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3 years ago

A real image is four times as far from a lens as is theobject.

Physics

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

1.25 focal lengths

Explanation:

The lens equation states that:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the object distance

q is the image distance

In this problem, the image is 4 times as far from the lens as is the object: this means that

$q=4p$

If we substitute this into the lens equation and we rearrange it, we get

$\frac{1}{f}=\frac{1}{p}+\frac{1}{4p}=\frac{4+1}{4p}=\frac{5}{4p}\\p=\frac{5}{4}f=1.25 f$

so, the object distance measured in focal lengths is

1.25 focal lenghts

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Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m

Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0

2 years ago

Read 2 more answers

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

Harman [31]

The magnitude and direction of the net force is 9 newtons to the right ⇒ answer A

Explanation:

The net force is the vector sum of all the forces that act upon an object

If two forces are horizontal where,

$F_{1}$ > $F_{2}$
They are in opposite direction
Net force F = $F_{1}$ - $F_{2}$ in the direction of $F_{1}$

An object is acted upon by a force of 22 newtons to the right

and a force of 13 newtons to the left

We need to find the magnitude and direction of the net force

→ $F_{1}$ = 22 newtons to the right

→ $F_{2}$ = 13 newtons to the left

$F_{1}$ > $F_{2}$ and they are in opposite direction

Substitute these values in the rule of the net force above

→ Net force F = 22 - 13 = 9

→ The direction of the net force is the same direction of the larger

force, the direction of the larger force is to the right

→ Net force is 9 newtons to the right

The magnitude and direction of the net force is 9 newtons to the right

Learn more:

You can learn more about the net force in brainly.com/question/4033012

#LearnwithBrainly

8 0

3 years ago

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

5 0

3 years ago

The gradual reduction of the amplitude

Lelu [443]

The phase and amplitude of rhythms in physiology and behavior are generated by circadian oscillators and entrained to the 24 h day by exposure to the light and dark cycle and feedback from the sleep wake cycle.the extent to which the phase and amplitude of multiple rhythms are similarly affected during altered timing of light exposure and the sleep wake cycle has not been fully characterized.

4 0

3 years ago

A 43 kg box is being pushed and pulled by

m_a_m_a [10]

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

$155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]$

The negative sign means that the box accelerates in a negative direction to the left.

4 0

3 years ago