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3 years ago

What is the acceleration at the apex of a vertical up and down problem?

Physics

1 answer:

Natalija [7]3 years ago

7 0

Answer:

The correct option is +9.81 m/s²

Explanation:

The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².

At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).

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3 years ago

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3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

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4 years ago

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha

zaharov [31]

Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²

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3 years ago