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1 year 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?

1 answer:

Ainat [17]1 year 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.

You might be interested in

The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms

aalyn [17]

The question to the above information is;

What is the best use of an atomic model to explain the charge of the particles in Thomson's beams?

Answer;

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation;

-Atoms are comprised of a nucleus consisting of protons (red) and neutrons (blue). The number of orbiting electrons is the same as the number of protons and is termed the "atomic number" of the element.

J.J. Thomson discovered the electron. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

5 0

2 years ago

Read 2 more answers

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is $\tau = 48t \r k$

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

A 2.0 g metal cube and a 4.0 g metal cube are 6.0 cm apart, measured between their centers, on a horizontal surface. For both, t

emmasim [6.3K]

Answer:

a) t=10.2s

b) The 2g-cube moves first

Explanation:

Since the electric force is the same on both cubes and so is the coefficient of static friction, the first one to move will be the one with less mass.

So, on the 2g-cube the sum of forces are:

$\left \{ {{Ff-Fe=0} \atop {N-m*g=0}} \right.$

Replacing the friction on the first equation:

$\mu*m*g-Fe=0$ Thus $Fe=\mu*m*g=12.74*10^{-3}N$

The electric force is:

$Fe = \frac{K*q^2}{d^2}$ Solving for q:

q=71.44nC

This amount divided by the rate at which they are being charged:

t = 71.44nC / 7nC/s = 10.2s

7 0

1 year ago

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

1 year ago

A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient

Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0

2 years ago