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3 years ago

5

Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun

dary?

1 answer:

fiasKO [112]3 years ago

6 0

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

$\alpha+\rho+\tau=1$

where:

$\alpha=$ absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

$\rho=$ reflectivity is defined as the ratio of reflected radiation to the total irradiation

$\tau=$ transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

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What is another name for the breathing mechanism?

aleksandrvk [35]

Aspirate or inhale or respire or

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4 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$

where:

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the hopper

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0$ is the initial height

u is the launch speed of the hopper

$h_f = 1.20 m$ is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

$u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s$

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4 0

3 years ago

12 points , just need help with this question

tigry1 [53]

Answer:f 30

Explanation: I am not really sure but try this

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3 years ago

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

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3 years ago

Free tingsssssssssss

il63 [147K]

Explanation:

yayyy thx u soo muchihuhuhu

5 0

2 years ago

Read 2 more answers