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3 years ago

A heavy object is dropped from a vertical height of 8.0 m. What is its speed when it hits the ground?

1 answer:

5 0

What do you know:
d=-8.0mm (starts higher and ends lower so it is negative)
Vi=0m/s (because it was dropped, not thrown)
A=-9.8m/s^2 (gravity)
Vf=???
Use the equation:
Vf^2=Vi^2+2A(d)
Vf^2=2(-9.8)(-8.0)
Vf^2= 156.8 -->find the square root of this
Final answer: Vf= 12.5m/s

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Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v) / (displacement² + distance²) .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity = velocity ( v) / (displacement² + distance²)

angular velocity = 2.2 / (0.22² + 0.51²)

angular velocity = 2.2 / 0.3085

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3 years ago

What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed

RoseWind [281]

<h3><u>Minimum uncertainty in the vertical component of the momentum of each photon:</u></h3>

According to Heisenberg's Uncertainty principle, both the “position and velocity of the particle” cannot be measured exactly at the same time. The momentum of the particle equals the product of its mass and velocity. And it can be inferred that the “product of the uncertainties” in the “momentum and the position” of a particle equals $\frac{h}{4 \pi}$ .

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3 years ago

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xxMikexx [17]

To develop the problem, we require the values concerning the conservation of momentum, specifically as given for collisions.

By definition the conservation of momentum tells us that, $m_1V_1+m_2V_2 = (m1+m2)V_f$

To find the speed at which the arrow impacts the apple we turn to the equation of time, in which,

$t= \sqrt{\frac{2h}{g}}$

The linear velocity of an object is given by

$V=\frac{X}{t}$

Replacing the equation of time we have to,

$V_f = \frac{X}{t}\\V_f =\frac{X}{\sqrt{\frac{2h}{g}}}\\V_f = \frac{6.9}{\sqrt{\frac{2(1.85)}{9.8}}}\\V_f = 11.23m/s$

Velocity two is neglected since there is no velocity of said target before the collision, thus,

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Clearing for m_2

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