did ernest rutherford make his discovery

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Answer:

1911

Explanation:

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Which is a primary energy source used by power plants to generate electricity?

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a lamp rated 100w 220 v is connected to mains electric supply . what current is drawn from the supplyy line, if the voltage is 2

melomori [17]

Power = (voltage) x (current)

100w = 220v x current

Current = 100w / 220v = <u>0.455 Ampere</u> (rounded)

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3 years ago

In a physics lab a 90 kg student runs up a 1.5 meter tall flight of stairs in 3.0 seconds. The students power rating is approxim

wariber [46]

The power generated by the students will be .
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3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

Marysya12 [62]

Answer:

Mininmum acceleration required = $2.74 m/s^{2}$

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is $\frac{5}{18}$ ,

Therefore,

Lift off speed required = $125 \times\frac{5}{18}$ m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

$v^{2}-u^{2}=2as$ ,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

$(34.72222)^{2}-0 = 2\times a \times 220$

a = $\frac{(34.72222)^{2} }{2\times220}$

a = $2.74 m/s^{2}$ .

Therefore,the minimum acceleration of the plane required = $2.74 m/s^{2}$ .

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3 years ago