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2 years ago

Why do mirrors form inverted images?

2 answers:

7 0

Answer: Reflection from a Concave Mirror
When the object is far from the mirror, the image is inverted and at the focal point. The image is real light rays actually focus at the image location). ... As the object moves towards the mirror inside the focal point the image becomes virtual and upright behind the mirror.

marysya [2.9K]2 years ago

3 0

Answer:

Explanation:

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted).

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The inner planets or call terrestrial because they

AnnyKZ [126]

They are similar to the earth. Not relative to the outer planets

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3 years ago

Which career would be the best for someone who wants to work in entertainment but does not want to be in front of any type of au

GalinKa [24]

directing, your out of the spotlight

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3 years ago

Read 2 more answers

4 NaF + Br2 + Al(H3Li)2 -->

Agata [3.3K]

Answer:

Molecules are substances that contain two or more elements, if these elements are different, then the substance is also a compound.

However, considering only the definition of a molecule, it is evident that both CH4 and NaF are molecules as they contain more than one element.

Explanation:

hope this helps!!

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3 years ago

A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in

timama [110]

Answer:

$KE2/KE1=0.71$

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

$V2 = \frac{m1}{m1+m2}*V1$

The kinetic energies are:

$KE1=1/2*m1*V1^2$

$KE2 = 1/2*(m1+m2)*V2^2$

$KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2$

Simplifying:

$KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2$

The ratio will be:

$KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}$

$KE2/KE1=0.71$

5 0

3 years ago

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

slega [8]

Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

Explanation:

Given that,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

The magnitude of the is

$|\vec B|=\sqrt {55^2+ (-12)^2}$

$\Rightarrow |\vec B|=\sqrt {3169}$

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

6 0

2 years ago