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3 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]3 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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Determine the mechanical energy of this object; a 1-kj ball rolls on the ground at 2m/s

Daniel [21]

If you still need the answer it is B. 2 J

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3 years ago

How t calculate gravitational force

lions [1.4K]

Answer:

F=G(m1m2)/Rsquare if radius is given

F=G(m1m2)/dsquare if distance is given

where,

f =gravitational force

G =gravitational constant

m1=mass of one object

m2=mass of another object

d=distance between two object from their center r=radius of earth/planet

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3 years ago

According to Dr. paul Narguizian professor of Biology and Science Education at California State University, ______ are generaliz

Mazyrski [523]

Answer:

I believe the correct answer would be A :)

Explanation:

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2 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago

A tightrope walker more easily balances on a tightwire if his pole

cestrela7 [59]

B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~

4 0

3 years ago