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3 years ago

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What wavelength of light is reflected from a red shirt

1 answer:

kakasveta [241]3 years ago

6 0

Answer:

Between 590 and 750 nm

Explanation:

When you look at someone's red shirt, for instance, that shirt will be absorbing or scattering wavelengths of light lower than 590 nm, so those waves will not reach your eyes. But a red shirt will be reflecting some wavelength between 590 and 750 nm, which your eyes process as red.

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Explain why the speed of a sled is increases as it moves down a snow -covered hill, even though no one is pushing the sled

Andrews [41]

Hope this helps you.

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3 years ago

Un carro de montaña rusa parte del reposo desde una primera cumbre, desciende una distancia vertical de 45 metros y luego sube u

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A roller coaster car starts from the rest from a first summit, descends a vertical distance of 45 meters and then climbs a second summit, reaching the top with a speed of 15m / s. How high is the second summit? Do not consider friction

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3 years ago

Suppose you made 5 measurements of the speed of a rocket:10.2 m/s, 11.0 m/s, 10.7 m/s, 11.0 m/s and 10.5 m/s. From these measure

Sav [38]

Answer:

mean = 10.68 m/s

standard deviation 0.3059

[/tex]\sigma_m = 0.14[/tex]

Explanation:

1) $Mean = \frac{ 10.2+11+10.7+11+10.5}{5}$

mean = 10.68 m/s

2 ) standard deviation is given as

$\sigma = \sqrt{ \frac{1}{N} \sum( x_i -\mu)^2}$

N = 5

$\sigma =\sqrt{ \frac{1}{5} \sum{( 10.2-10.68)^2+(11-10.68)^2 + (10.7- 10.68)^2+ (11- 10.68)^2++ (10.5- 10.68)^2$

SOLVING ABOVE RELATION TO GET STANDARD DEVIATION VALUE

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3 years ago

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A 24-cm circumference loop of wire has a resistance of 0.14 Ω. The loop is placed between the poles of an electromagnet, and a f

Triss [41]

Answer:

The induced current in the loop is 1.2 A

Explanation:

Given;

length of the wire, L = 24 cm = 0.24 m

resistance of the wire, R = 0.14 Ω.

magnetic field strength, B = 0.55 T

time, t = 15 ms = 15 x 10⁻³ s

Circumference of a circle is given as;

L = 2πr

0.24 = 2πr

r = 0.24 / 2π

r = 0.0382 m

Area of a loop is given as;

A = πr²

A = π (0.0382)²

A = 0.004585 m²

Induced emf is given as;

$emf = -\frac{\delta \phi}{dt}$

Ф = ΔB x A

Ф = ( 0 - 0.55 T) x 0.004585 m²

Ф = -0.002522 T.m²

$emf = - (\frac{-0.002522}{15*10^{-3}} )\\\\emf = 0.168 \ V$

According to ohm's law;

V = IR

Where;

I is current

The induced current in the loop is calculated as;

I = V / R

I = 0.168 / 0.14

I = 1.2 A

Therefore, the induced current in the loop is 1.2 A

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3 years ago

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melisa1 [442]

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3 years ago

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