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4 years ago

Hi

Physics

2 answers:

vaieri [72.5K]4 years ago

5 0

<h2>Explanation:</h2>

m = h₂/h₁

Here,

m = Magnification
h₁ = Height of the object
h₂ = Height of the image formed by the concave mirror

Magnification given is (-3). The linear magnification here is negative.

This implies that h₂ < h₁ because m < 1.

Therefore, the image formed is real and inverted. It is smaller than the object. So, the image is formed between the Centre of Curvature & the Focus.

[NOTE: If the image is highly reduced, it is formed at the focus.]

romanna [79]4 years ago

3 0

Answer:HERE IS YOUR ANSWER

THE POINTS ARE

IF WE PUT THE OBJECT BETWEEN THE FOCUS

AND THE POLE

THEN THE IMAGE FORMED WILL be MAGNIFIED

If that’s not what you are looking for, try this one:

For concave mirror the virtual image is formed when the object is kept in between the pole and the focus.

Given here the "size of the image" is twice to that of the object.

Hence, it is consider that the magnification is +2.

So, the magnification value is positive and the image formed will be "virtual and erect". Thus, the object should be kept in between the "pole and the focus" in concave mirror."

Explanation:

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4 years ago

Is the following statement true or false? Nuclear fission occurs naturally in stars .

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4 years ago

A spiral staircase in a building is in the shape of a helix of radius 5 meters. Between two oors of the building, the stairs mak

Vitek1552 [10]

Answer:

2744Nm

Explanation:

The gravitational field is conservative. The gravitational force only acts on a vertical direction. Hence, independently of the form of the trajectory, the work done by the gravity against the person demands the calculation of the integral of the force multiplied by the vertical displacement of the person:

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5 0

3 years ago

A racecar driver is driving her car down the drag strip at 120 m/s. What is the shortest distance in which she can brake and sto

DaniilM [7]

Answer:

1034.78 m

Explanation:

The shortest distance is the displacement of the car from initial position to final position.

Displacement of body is given using Newton's equations of motion.

Given:

Initial velocity, $u = 120$ m/s

Final velocity, $v = 0$ m/s( As the car stops in its final position)

Coefficient of static friction, $\mu=0.71$

Acceleration due to gravity, $g=9.8\textrm{ } m/s^{2}$

Now, when brakes are applied, only friction acts on the body in a direction opposite to that of its motion.

The acceleration of the car when friction acts the stopping force is given as:

$a=- \mu g$ .

The acceleration is negative as it reduces the velocity of the motion and acts in the direction opposite to that of the motion.

Plug in 0.71 for $\mu$ and 9.8 m/s² for $g$ . Solve for $a$ .

So, $a=-0.71\times 9.8=-6.958\textrm{ }m/s^{2}$ .

Now, displacement of the car is given using the following equation of motion:

$v^{2}=u^{2} +2aS$

Here, $S$ is the displacement of the racing car.

Plug in 120 m/s for $u$ , 0 m/s for $v$ , -6.958 m/s² for $a$ . Solve for $S$ . This gives,

$0^{2} =(120)^{2}+2(-6.958)S\\13.916S=14400\\S=\frac{14400}{13.916}=1034.78\textrm{ m}$

Therefore, the shortest distance in which she can brake and stop is 1034.78 m.

5 0

4 years ago

Read 2 more answers

A + 32.2 μC charge feels a 0.544 N force from a + 12.3p μC charge. How far apart are they? (u stands for micro.) [?] m

love history [14]

Answer:

r = 2.55 m

Explanation:

Given that,

First charge, q₁ = 32.2 μC

Second charge, q₂ = + 12.3 μC

The force between charges, F = 0.544 N

We need to find the distance between charges. The force between two charges is given by the formula as follows :

$F=k\dfrac{q_1q_2}{r^2}\\\\r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 32.2\times 10^{-6}\times 12.3\times 10^{-6}}{0.544}} \\\\r=2.55\ m$

So, the charges are 2.55 m apart.

5 0

3 years ago