All categories

3 years ago

Explain the term "avalanche state"

Physics

1 answer:

vagabundo [1.1K]3 years ago

5 0

Answer:

The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.

Explanation:

You might be interested in

A heat pump is used to maintain a house at 22°C by extracting heat from the outside air on a day when the outside air temperatur

Ilia_Sergeevich [38]

Answer:

It is enough

Explanation:

To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.

The two ways in which the performance coefficient can be expressed are given by:

$COP_p = \frac{T_H}{T_H-T_L}$

Where,

$T_H =$ High Temperature

$T_L =$ Low Temperature

And the other way is,

$COP_p = \frac{\dot{Q}}{W}$

Where $\dot{Q}$ is heat rate and W the power consumed.

We have all our terms in Celsius, so we calculate the temperature in Kelvin

$T_H = 22+273 = 295K$

$T_L = 2+273 = 275k$

The rate at which heat is lost is:

$\dot{Q} = 110000kJ/h$

The power consumed by the heat pump is

$\dot{W} = 5kW$

And the coefficient of performance is

$COP_p = \frac{T_H}{T_H-T_L}$

$COP_p = \frac{295}{295-275}$

$COP_p = 14.75$

With this value we can calculate the Power required,

$COP_p = \frac{\dot{Q}}{W}$

$14.75 = \frac{110000}{W}$

$W = \frac{110000}{3600*14.75}$

$W = 2.07kW$

The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.

4 0

3 years ago

A constant 75-N force acts on a 6.00 kg box. The box initially is moving at 20 m/s in the direction of the force, and 3.0 s seco

harkovskaia [24]

Answer: 299,427.84 J, 99,809.28 W

Explanation: we will assume the acceleration of the body to be constant, hence newton's laws of motion is applicable.

From the question

Force (f) = 75 N

Mass of object (m) = 6kg

Initial velocity (u) = 20 m/s

Final velocity (v) = 58 m/s

Time taken (t) = 3s

Recall that

F = ma

75 = 6a

a = 75/6 = 12. 5 m/s²

We need to get the distance before we can get the work done.

Recall that v² = u² + 2as, where s = distance covered

56² = 20² + 2(12.5)s

56² - 20² = 25s

2736 = 25s

s = 2736/ 25 = 109.44m

Work done = force × distance

Work done = 75 × 109.44 = 299,427.84 j

Power = work done / time

Power = 299,427.84 / 3

Power = 99,809.28 W

7 0

4 years ago

Read 2 more answers

6 A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 2

makvit [3.9K]

To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states

State 1

$T_1 = 700\°C$

$P_1 = 4 Mpa$

From steam table

$h_1 =3906.41 KJ/Kg$

$s_1 = 7.62 KJ/Kg.K$

Now

$s_1 = s_2 = 7.62 KJ/Kg.K$ As 1-2 is isentropic

State 2

$P_2 = 20 Kpa$

$s_2 = 7.62 KJ/Kg \cdot K$

From steam table

$h_2 = 2513.33 KJ/Kg$

PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then

$Power = m \times (h_1-h_2)$

$P = (50)(3906.41 - 2513.33)$

$P = 69654kW$

b) Pump Work

State 3

$P_3 = 20 Kpa$

$\upsilon= 0.001 m^3/kg$

The Work done by the pump is

$W= m\upsilon \Delta P$

$W = (50)(0.001)(4000-20)$

$W = 199kJ$

5 0

3 years ago

Which structure is represented by the letter D? Choose 1 answer: (Choice A) A Chloroplast (Choice B) B Vacuole (Choice C) C Mito

Otrada [13]

Answer:

vacuole

Explanation:

Remember vacuoles in plant cells are a lot bigger than there animal cell counterparts.

6 0

3 years ago

A couch of mass 210 kg is being pushed across the floor with an applied force of 4100 N. The coefficient of kinetic friction bet

Dennis_Churaev [7]

Let us use the formula for Newton's Second Law of Motion:

Net force = Mass*Acceleration
Net force = Applied Force - μ*Normal Force
where μ is the coefficient of kinetic friction

Normal Force = Force due to gravity = mass*gravity
Normal Force = (210 kg)(9.81 m/s²) = 2,060.1 N

Then,
Net force = 4100 - 0.38*2060.1 = 3317.162 N

3317.162 N = (210 kg)(a)
Solving for acceleration,
a = 15.796 m/s²

8 0

4 years ago