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RideAnS [48]
4 years ago
15

Secondary waves cannot travel through

Physics
1 answer:
7nadin3 [17]4 years ago
7 0
S-Waves cant travel through liquids. ;)
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Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig
MissTica

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

         

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

6 0
3 years ago
A student writes the following script for a scene in a futuristic movie: A rocket ship is landing on the Moon where many astrona
Sav [38]

Answer: No crash landing on the surface of the moon and there is no vibration due to low acceleration due to gravity.

Explanation:

The rocket ship can't land on the surface of the moon because it depends on winged flight for its controlled descent to the surface of the Earth.  Since the moon has no real atmosphere, the wings would not be able to glide the craft smoothly to the moon's surface.

The acceleration due to gravity in the moon is very low compared to the earth. The astronauts cannot experience vibration ripple through the Moon’s surface beneath their feet because the low level acceleration due to gravity will drastically affect the weight of the ship.

6 0
3 years ago
Suppose I produce radio waves with an antenna that have a peak electric field amplitude E and peak magnetic field amplitude B. I
denis23 [38]

Answer:

Correct answer is 2B

Explanation:

The electric field magnitude and magnetic field magnitude in an electromagnetic waves are related as under

E=cB

where 'c' is velocity of light in the medium of transmission

According to the given question if we double the electric field we have

2E=cB'\\\\2cB=cB'\\\\B'=2B

Thus the magnetic field also doubles

3 0
3 years ago
¿Cuál será la potencia de una bombilla conectada a una red de energía eléctrica de 440V, si la corriente que circula por el circ
goblinko [34]

Answer:

La potencia de la bombilla es de 1056 W.

Explanation:

La potencia de la bombilla se puede calcular usando la siguiente ecuación:

P = I*V

En donde:

P: es la potencia

I: es la corriente = 2,4 A

V: es la diferencia de potencial = 440 V

Entonces, la potencia es:

P = I*V = 2,4 A*440 V = 1056 W

Por lo tanto, la potencia de la bombilla es de 1056 W.

Espero que te sea de utilidad!

3 0
3 years ago
A pumpkin weighs 5.4 pounds what is its mass in grams?
olga nikolaevna [1]
THe answer would be 2,449.4 grams
8 0
3 years ago
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