Cindy runs 500 meters every morning. It takes her 2 hours to complet

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

Dafna1 [17]

Answer:

(a) W = 3293 J = 3.293 KJ

(b) W = 0 KJ

(c) W = -506.6 J = -0.507 KJ

(d) W = 0 KJ

Net Work = 2.786 KJ = 2786 J

Explanation:

(a)

The work done is given as:

$W = P\Delta V\\W = P(V_2-V_1)$

where,

P = Constant Pressure = (6.5 atm)(101325 Pa/1 atm) = 6.59 x 10⁶ Pa

V₁ = initial volume = (1 L)(0.001 m³/1 L) = 0.001 m³

V₂ = final volume = (6 L)(0.001 m³/1 L) = 0.006 m³

Therefore,

$W = (6.59\ x\ 10^6\ Pa)(0.006\ m^3-0.001\ m^3)$

W = 3293 J = 3.293 KJ

(b)

Since the volume is constant in this stage. Therefore,

ΔV = 0

As a result:

W = 0 KJ

(c)

The work done is given as:

$W = P\Delta V\\W = P(V_2-V_1)$

where,

P = Constant Pressure = (1 atm)(101325 Pa/1 atm) = 1.01 x 10⁵ Pa

V₁ = initial volume = (6 L)(0.001 m³/1 L) = 0.006 m³

V₂ = final volume = (1 L)(0.001 m³/1 L) = 0.001 m³

Therefore,

$W = (1.01\ x\ 10^5\ Pa)(0.001\ m^3-0.006\ m^3)$

W = -506.6 J = -0.507 KJ

negative sign show that the work is done on the gas by the environment.

(d)

Since the volume is constant in this stage. Therefore,

ΔV = 0

As a result:

W = 0 KJ

Net work will be the sum of all the works:

Net Work = 3.293 KJ + 0 KJ - 0.507 KJ + 0 KJ

Net Work = 2.786 KJ = 2786 J

4 0

3 years ago

a circuit has a total resistance of 5 ohms and a current of 0.3 amperes. What voltage must the battery supply?

iren [92.7K]

Answer:

1.5volts

Explanation:

V=IR =0.3×5=1.5volts

5 0

4 years ago

A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radiu

Papessa [141]

Answer:

2.5 m/s²

Explanation:

The given parameters are;

The mass of the car, m = 2,000 kg

The radius of the car, r = 40.0 m

The coefficient of friction between the car tires and the road, μ = 0.500

The constant speed with which the car moves, v = 10.0 m/s

The normal reaction of the road on the car, N = The weight of the car;

∴ N = m × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

N ≈ 2,000 kg × 9.81 m/s² = 19,620 N

The frictional force, $F_f$ = μ × N

The centripetal force, $F_c$ = m·v²/r

The car moves without slipping when $F_f$ = $F_c$

Therefore, $F_f$ = 0.500 × 19,620 N = 2,000 kg × $v_{max}$ ²/40.0 m

∴ $v_{max}$ = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s

Therefore, the velocity with which the car moves, v < $v_{max}$

The cars centripetal acceleration, $a_c$ = v²/r

∴ $a_c$ = (10.0 m/s)²/40.0 m = 2.5 m/s²

The cars centripetal acceleration as it goes round the turn, $a_c$ = 2.5 m/s².

7 0

3 years ago

The cable of a crane is lifting a 950 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/sin a distance of 5.5m

Umnica [9.8K]

Explanation:

a) How much work is done by gravity?

w = f x d
w = 950 x 10 x 5.5 = 52250j

b) How much work is done by tension?

v²=u²+2as
0.75²=0.25²+2a x5.5
0.56=0.06+2a x5.5
2a x5.5 = 0.56 - 0.06
2a x 5.5 =0.5
11a=0.5
a = 0.5/11 = 0.05m/s²

w = f x d

w = 950 x 0.05 x 5.5 = 261.25j

7 0

3 years ago

1)How fast is a ball moving when falling from rest and hits the ground 1.70 seconds later?

NemiM [27]

The final velocity of the ball is16.66 m/s.

Given the following data:

Initial velocity, U = 0 m/s (since the ball is falling from rest).

Time, t = 1.70 seconds.

We know that acceleration due to gravity (a) for a ball in free fall is equal to 9.8 meter per seconds square.

To find how fast (velocity) the ball is moving, we would use the first equation of motion:

Mathematically, the first equation of motion is calculated by using the formula;

$V = U + at$

Where:

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$V = 0 + 9.8(1.7)$

$V = 9.8$ × $1.70$

Final velocity, V = 16.66 m/s.

Therefore, the final velocity of the ball is16.66 m/s.