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2 years ago

What kind of waves are present during an earthquake? ...

Physics

1 answer:

Brrunno [24]2 years ago

6 0

Answer:

1. The two main types of waves are body waves and surface waves. Body waves can travel through the earth's inner layers, but surface waves can only move along the surface of the planet like ripples on water. Earthquakes radiate seismic energy as both body and surface waves.

2. potential energy

3. Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

4. Refraction is the bending of light

5. Density uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). Density is defined as mass per unit volume.

Explanation:

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When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

3 years ago

Read 2 more answers

A soccer player carries the ball for a distance of 40.0 m in the direction 42.0° west of south. find the westward component of t

Maksim231197 [3]

Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

sin θ = side opposite to the angle / hypotenuse of the triangle

sin 42 = x / 40 m

x = (40 m) sin 42

x = 26.77 m

Therefore the ball has a westward displacement of about 26.77 m

3 0

3 years ago

(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en

V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0

2 years ago

A molecule that moves across a cell membrane without using the cell’s energy tends to move

zaharov [31]

Answer:answer is b

Explanation:

4 0

2 years ago

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0

2 years ago