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3 years ago

Rock breaking apart due to constant freezing and thawing is a mechanical process. true or false

1 answer:

6 0

The mechanical process<span> is the positing of what is contained in the Notion of mechanism, and therefore, in the first instance, of a contradiction. 1. It follows from the Notion just indicated that the interaction of objects takes the form of the positing of the identical relation of the objects. (thats the definition of this process and from reading this i would say... false? because rain isn't really an object but i dont know) </span>

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Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago

A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands

seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV= ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0

3 years ago

What is acceleration measured in?

riadik2000 [5.3K]

Calculating acceleration involves dividing velocity by time — or in terms of units, dividing meters per second [m/s] by second [s]. Dividing distance by time twice is the same as dividing distance by the square of time. Thus the SI unit of acceleration is the meter per second squared.

This I what it showed when I looked it up so if your writing this down try and put some in your own words

6 0

3 years ago

Read 2 more answers

How is energy sent from Earth to space?

Nostrana [21]

It might be radiation and reflection but I’m not sure

8 0

3 years ago

Read 2 more answers

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti

Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0

3 years ago