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3 years ago

Question 1 (1 point)

Physics

1 answer:

IrinaK [193]3 years ago

8 0

momentum
Yes, if the elephant is standing still.
Fullback
impulse acting on it.
2.25 N∙s
A cannon firing.
Inelastic
it stays the same
When the cue ball contacts the other balls, momentum is transferred causing them to gain momentum and speed.
less than 3 m/s

<h3><u><em>these are all correct i got an 100%</em></u><em><u> </u></em></h3>

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If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low

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At the highest point: kinetic energy is 0 due to the speed is 0

So the total mechanical energy is 20

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So at the lowest point, kinetic energy = mechanical energy - potential energy

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3 years ago

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from

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Answer:

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

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3 years ago

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

$a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}$

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

$v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}$

Finally, since $\sin60\°=\frac{\sqrt{3} }{2}$ and $\sin45\°=\frac{\sqrt{2} }{2}$ , we can conclude that:

$t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}$

In words, the cannonball b spends more time in the air than cannonball a.

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3 years ago