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3 years ago

A car leaves the pit after a refueling stop and accelerates uniformly to a speed of 62.2 m/s in 5.0 s to rejoin the race.

Physics

1 answer:

MakcuM [25]3 years ago

5 0

Answer:

The car's acceleration is: a = 12.44 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} = v_{o}+a*t$

where:

Vf = final velocity = 62.2 [m/s]

Vo = initial velocity = 0 (because the car stars from the rest)

a = acceleration [m/s²]

t = time = 5 [s]

62.2 = 0 + a*5

a = 12.44 [m/s²]

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vector ????⃗ has a magnitude of 17.9 and its direction is 80∘ counter‑clockwise from the x- axis. what are the x- and y- compone

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An object with both magnitude and direction is referred to be a vector. A vector can be visualized geometrically as a directed line segment, with an arrow pointing in the direction and a length equal to the magnitude of the vector. The vector points in a direction from its tail to its head.

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2 years ago

A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environ

jeka57 [31]

Answer:

(a) W₁ = 3293.06 J = 3.293 KJ

(b) W₂ = 0 J

(d) W₄ = 0 J

(e) W = 2786.435 J = 2.786 KJ

Explanation:

The complete question has following parts:

(a) First gas expands from a volume of 1 L to 6 L at a constant pressure of 6.5 atm

(b) Second, the gas is cooled at constant volume until the pressure falls to 1 atm

(c) Third, the gas is compressed at a constant pressure of 1 atm from a volume of 6 L to 1 L.

(d) Finally the gas is heated until its pressure from 1 atm to 6.5 atm at constant volume

(e) what is the net work?

ANSWERS:

(a)

The work done by a gas at constant pressure is given as follows:

W = PΔV

where,

W = Work done by the gas

P = Constant Pressure of the Gas

ΔV = Change in Volume of The gas

Therefore, for the first step:

P = P₁ = (6.5 atm)(1.01325 x 10⁵ Pa/1 atm) = 6.58613 x 10⁵ Pa

ΔV = ΔV₁ = 6 L - 1 L = (5 L)(0.001 m³/1 L) = 5 x 10⁻³ m³

W = W₁

Therefore,

W₁ = (6.58613 x 10⁵ Pa)(5 x 10⁻³ m³)

W₁ = 3293.06 J = 3.293 KJ

(b)

The work done at a constant volume by a gas is always zero due to no change in volume:

W₂ = P₂ΔV₂

W₂ = P₂(0)

W₂ = 0 J

(c)

For the third step:

P = P₃ = (1 atm) = 1.01325 x 10⁵ Pa

ΔV = ΔV₃ = 1 L - 6 L = (- 5 L)(0.001 m³/1 L) = - 5 x 10⁻³ m³

W = W₃

Therefore,

W₃ = (1.01325 x 10⁵ Pa)(-5 x 10⁻³ m³)

W₃ = - 506.625 J = - 0.506 KJ

(d)

The work done at a constant volume by a gas is always zero due to no change in volume:

W₄ = P₄ΔV₄

W₄ = P₄(0)

W₄ = 0 J

(e)

Hence, the net work is given as follows:

W = W₁ + W₂ + W₃ + W₄

W = 3293.06 J + 0 J + (- 506.625 J) + 0 J

W = 2786.435 J = 2.786 KJ

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