What is the name of the process in which oceanic crust moves apart at mid-ocean ridges due to convention currents in the mantle?

mojhsa [17]

Answer:

Explanation:

There are 3 main forces at work here, gravity, normal and friction. The gravity pulls the car straight down and is what keeps the car on the ground. Normal force is straight up from the points where the car is touching, so since the wheels are the only parts of the car touching the street, this is where all the normal force is. Friction force opposes any and all motion, the car wants to slide down the hill and would slide down the hill if there was no friction, so the friction force is in the opposite direction of the cars intended motion.

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8 0

3 years ago

Humans can typically perceive frequencies that range from 20 Hz–20 kHz. Dogs can hear sounds up to 45 kHz. Dog whistles produce

Aleksandr [31]

Pitch is directly related to the frequency of the sound. In this item, we are given that the frequency of the sound is higher compared to those which are audible to the human being's ears. The pitch therefore of the dog's whistle is high.

On the other hand, the frequency and the wavelength of a certain wave are inversely proportional. This means that the high frequency wave will have a short wavelength.

Hence, the answer to this item would have to be "high pitch with a short wavelength"

The answer to this item is the second option.

6 0

3 years ago

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A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago

A 72-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.61 m,

chubhunter [2.5K]

Answer:

2849.98 J

Explanation:

From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

Given: m = 72 kg, Δh = 1.61 m, Δv = 8.5-1.6 = 6.9 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 72(9.8)(1.61)+1/2(72)(6.9²)

W = 1136.016+1713.96

W = 2849.98 J

8 0

3 years ago

How many type of expansion physic

Yuki888 [10]

Answer:

Three types of thermal expansion are linear expansion,s superficial expansion,cubical expansion

4 0

3 years ago