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4 years ago

15

a seismic wave has an amplitude of 0.012 Meters.If the amplitude of this wave reduces to 0.006 meters, what happens to the energ

y associated with this wave?

1 answer:

irina1246 [14]4 years ago

5 0

Answer:The energy of the wave by a factor of 4

Explanation:

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a plane is flying 120 m above the ground at an angle of 30 degrees to the horizontal, when the pilot released 2 fuel tanks to de

charle [14.2K]

Answer:

<u></u>

<u>2.26 seconds</u>

<u>97m/s</u>

<u></u>

Explanation:

1. Fall time

<u>i) Find the vertical speed of the plane when the tanks were released</u>

$V_{y,0}=84m/s\times sin(30\º)=42m/s$

That is the same initial vertical speed of the tanks.

<u />

<u>ii) Find the fall time</u>

$y-y_0=V_{y,0}\cdot t+g\cdot t^2/2$

$120=42t+4.9t^2$

$4.9t^2+42t-120=0$

$t=\dfrac{-42\pm\sqrt{(42)^2-4(4.9)(-120)}}{2\times 4.9}$

Only the positive value has aphysical meaning: t = 2.26 seconds.

2. Speed when they hit the ground

<u>i) The horizontal speed is constant:</u>

$V_x=84m/s\times cos(30\º)\approx72.5m/s$

<u />

<u>ii) The vertical speed is:</u>

$V_y=V_{y,0}+g\cdot t$

$V_y=42m/s+9.8m/s^2\cdot (2.26s)\approx64.1m/s$

<u>iii) Total speed</u>

$V=\sqrt{V_x^2+V_y^2}$

$V=\sqrt{(72.5m/s)^2+(64.1m/s)^2}\approx97m/s$

6 0

4 years ago

Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what

hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² = k*1.6²

$k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}$

The value of the spring constant is 187,500N/m

7 0

3 years ago

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

So

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

3 years ago

What are the two most common ways to produce hydrogen gas used in fuel cells?

Natali5045456 [20]

Answer:

Steam-methane reforming

Electrolysis of water

Explanation:

Steam methane reforming involves reaction of methane with water in the presence of a catalyst such as nickel to form Carbon oxides and Hydrogen.

CH4 + H2O ⇌ CO + 3 H

Electrolysis of water involves splitting of water through application of electric current to give Hydrogen and Oxygen gas.

2 H2O(l) → 2 H2(g) + O2(g)

8 0

3 years ago