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3 years ago

A pencil is dropped from the roof of the school what is the speed after 3s down is the negative direction.

2 answers:

8 0

On Earth, the acceleration due to gravity is (-9.8 meters per second) per second.

If the school, the roof, and the pencil are on Earth,
then after 3 seconds, the pencil's speed is

[ (-9.8 meters per second)/second ] times [ 3seconds ] = <u>-29.4 meters per second </u>

uysha [10]3 years ago

4 0

You have to use 1st equation of motion v = u +at as distance(s) is not given.....
as pencil is dropped from rest therefore u = 0m/s and accleration due to gravity is = -9.8 m/s
v = 0 - (9.8)(3)
v = -29.4 m/s

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3 years ago

Read 2 more answers

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$

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