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2 years ago

Which of the following statements is FALSE?

Physics

1 answer:

Levart [38]2 years ago

5 0

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

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Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and

masha68 [24]

Answer:

F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

fr = μ N

We define a reference system parallel to the floor

block B ( lower)

Y axis

N - W₁-W₂ = 0

N = W₂ + W₂

N = (M + m) g

X axis

F -fr = M a

for block A (upper)

X axis

fr = m a (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

F = (M + m) a (3)

a = $\frac{F}{ M+m}$

the friction force has the formula

fr = μ N

fr = μ (M + m) g

let's calculate

fr = 0.34 (2.0 + 0.250) 9.8

fr = 7.7 N

we substitute in equation 2

fr = m a

a = fr / m

a = 7.7 / 0.250

a = 30.8 m / s²

we substitute in equation 3

F = (2.0 + 0.250) 30.8

F = 69.3 N

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2 years ago

Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve

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If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward. So since the current is going at an angle then it has a x and y component. So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

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2 years ago

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Select examples of simple harmonic motion that can be observed in everyday life.

liberstina [14]

A. a child gently swinging on a swing at small angles all the time

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2 years ago

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

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2 years ago

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When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h

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Answer:

The correct answer is c. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

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2 years ago