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3 years ago

A virtual image seems to come from

Physics

2 answers:

Alex_Xolod [135]3 years ago

4 0

behind the mirror

Contrary to real images, virtual images cannot be illustrated on a screen. The optical image will appear at the point of divergence ( of outgoing rays from a point). It is an optical image that seems to be projected from behind the mirror.

adell [148]3 years ago

3 0

The answer is A. behind the mirror

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What will the north poles of two bar magnets do when brought together?

Reika [66]

Answer:

They will repel each other since they have the same charge specifically negative

Explanation:

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3 years ago

Read 2 more answers

A spring has a constant k= 10.32 N/m and is hung vertically. A 0.400kg mass is suspended from the spring. What is the displaceme

alukav5142 [94]

Answer:

displacement (x) = 0.003798 meters

Explanation:

from the fact that the string is hung vertically we can deduce that:

Total force acting on the mass = Fs (by spring) + Fg (by gravity)

where

Fs = k*x , x is the displacement..

Fg = m*g

then:

Ftot = m*a, but a = 0 m/(s^2) because the mass becames stationary.

Ftot = 0

Fs + Fg = 0

by direction, take down as negative.

Fs - Fg = 0

k*x = m*g

x = m*g/k = [(0.400)(9.8)]/(10.32)

= 0.3798 meters

4 0

3 years ago

A solenoid having an inductance of 6.95 μh is connected in series with a 1.24 kω resistor. (a) if a 12.0 v battery is connected

Serggg [28]

In electrical circuit, this arrangement is called a R-L series circuit. It is a circuit containing elements of an inductor (L) and a resistor (R). Inductance is expressed in units of Henry while resistance is expressed in units of ohms. The relationship between these values is called the impedance, denoted as Z. Its equation is

Z = √(R^2 + L^2)
Z = √((1.24×10^3 ohms)^2 + (6.95×10^-6 H)^2)
Z = 1,240 ohms

The unit for impedance is also ohms. Since the circuit is in series, the voltage across the inductor and the resistor are additive which is equal to 12 V. Knowing the impedance and the voltage, we can determine the maximum current.
I = V/Z=12/1,240 = 9.68 mA
But since we only want to reach 73.6% of its value, I = 9.68*0.736 = 7.12 mA. Then, the equation for R-L circuits is
$I= \frac{V( 1- e^{-t/τ} )}{R}$ , where τ = L/R = 6.95×10^-6/1.24×10^3 = 5.6 x 10^-9
Then,
$7.12x 10^{-3} = \frac{12( 1- e^{-t/5.6x 10^{-9} } )}{1240}$

t = 7.45 nanoseconds
Part B.) If t = 1.00τ, then t/τ = 1. Therefore,
$I= \frac{12( 1- e^{-1 } )}{1240}$

I = 6.12 mA

3 0

3 years ago

Kepler's laws, satellites motion and weightlessness

Anon25 [30]

Answer:

First Kepler law states that Each planet describes an ellipsoidal motion about the sun as its single focus.

Second Kepler law states that An imaginary line joining a planet to the Sun sweeps out equal areas in equal time intervals.

Third Kepler law states that The squares of period of revolution of the planet around the sun are proportional to the cubes of mean distance between the planet and the sun.

Weightlessness is the condition where the body has zero gravity ( its acceleration is equal to the acceleration due to gravity )

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3 years ago

A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the

LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

f = Focal length of the ornament
u = image distance
v = object distance.

make u the subject of the equation

u = fv/(f+v)................ Equation 2

From the question,

Given:

f = 8/2 = 4 cm
v = 12 cm

Substitute these values into equation 2

u = (12×4)/(12+4)
u = 48/16
u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

M = u/v.................. Equation 3

Where

M = magnification of the ornament.

Substitute these values above into equation 3

M = 3/12
M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0

3 years ago