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2 years ago

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1. What are electromagnetic waves?

1 answer:

Wewaii [24]2 years ago

8 0

las ondas electromagnéticasSon aquellas ondas que no necesitan un medio material para propagarse. Incluyen, entre otras, la luz visible y las ondas de radio, televisión y telefonía. ... Las ondas electromagnéticas se propagan mediante una oscilación de campos eléctricos y magnéticos.

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A ball is thrown straight up. At the top of its path its acceleration is

timurjin [86]

C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)

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2 years ago

How many significant figures does the value 0.080 have? 1 2 3 4

lara [203]

2. The zeros in front do not matter

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3 years ago

I don't get this stuff!!! help me!!!

Harrizon [31]

The answer is A.) igneous rocks

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2 years ago

A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron

alex41 [277]

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

$E=\dfrac{hC}{\lambda }$

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$

5 0

2 years ago

in order to qualify for the finals in a racing event, a race car must achieve an average speed of 278 km/h on a track with a tot

Pachacha [2.7K]

The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

<h3>What is average speed?</h3>

The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.

<h3>Total time taken by the car during the entire race</h3>

time = distance/average speed

time = (1.41 km) / (278 km/h)

time = 0.0051 hr

The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;

d = 705 m = 0.705 km

t1 = 0.705/210

t1 = 0.0034 hr

<h3>time for the second half</h3>

t2 = 0.0051 - 0.0034 hr

t2 = 0.0017 hr

<h3>minimum average speed of the second half</h3>

v = d/t

v = 0.705 km / 0.0017 hr

v = 414.7 km/hr

Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

Learn more about average speed here: brainly.com/question/4931057

#SPJ1

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8 months ago