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valkas [14]
1 year ago
7

How will sunlight most likely affect a black shirt on a hot summer day? the temperature of the shirt will depend on how much sun

light is diffracted around the shirt. the temperature of the shirt will not change because the sunlight is transmitted through the shirt to the body. the temperature of the shirt will decrease because most of the light is reflected off of the shirt. the temperature of the shirt will increase because all wavelengths of light are absorbed by the shirt.
Physics
1 answer:
vladimir2022 [97]1 year ago
5 0

The correct answer is option (C) the temperature of the shirt will increase because all wavelengths of light are absorbed by the shirt.

The relationship of heat and light

  • Heat is a measure of the movement of particles in the body, the more particles move, the warmer the body becomes.
  • When the body absorbs light radiation, its particles vibrate in accordance with the electromagnetic radiation's wavelengths, which causes an increase in the temperature with the increase in particle movement.
  • The more wavelengths of radiation absorbed by an object, produces more heat.

Learn more about the Wavelength of light with the help of the given link:

brainly.com/question/13961990

#SPJ4

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if a bicyclist, with initial velocity of zero, steadily gain speed until reaching a final velocity of 26m/s, how far away did sh
Alja [10]

Answer:

<h2><em>34.46m</em></h2>

Explanation:

Using one of the equation of motion to solve the question. According to the equation v² =u²+2as where;

v is the final velocity of the  bicyclist = 26m/s

u is the initial velocity of the  bicyclist = 0m/s

a is the acceleration due to gravity = 9.81m/s

s is the distance covered during travel

Substitute the given parameters into the formula above to get the distance traveled

26² = 0² + 2(9.81)s

676 = 19.62s

Divide both sides  by 19.62

676/19.62 = 19.62s/19.62

s = 34.46m

<em>The distance traveled by the  bicyclist during the race is 34.46m</em>

5 0
4 years ago
The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.
mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is r_1=1.496\times 10^{11}\ m

The orbital radius of the Mercury is r_2=5.79 \times 10^{10}\ m

The orbital radius of the Pluto is r_3=5.91 \times 10^{12}\ m

We need to find the time required for light to travel from the Sun to each of the  three planets.

(a) For Sun -Earth,

Kepler's third law :

T_1^2=\dfrac{4\pi ^2}{GM}r_1^3

M is mass of sun, M=1.989\times 10^{30}\ kg

So,

T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s

(b) For Sun -Mercury,

T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s

(c) For Sun-Pluto,

T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s

8 0
3 years ago
The lowest energy state of a quantized system is
son4ous [18]

A state in which an atom has more energy than it does at its ground state.

Hopes this helps!

3 0
3 years ago
. A student claims that if lighting strikes a metal flagpole, the force exerted by the Earth’s magnetic field on the current in
Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

The magnetic force is given as

F = iLB

The magnetic field on the earth surface is

B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0
4 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
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