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4 years ago

Which statement is true for particles of the medium of an earth quake p wave

1 answer:

8 0

I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.

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A model plane has a mass of 0.75 kg and is flying 12 m above the ground

Grace [21]

Answer:

Option C. 210 J.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Velocity (v) = 18 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Total Mechanical energy (ME) =?

Next, we shall determine the potential energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 0.75 × 9.8 × 12

PE = 88.2 J

Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Velocity (v) = 18 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.75 × 18²

KE = ½ × 0.75 × 324

KE = 121.5 J

Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:

Potential energy (PE) = 88.2 J

Kinetic energy (KE) = 121.5 J

Total Mechanical energy (ME) =?

ME = PE + KE

ME = 88.2 + 121.5

ME = 209.7 J

ME ≈ 210 J

Therefore, the total mechanical energy of the plane is 210 J.

8 0

3 years ago

How does newton's second law describe the motion of an object?

vodomira [7]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

4 0

3 years ago

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For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2

Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

$n = \frac{c}{v}$

The speed of violet light in the object is $1.9 * 10^8 m/s$ .

The speed of red light in the object is $2 * 10^8 m/s$

Hence, the refractive index for violet light is:

$n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56$

and for red light, it is:

$n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5$

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

$n = \frac{sin(i)}{sin(r)}$

The angle of incidence is 30°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56} = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5} = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o$

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56} = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5} = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o$

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0

3 years ago

Help !!!!!!Estimate the number of times your heart beats in a day?

Nastasia [14]

Roughly 115,200 times a day

6 0

3 years ago

Read 2 more answers

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy whe

boyakko [2]

Answer:

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, the water undergoes 3 stages:

Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C

Given Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)

H = 4186 × 81.5 + 2. 256 ×10⁶ + 2020× 185

= 2.696 ×10⁶ J/kg

The amount of heat in J/L of water needed can be calculated as follows:

Density of water = 1000kg/m³ =

1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)

Let the volume of water needed be V litres.

Then the mass of water that must be expended = Density × Volume

= 1kg/L × V L = Vkg

The heat that would be absorbed by the water when 1L of crude oil is burned is V×H

= 2.696×10⁶ × V

This is also equal to 2.80×10⁷ J of energy (given).

So,

2.696×10⁶V = 2.8×10⁷

V = (2.80×10⁷)/(2.696×10⁶) = 10.4L of water.

3 0

3 years ago

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