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3 years ago

Which is a characteristic of the image formed between F and the center of the lens?

Physics

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer: The image is upright

Explanation:

The lens shown in the diagram is concave lens. It is a diverging lens which means that the light rays which fall from the object onto the lens diverge.

Here, the object is kept between focus and center of lens. The image would form between focus and center of the lens at the same side of the object. The image would virtual, diminished and upright. Thus, the correct option is: The image is upright.

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As the frequency of an electromagnetic wave increases, its _____ also increases.

pochemuha

I believe it is velocity. But not to sure

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3 years ago

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Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7

mafiozo [28]

Answer:

1.5 × 10³⁶ light-years

Explanation:

A certain square region in interstellar space has an area of approximately 2.4 × 10⁷² (light-years)². The area of a square can be calculated using the following expression.

A = l²

where,

A is the area of the square

l is the side of the square

l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

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3 years ago

Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you

dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = $\frac{Kq1q2}{r^{2} }$

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

then

q2 = Q-q

Product of Charges = q1 x q2

Now, it is:

Product of Charges = q x (Q-q)

So,

Product of Charges = qQ - $q^{2}$

And the expression qQ - $q^{2}$ is clearly a quadratic expression. And clearly its roots are 0 and Q.

So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.

So, the midpoint is:

q = $\frac{Q + 0}{2}$

q = Q/2 and it is the highest value of each charge in order to get the greatest force.

8 0

4 years ago

A 1.40-kg block is on a frictionless, 25 ∘ inclined plane. The block is attached to a spring (k = 30.0 N/m ) that is fixed to a

rosijanka [135]

Answer: The small mass will drop below 11.6cm before coming to rest

Explanation:

Equating the Tension of both the two masses, and inputting the given parameters adequately...

Kindly check ATTACHED PICTURE for complete solution.

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3 years ago

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Can someone help meeeeee... show how to solve it plzzzzzzzz

liubo4ka [24]

<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have a gravitation force $F_{1}=16units$ , given by the formula written at the beginning. Let’s rename the distance $r$ as $d$ :

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$ (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$ :

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$ (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations: