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2 years ago

Which describes the adiabatic process? Check all that apply.

Physics

2 answers:

Colt1911 [192]2 years ago

8 0

Answer:

Heat is not absorbed by the system.

A rapid shift occurs between gas compression and expansion.

Explanation:

As we know by first law of thermodynamics

$\Delta Q = \Delta U + W$

here in case of adiabatic process we know that transfer of heat from the system must be zero

So here in that case

$\Delta Q = 0$

so here we will have

$\Delta U + W = 0$

so we have

$\Delta U = - W$

so the process must be very fast so that transfer of heat is not possible

so correct answer is

Heat is not absorbed by the system.

A rapid shift occurs between gas compression and expansion.

kiruha [24]2 years ago

7 0

Answer:

heat is not absorbed by the system

A rapid shift occurs between gas compression and expansion.

All heat is transformed to work done by the system.

Explanation:

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A steel plate shine but wooden vessel desnt

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Answer :

A new steel plate shines but an wooden vessel does not shine because new steel plate has lustre. ... Because steel absorbs very little amount of light and reflects back the light but in case of wooden vessel it has dull appearence and reflects very little amount of light so it does not shine.

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2 years ago

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As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl

Brilliant_brown [7]

Answer:

Thermal Power = 460W

Explanation:

From Stephan-Boltzmann Law Formula;

P = єσT⁴A

Where,

P = Radiation energy

σ = Stefan-Boltzmann Constant

T = absolute temperature in Kelvin

є = Emissivity of the material.

A=Area of the emitting body

Now, σ = 5.67 x 10^(-8)

є = 0.6

Temperature = 30°C and coverting to kelvin = 30 + 273 = 303K

Area ; since we are to consider the sides of the human body as 2m and 0.8m,thus area = 2 x 0.8 = 1.6

Thus thermal power = 0.6 x 5.67 x 10^(-8) x303⁴ x 1.6 = 458. 8W

Normally, we approximate to the nearest 10W. Thus, thermal power is approximately 460W

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2 years ago

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

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2 years ago

A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

Answer:

Maximum height reached = 35.15 meter.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

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2 years ago

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

MAVERICK [17]

Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

3 0

2 years ago

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