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guajiro [1.7K]
3 years ago
8

Why are the transverse waves produced by an earthquake known as secondary wave?

Physics
1 answer:
zheka24 [161]3 years ago
8 0
The transverse waves show up second in time, they are slower than the primary waves
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Which is greater, the gravitational force between earth and the moon, or the force between earth and the sun?
Karo-lina-s [1.5K]
Earth and the Sun
GF = 3.647x10^22 N
8 0
4 years ago
What is photoelectron ​
Korvikt [17]

Answer:

The answer is:-

Explanation:

An electron emitted from an atom by enteraction with an photon, especially an eclectron emitted from a solid surface by the action of light.

8 0
3 years ago
Read 2 more answers
A 6.25 L tank holds helium gas at a pressure of 1759 psi. A second 6.25 L tank holds oxygen at a pressure of 467.7 psi. The two
rodikova [14]

Answer:

P=1113.35 psi

Explanation:

For tank 1

V₁= 6.25 L

P₁=1759 psi

For tank 2

V₂=6.25 L

P₂=467.7 psi

Lets take final pressure is P

The final volume  V= 6.25 + 6.25 L = 12.5 L

P V = V₂P₂+V₁P₁

Now by putting the values

P V = V₂P₂+V₁P₁

P x 12.5 = 6.25 x 467.7+6.25 x 1759

P=1113.35 psi

So the final volume of the system will be 1113.35 psi.

8 0
3 years ago
Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t
IrinaK [193]
 t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec 
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec 

In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)

d1 = 1*0.319 = 0.319 m 
d2 = 0.5*0.451 = 0.225 m 

in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225    

 = 3.19 : 2.25
4 0
3 years ago
Read 2 more answers
Una flecha tiene una rapidez de lanzamiento inicial de 18 m/s. Si debe dar en un blanco a 31 m de distancia, que está a la misma
maks197457 [2]

Answer:

34.8 and 55.2º

Explanation:

This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation

         

The scope equation is

         R = v₀² sin 2θ /g

         sin 2 θ = R g / v₀²

         sin 2 θ = 31 9.8 / 18²

         2 θ = sin⁻¹ 0.93765

          θ = 34.8º

At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º

4 0
4 years ago
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