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1 year ago

Why do the waves have different speeds in different layers of Earth's surface?

Physics

1 answer:

Genrish500 [490]1 year ago

6 0

Answer:

Material's density

Explanation:

Seismic waves travel at different rates of speed based on a material's density. Hopefully, you understand that the Earth has three main layers: the crust, mantle, and core. Earthquake waves move faster through solids.

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Blake is setting up his tent at a renaissance fair. If the tent is 8 feet tall and the tether can be staked no more than 2 feet

Mariana [72]

The stake, height and tether length of the tent form a right angle triangle where the tether length is the hypotenuse.
Applying Pythagoras theorem:
length² = height² + (stake distance)²
length = √(8² + 2²)
length = 8.5 feet

3 0

2 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing

Semmy [17]

If Juan used a Celsius thermometer, it would tell him the Celsius temperature.

If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.

7 0

2 years ago

What is the Doppler Effect equation?

MAXImum [283]

Fperson =[( velocity of wind )+ or - (velocity of person)] / [(velocity of wind) + or - (velocity of sounds)] x frekuency of sounds

7 0

2 years ago

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed o

adell [148]

Answer:

Part a)

$v_f = 7.99 m/s$

Part b)

$y = 3.25 m$

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

$v_f^2 - v_i^2 = 2 a y$

here we have

$v_i = 2.22 m/s$

$y = -3 m$

$v_f^2 - (2.22)^2 = 2(-9.81)(-3)$

$v_f = 7.99 m/s$

Part b)

at highest point of his motion the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a (\Delta y)$

$0 - 2.22^2 = 2(-9.81)(y - 3)$

$y = 3.25 m$

7 0

2 years ago

The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a

rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0

2 years ago