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3 years ago

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I need help, please help an object starts from rest at time t=0 and moves in the with constant acceleration. The object travels

3m from time t=1 to time t=2. what is the acceleration of the object?

1 answer:

vichka [17]3 years ago

4 0

<span>a=8(m/<span>sec2</span>).. i answered this earlier i think.</span>

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Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.

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3 years ago

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

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Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

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object distance u = 18cm

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Since f = R/2

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Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

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Read 2 more answers

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

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3 years ago