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3 years ago

Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3

Physics

1 answer:

kobusy [5.1K]3 years ago

8 0

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

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mamaluj [8]

Answer:

i think the answer is B but im not sure

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3 years ago

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Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy

Setler79 [48]

Answer:

The correct option is;

Raymond: I think the skateboarder has the same total energy at all points on the ramp

Explanation:

The total energy, also known as the total mechanical energy, is the sum of the kinetic and potential energies of the skateboarder

Given that the potential energy is the energy gained due to elevation, the maximum potential energy is obtained at the top of the ramp, while the maximum kinetic energy, which is the energy due to motion, is at the bottom of the ramp where the skateboarder moves fastest.

However, by the energy conservation principle, the kinetic energy of he skateboarder comes from the conversion of the potential energy, such that the total energy is the same at any particular point on the ramp.

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2 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

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3 years ago

The table represent the thickness, top density, and bottom density of the different layers of the Earth. In most of the layers,

Nostrana [21]

Answer:inner core?

Explanation:

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3 years ago

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ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
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2 years ago

Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3​

Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3