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erik [133]
3 years ago
13

Geologists apply various methods to study the layers of the Earth. Which of the following is NOT a method used to study the Eart

h’s layers?
A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust

B. Scientists map seismic waves to determine the densities of materials that lie beneath Earth’s surface.

C. Scientists measure vibrations on Earth's surface using a seismograph (an instrument that measures seismic waves).

D. Scientists measure the time it takes for seismic waves to reach a seismograph (an instrument that measures seismic waves).
Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust

Explanation:

why would use atmosphere to study the layers of earth? dont think thats possible

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A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s
irga5000 [103]
14 m/s in the direction of the truck
5 0
3 years ago
a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the
erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

W=Fx

This is equal to the change in the bullet's kinetic energy.

W=Fx=\frac{1}{2} m(u^2-v^2)......(1)

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s

Thus the speed of the bullet is 71 m/s


3 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
What is net force?
kramer

Answer:

A. The sum of all the forces acting on an object.

3 0
3 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
3 0
3 years ago
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