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4 years ago

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An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above

the launch point. Determine this maximum height.
A) 96m
B) 32m
C) 48 m
D) 75m
E) 64 m

1 answer:

AlekseyPX4 years ago

4 0

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

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Burka [1]

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction μ = 0.40

coefficient of kinetic friction = 0.20

a). when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

friction force =.Static friction coefficient × weight of the box

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A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s

expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

$\vec v_f=\vec v_o+\vec a.t$

$\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}$

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is $2.7 m/s^2$ to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

$v_f=v_o+a.t$

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Solving the first one for t

$\displaystyle t=\frac{v_f-v_o}{a}$

We have

$v_o=-6.5,\ v_f=-1.5,\ a=2.7$

Using these values

$\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s$

We now compute x

$\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}$

$x=-7,41\ m$

The motorboat ends up 7.41 meters to the west of the initial position

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3 years ago