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crimeas [40]
3 years ago
10

1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?

Physics
1 answer:
Korvikt [17]3 years ago
3 0

For this case we have that by definition, the momentum equation is given by:

p = m * v

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

m = 0.15 \ kg\\v = 120 \frac {m} {s}

Substituting:

p = 0.15 * 120\\p = 18 \frac {kg * m} {s}

On the other hand, if we clear the variable "mass" we have:

m = \frac {p} {v}

According to the data we have:

p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg

Thus, the mass is 50 \ kg

Answer:

p = 18 \frac {kg * m} {s}\\m = 50 \ kg

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Which of the following is considered a calm area close to the equator resulting in little to no wind?
solong [7]
<h2>Right answer: Doldrums</h2>

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In this area periods of great calm occur when the winds virtually disappear completely, trapping the sailing ships for long periods (days or weeks). This is why the term <em>doldrum</em> became popular as a colloquial expression in the eighteenth century, to refer to "<em>the caprice of the wind that slows down the navigation to sail". </em>


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5 0
4 years ago
3 Below, someone is trying to balance a plank with
belka [17]

Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

8 0
3 years ago
If a person whirls an object horizontally in a counter clockwise direction and the string breaks at point P as shown, what path
Nata [24]

when an object is revolving in circular path then its velocity is always along the tangent of the circular path

so while moving in circular path if the string is break then due to law of inertia the object will always move in the direction of initial motion

As we know that as per law of inertia if an object will not change its state of motion or state of rest until some external force will act on it.

So here also the object will move along its tangential direction once the string will break

so here the correct path will be

Option B

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Explanation:

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