All categories

3 years ago

2. Compare and Contrast A fault cuts through

Physics

1 answer:

Elenna [48]3 years ago

6 0

Explanation:

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through. The fault labeled "E" cuts through all three sedimentary rock layers (A, B,and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen and known of.

You might be interested in

Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts

Mrac [35]

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

$T=2\pi\sqrt{\frac{L}{g} }$ where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

$T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }$

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

$T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec$

3) The string length is halved

Given that:

$T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }$

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

$T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec$

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.

6 0

3 years ago

Which of the following is not an example of Newton’s third law?

Mrrafil [7]

C) All of these are examples of Newton’s third law.

3 0

4 years ago

A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).

Tems11 [23]

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

$\sigma_{max}$ = Maximum stress

$\sigma_{min}$ = Minimum stress

$\sigma_m$ = Mean stress = 50 MPa

$\sigma_a$ = Stress amplitude = 225 MPa

Mean stress is given by

$\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}$

Stress amplitude is given by

$\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa$

$\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa$

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

$R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636$

The stress ratio is -0.63636

Stress range is given by

$\sigma_{max}-\sigma_{min}=450\ MPa$

Magnitude of the stress range is 450 MPa

8 0

3 years ago

Anna and Jon sit on a seesaw. Anna has a mass of 60 kg and sits 2 m from the center. Jon has a mass of 70 kg. How far from the c

Ronch [10]

Answer:

dJ = 1.7 m

Explanation:

The Equation of the Balancing the moments in the center of the seesaw is like this:

∑Mo = 0

Mo = F*d

Where:

∑Mo : Algebraic sum of moments in the center(o) of the balance

Mo : moment in the o point ( N*m)

F : Force ( N)

d : distancia of the force to the the o point ( N*m)

Data

mA = 60 kg : mass of the Anna

mJ = 70 kg : mass of theJon

dA = 2 m : Distance from Anna to the center of the seesaw

g: acceleration due to gravity

Calculation of the distance from Jon to the center of the seesaw (dJ)

∑Mo = 0 WA : Ana's weight , WJ : Jon's weight

W = m*g

(WA)(dA) - (WJ) (dJ) = 0

(mA*g)(dA) - (mJ*g)(dJ) = 0

We divide by g the equation:

(mA)(dA) - (mJ)(dJ)= 0

(mA)(dA) = (mJ)(dJ)

$d_{J} = \frac{m_{A} *d_{A}}{m_{J}}$

$d_{J} = \frac{60 kg *2 m}{70 kg}$

dJ = 1.7 m

5 0

3 years ago

State the basic assumptions of the kinetic theory.

My name is Ann [436]

Answer:

option a

Explanation:

a- All matter is composed of small particles.

6 0

2 years ago

Read 2 more answers