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11Alexandr11 [23.1K]
3 years ago
9

Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _______ and the plate _______.

Physics
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

Projected area= Diameter of the bolt* thickness.

Explanation:

Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _Diameter of the bolt______ and the plate ___thickness____.

Projected area= Diameter of the bolt* thickness.

Projected area is a 2-dimensional area measurement of a 3-dimensional body by projecting its surface on an arbitrary plane

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A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
A child pushes a 100 kg refrigerator with a force of 50 N, but the refrigerator does not move. Suppose the coefficient of static
faust18 [17]

Answer:

50 N

Explanation:

Since the refrigerator doesn’t move, that means the force of friction equals the amount of force the child exerts on the fridge. If the friction force were greater than the force by the child, the fridge would start accelerating towards the child. If it were less than the force the child exerted, the fridge would start accelerating away from the child. Therefore, the net force must be 0, in this case, the friction force is equal to the force the child exerted, for it to stay at rest (as Newton’s First Law stated).

I hope this helps! :)

8 0
3 years ago
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