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2 years ago

The seafloor spreading process at ridges produces what kind of faults?

Physics

1 answer:

irga5000 [103]2 years ago

4 0

Active transform faults are between two tectonic<span> structures or faults.</span>

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The windshield of a car has a total length of arm and blade of 9 inches, and rotates back and forth through an angle of 93degre

Sergio039 [100]

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7-in wiper blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

5 0

3 years ago

Read 2 more answers

F F= {mango, apple, banana, orange)

Stels [109]

Answer:

Explanation:

Cardinality of a set is the number of elements in that set. Given the set.

F= {mango, apple, banana, orange), we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).

Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4

6 0

3 years ago

El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno

77julia77 [94]

Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

7 0

2 years ago

A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential

BARSIC [14]

Answer:

$1.88\cdot 10^{-5} A$

Explanation:

The capacitance of a parallel plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$ (1)

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

$Q=CV$ (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

$J=\frac{Q}{t}$ (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

$J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}$

where we have

$A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2$

$d = 0.58 mm = 5.8\cdot 10^{-4} m$

$\frac{V}{t}=500,000 V/s$

Substituting into the equation, we find

$J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A$

6 0

3 years ago

A car is brought to rest uniformly in 6 seconds. The initial velocity of the car was 24 m/s. How far does the car travel while d

pav-90 [236]

Answer:

v=u+at

24=0+at

24=a×6

a=4m/s

hence

s=ut+at^2÷2

s=36m

Explanation:

since the car is brought to rest the u=0

4 0

2 years ago