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3 years ago

For a concave mirror, when the object is at focus the image is formed at

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

When the object is placed at the focus the image is formed at infinity.

Explanation:

When a ray passes through focus and incident on a concave mirror then it will travel parallel to principal axis after reflection.Hence the image is formed at infinity.

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Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as

Lesechka [4]

We are given:

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

Momentum of Tarzan before swinging:

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

__________________________________________________________

Speed of Tarzan after grabbing the bananas:

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v [where v is the velocity of Tarzan]

v = 5.9 m/s

__________________________________________________________

<h3>Finding the Initial and Final KE and PE:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

Initial and Final KE:

We know that KE = 1/2*(mv²)

Initial KE:

Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

Final KE:

Final KE = 1/2*(mv²)

[where v is the velocity at the maximum point of the swing]

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

Initial and Final PE:

We know that:

PE = mgh

[where g is the acceleration due to gravity and h is the height]

Initial PE:

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

Final PE:

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3>Finding the maximum height Tarzan will reach:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

h = 1653.5/950 [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey

6 0

3 years ago

As demonstrated by the Doppler Effect, why does sound increase in pitch as a sound source approaches you?

scZoUnD [109]

I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.

3 0

3 years ago

Read 2 more answers

So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.

bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at² Subtract ViT from both sides

D - ViT = 0.5at² Divide both sides by 0.5t²

$\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }$ I wrote this step out a little more to show how your fraction will cancel

$\frac{D - ViT}{0.5t^{2} }$ = a I like to flip these around so the single variable is on the right

a = $\frac{D - ViT}{0.5t^{2} }$

7 0

3 years ago

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

$X=V_{o}*t + \frac{a*t^{2}}{2}$ Replacing the known values:

$1500=6t+0.9*t^{2}$ Solving for t we get 2 possible answers:

t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0

3 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

3 years ago