Many arlines restrict maximum mass of a case to 23 kg why

Physics

1 answer:

Alexxandr [17]2 years ago

7 0

Answer:

weight

Explanation:

the weight of an object on an airline is one of the most important thing a pilot has to consider when prepping a flight and that is because if there is too much weight then the plane simply can't fly. imagine if everyone wanted to bring a 50 kg box. there are at least 200 people. that alone is 10,000 lg of weight than you have to factor in all the people, wires on the plane, and certain appliances that some planes have.

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Two male moose charge at each other with the same speed and meet on a icy patch of tundra. as they collide, their antlers lock t

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Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.

m₁u + m₂u = (m₁ + m₂) x 1/3 u
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3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2

The ratio of their inertias is 2

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Explain how sociology will contribute to the understanding of daily life and what is happening within it?

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Answer:

Explain how sociology will contribute to the understanding of daily life and what is happening within it?

Sociology is the study of society of human being, how they develop and those things put in place for such development

Explanation:

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2 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .