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2 years ago

9

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing

a white shirt. Which person will be warmer after the game? Explain your answer.

2 answers:

melomori [17]2 years ago

7 0

Answer: Anthony will be warmer after the game.

Explanation :

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.

The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

astra-53 [7]2 years ago

3 0

Answer:

Anthony will be warmer after the game because he is wearing a black shirt. Dark colors absorb more thermal energy than light colors.

Explanation:

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tiny-mole [99]

Explanation:

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2 years ago

An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which

katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

$\lambda=80 cm = 0.80 m$ (wavelength)

So the proper frequency is

$f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz$

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

$v_s = +35.0 m/s$ (velocity of the ambulance)

$v_o = 0$ (velocity of the staff)

Substituting,

$f'=\frac{343+0}{343+35}(429)=389 Hz$

Learn more about frequency and wavelength:

brainly.com/question/5354733

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1 year ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

1 year ago

Read 2 more answers

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

2 years ago

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = <em>1.55 x 10^-8 v/m</em>

6 0

2 years ago