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2 years ago

Under which condition does a light ray striking a concave mirror reflect off parallel to the principle axis ?

Physics

2 answers:

asambeis [7]2 years ago

5 0

The answer would be c if u need to know

OleMash [197]2 years ago

4 0

Answer:

When the incident ray passes through the focal point

Explanation:

A concave mirror has an inward curved surface that reflects the light making it converge at a focal point. It is characterized by the fact that, unlike convex mirrors, concave mirrors show images of different shapes depending on the distance between the object and the mirror.

When the object is at the same focal point, the reflected rays are parallel and never cross, so no image is formed, or in other words, it is reflected parallel to the main axis.

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What is precision?what is precision?

RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0

1 year ago

Read 2 more answers

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

2 years ago

Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?

Roman55 [17]

Answer:

less than stating velocity due to friction and air resistance.

Explanation:

3 0

2 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

$F =326.7 \ N$

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

2 years ago

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has

AlladinOne [14]

Given Information:

Angular displacement = θ = 51 cm = 0.51 m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

3 0

2 years ago