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Anastasy [175]
3 years ago
7

Heavy rains in late July 2010 triggered the mass wasting that occurred in this mountain valley near Durango, Colorado. Heavy equ

ipment is clearing away material that blocked railroad tracks and significantly narrowed the stream channel. What type of mass wasting likely occurred here?
Physics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

Land Slides

Explanation:

The type of mass wasting that most likely would have occurred in the mountain valley near Durango, Colorado is land slide or land slump.

The heavy rain might have loosened the soil and the rock strata of the mountain valley.And this Loosened mud, rock would have slide down along the inclined slops of the mountain leading to land slides.

You might be interested in
The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant
AlexFokin [52]

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m =  6,370,000 x 0.001

                       = 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

      = 6.37 x 10³ Km

7 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is the acceleration?
fredd [130]

Answer:

a = 0.83\ m/s^2

Explanation:

<u>Uniform Acceleration </u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

Substituting:

\displaystyle a=\frac{25-0}{30}

\boxed{a = 0.83\ m/s^2}

4 0
3 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
3 years ago
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