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Schach [20]
3 years ago
6

A ____fault forms as a result of horizontal compression.

Physics
1 answer:
kirill [66]3 years ago
4 0

Answer:

Fracture or system of fractures in Earth's crust that occurs when stress is applied too quickly or stress is too great; can form as a result of horizontal compression (reverse fault), horizontal shear (strike-slip fault), or horizontal tension (normal fault).

Explanation:

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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

#SPJ4

7 0
1 year ago
Per person, the energy consumption in the United States is about 3.3 × 10^8 Btu/yr. This is equivalent to a power of:________. A
zhuklara [117]

Answer:

3.3\times 10^8\ btu/yr=11\ kW

Explanation:

Btu of British thermal unit is a unit of heat. The relation between btu/year and watts is given by :

1\ btu/yr=3.343\times 10^{-5}\ W\\\\3.3\times 10^8\ btu/yr=3.343\times 10^{-5}\times 3.3\times 10^8\ W\\\\3.3\times 10^8\ btu/yr=11031.9\ W\\\\3.3\times 10^8\ btu/yr=11\ kW

So,  3.3\times 10^8\ btu/yr is equal to 11 kW. Hence, the correct option is (d).                                                                      

4 0
3 years ago
An ambulance is traveling north at 55.9 m/s, approaching a car that is also traveling north at 28.4 m/s. The ambulance driver he
wlad13 [49]

Answer:

915 Hz

Explanation:

The observed frequency from a sound source is given as

f₀ = f [(v + v₀)/(v+vₛ)]

where

f₀ = observed frequency of the sound by the observer = ?

f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

v₀ = velocity of the observer = 28.4 m/s

f₀ = 983 [(343+28.4)/(343+55.9)]

f₀ = 915.2 Hz = 915 Hz

6 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.
evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

where

(A) a = -k × v

 at v= 910 ft/s

     a = - 1.987 × 10⁶ ft/s²

(B)  at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{dv}{v} = -kdt

\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt

ln(200.2)-ln(900) = -kt

t = 6.84 × 10⁻⁴ s

3 0
3 years ago
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