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ivolga24 [154]
2 years ago
7

The waves produced on the earth’s surface is called

Physics
2 answers:
Maslowich2 years ago
6 0

Answer:

seismic wave

Explanation:

  • These waves are produced because of plate tectonics
  • These helps are useful in measurements of earthquake.
  • Used in seismograph.
kotykmax [81]2 years ago
6 0

Answer:

Option A. Seismic wave

Explanation:

The waves produced on the earth's surface during earthquakes are called seismic waves

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The plates on a capacitor have a radius of 1.5 mm and are separated by a distance of 0.43 mm. The space between the plates is fi
zhenek [66]
So, C = kE°A/d

putting the values,

C
= 3.8 × 8.85×10^(-12) × 3.14×1.5×1.5 × 10^(-6)/0.43 × 10^(-3)

so, 1.02 × 10^(-13)

so the most appropriate answer is 2 ...that is
1.4 × 10^(-13) ....answer !!
3 0
3 years ago
A 66​-foot-tall woman walks at 55 ​ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th
I am Lyosha [343]

Answer:

a. \frac{dx}{dt}=20ft/s

b. \frac{d(x+L)}{dt}==25ft/s

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}

Solve to find the rate relation

x=\frac{24}{6}*L

x=4*L

Now the rate of the change rate

\frac{dx}{dt}=4*\frac{dL}{dt}

\frac{dx}{dt}=4*5ft/s=20ft/s

Finally the rate of her shadow moving

\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}

\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s

5 0
2 years ago
Bob is pushing a box across the floor at a constant speed of 1.5 m/s, applying a horizontal force whose magnitude is 60 n. alice
earnstyle [38]

120n

since the speed is doubled, her force is doubled

7 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
3 years ago
If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun
inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0
3 years ago
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