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3 years ago

Breaks in the Earths crust called faults from where plates meet true or false

Physics

2 answers:

dezoksy [38]3 years ago

7 0

True. Whenever there is a rock shifting on the crust it is a fault

KonstantinChe [14]3 years ago

5 0

False theyre called plate boundries

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PART A) The acceleration of gravity is 9.8 m/s^2 What is the magnitude of the net force on a(n) 82 kg driver operating a dragste

baherus [9]

Answer:

0.654 kN

Explanation:

If the driver accelerates from rest to 67 m/s in 8.4 seconds, then the magnitude of the acceleration is:

a = (67 - 0)/ 8.4 = 7.976 m/s^2

Then the net force on the 82 kg driver is:

F = m * a = 82 * 7,976 = 654 N

Therefore, in kN the force's magnitude is: 0.654 kN

8 0

2 years ago

Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once

GaryK [48]

NOTE: The diagram is attached to this solution

Answer:

The acceleration of point A = 14.64 ft/s²

Explanation:

By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)

$a_{A} = \bar{a} + \frac{\alpha L}{2}$

Weight, W = mg

g = 32.2 ft/s²

m = W/g

$p = m \bar{a}\\p = w \bar{a} /g$

$\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66$

$\frac{pL}{2} = I \alpha$

but $I = \frac{wL^{2} }{12g}$

$\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L$

$a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}$

6 0

3 years ago

The drawing shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel a

Pavel [41]

Answer:

$F=27.39N$

Explanation:

Take sum of torques at the point the step touches the wheel, that eliminates two torques

Σ $T=T_{N}+T_{f}+T_{W}$

Since we are looking for when the wheel just starts to rise up N-> 0 so no torque due to normal force

$T_{N}=0$

The perpendicular lever arm for the F force is R-h

$T_{f}=F*(r-h)$

And the T of gravity according to the image

$T_{W}=W*(\sqrt{r^2-(r-h)^2}$

Σ $T=0$

$T_{N}+T_{f}+T_{W}=0$

$F*(r-h)+W*(\sqrt{r^2-(r-h)^2}=0$

$F=\frac{W*(\sqrt{r^2-(r-h)^2}}{r-h}$

$F=\frac{24.9 N*(\sqrt{0.336^2-(0.336-0.110)^2}}{(0.336-0.11)}$

$F=27.39N$

4 0

2 years ago

49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner

faust18 [17]

Answer:

$\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}$

Explanation:

For this interesting problem, we use the definition of centripetal acceleration

a = v² / r

angular and linear velocity are related

v = w r

we substitute

a = w² r

the rectangular body rotates at an angular velocity w

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A

the distance OB = L₂

the distance AB = L₁

the sides of the rectangle

It is indicated that the acceleration in in A and B are related

$a_A = n \ a_B$

we substitute the value of the acceleration

w² r_A = n r_B

the distance from the each corner is

r_B = L₂

r_A = $\sqrt{L_1^2 + L_2^2}$

we substitute

\sqrt{L_1^2 + L_2^2} = n L₂

L₁² + L₂² = n² L₂²

L₁² = (n²-1) L₂²

4 0

3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

6 0

3 years ago