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3 years ago

How can density be determined in a lab of a rectangular solid?

1 answer:

6 0

We will first record its mass and then its volume by measuring its dimensions
then divide mass by volume and will get density of regular solid

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Use the drop-down menu to complete the statement. Based on the data in the line graph, you can predict there will likely be visi

Dvinal [7]

Answer:what are the options?

Explanation:

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3 years ago

Read 2 more answers

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by

$y_2 = v_0t - \frac{1}{2}gt^2$ (2)

At the instant the two balls collide, they will have the same displacement, therefore

$y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2$

$v_0t = 10\:\text{m}$

Solving for t, we get

$t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}$

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

$y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2$

$\:\:\:\:\:\:\:= 5.1\:\text{m}$

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3 years ago

If two charged balloons are 24cm apart and they feel a force of electrical repulsion of 20N, what would the force of electrical

OverLord2011 [107]

Answer:

Soory

Explanation:

I really dont know but i will send you wait

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3 years ago

What do you think that astronomers mean when they use the term observable universe?

iren2701 [21]

The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us.

Hope this helps :)

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3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago