6. A light ray strikes a reflective plane surface at an angle of 560 with the surface.

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

2 years ago

Dans car has broken down. He decides to push it to a garage 800m away qlong a flat road. He pushes the car with qn average force

Ede4ka [16]

Answer:

Answer = 480000 joule.

mark me the Brainliest.. pls and follow me

3 0

2 years ago

Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a

geniusboy [140]

Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

a sin θ = m λ

sin θ = m λ/ a

the first zero of the diffraction occurs for m = 1

sin θ = λ / a

angles are generally very small and are measured in radians

sin θ = θ = y / x

we substitute

$\frac{y}{x} = \frac{\lambda}{a}$

the width of the central maximum is twice the distance to zero

w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

W = 2x

2y = 2x

we substitute

1 = λ/ a

λ = a

we see that the width of the slit is equal to the wavelength used.

3 0

3 years ago

A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co

Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W$

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = $-F_Dd$

Where:

d = Distance traveled = 450 m

Therefore,

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd$ and

$v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh -F_Dd}{ \frac{1}{2}m} = v_1^2 + 2gh -\frac{ 2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}$

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0

2 years ago

If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he disting

iVinArrow [24]

Answer:

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

8 0

2 years ago