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3 years ago

Hi

Physics

2 answers:

vaieri [72.5K]3 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]3 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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3 years ago

Read 2 more answers

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

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