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Allushta [10]
2 years ago
7

Let x define the position of an object such that x =

Physics
1 answer:
timama [110]2 years ago
7 0
See the attached picture for answers

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HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0
3 years ago
Read 2 more answers
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
Rafe is placing his car on a lift to change the oil and oil filter. He needs to drive up the lift 1.5 meters and the lift raises
jeyben [28]

Answer:

104 IMA

Explanation:

7 0
2 years ago
What would happen to the function or purpose of flower if the stigma wasn't sticky?
sashaice [31]

Answer:

The plant would not reproduce because the flower uses the stigma to catch the pollen

8 0
2 years ago
Read 2 more answers
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