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laiz [17]
2 years ago
10

Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b

e warmer, and why?
Physics
1 answer:
Ainat [17]2 years ago
4 0

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

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a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
Effectus [21]
Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer:  11.025 m
7 0
2 years ago
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If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus
anastassius [24]

Answer:

E=477.92\ W.m^{-2}

Explanation:

Given that:

Absolute temperature of the body, T=273+30=303\ K

  • emissivity of the body, \epsilon=1

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

E=\epsilon. \sigma.T^4

where:

\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}   (Stefan Boltzmann constant)

Now putting the respective values:

E=1\times 5.67\times 10^{-8}\times 303^4

E=477.92\ W.m^{-2}

5 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
What conclusion can be derived by comparing the central tendencies of the two data sets?
zhannawk [14.2K]
The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
3 0
2 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
4 0
2 years ago
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