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2 years ago

The corner store sells candy in ₵20, ₵30 and ₵50 packages. List all the ways in which the Candyman

Engineering

1 answer:

Novosadov [1.4K]2 years ago

6 0

Answer: I will list them down below!

Explanation:

He can buy 6, 50 cent candies.

He can buy 30, 20 cent candies.

He can buy 6, 30 cent candies and 6, 20 cent candies.

He can buy 15, 20 cent candies and 3, 50 cent candies.

He can by 3, 20 and 30 cent candies and 3, 50 cent candies.

That's it.

Hope this helps!

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20 POINTS

Kobotan [32]

Answer:

True ...................

5 0

3 years ago

An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output

GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output $\dot W$ = $\dot m$ × $c_p$ ×(T₂ - T₁)

Where;

$c_p$ = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

$\dot m$ = The mass flow rate = 2.0 kg/s

Substituting the values, we have;

$\dot W$ = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

$\dot W$ = -396.17 kJ/s

7 0

4 years ago

An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat

IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature $T_L$ = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature $T_H$ = 99° F = ( 99 + 460 )R = 559 R

Cooling Load $Q_L$ = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

Coefficient of performance of the air conditioner;

$COP_{air-condition$ = Cooling Load $Q_L$ / power P

we substitute

$COP_{air-condition$ = 30000 Btu/h / 8396.619 Btu/h

$COP_{air-condition$ = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

- Power input required ( in hp )

$Q_L$ / $P_{required$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

30000 Btu/h / $P_{required$ = 530 R / ( 559 R - 530 R )

30000 Btu/h / $P_{required$ = 530 R / 29 R

we solve for $P_{required$

$P_{required$ = ( 30000 Btu/h × 29 R ) / 530 R

$P_{required$ = ( 870000 Btu/h / 530 )

$P_{required$ = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

$P_{required$ = ( 1641.5094 / 2544.43 ) hp

$P_{required$ = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

$COP_{rev-air-condition$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

$COP_{rev-air-condition$ = 530 R / ( 559 R - 530 R )

$COP_{rev-air-condition$ = 530 R / 29 R

$COP_{rev-air-condition$ = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0

3 years ago

A/an _____________ comes on if one of the dual hydraulic brake systems should fail or, in some vehicles, if the brake fluid is l

jolli1 [7]

Answer: Dash light

Explanation:

A dash light, usually labeled "BRAKE" will illuminated if a braking system fails, or if the brake fluid is low.

3 0

3 years ago

A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p

Stella [2.4K]

Answer:

Cut-off ratio $\dfrac{V_3}{V_2}=6.15$

Cxpansion ratio $\dfrac{V_4}{V_3}=3.25$

The exhaust temperature $T_4=1997.5R$

Explanation:

Compression ratio CR(r)=20

$\dfrac{V_1}{V_2}=20$

$P_2=P_3=920 psia$

$T_1=520 R ,T_{max}=T_3,T_3=3200 R$

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{520}=20^{1.4 -1}$

$T_2=1723.28R$

$\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}$

$\dfrac{V_3}{V_2}=\dfrac{3200}{520}$

So cut-off ratio $\dfrac{V_3}{V_2}=6.15$

$\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}$

Now putting the values in above equation

$\dfrac20=\dfrac{V_4}{V_3}\times 6.15$

$\dfrac{V_4}{V_3}=3.25$

So expansion ratio $\dfrac{V_4}{V_3}=3.25$ .

$\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}$

$\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}$

$T_4=1997.5R$

So the exhaust temperature $T_4=1997.5R$

3 0

4 years ago