All categories

3 years ago

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Engineering

1 answer:

Ann [662]3 years ago

6 0

The number of tubs that each of them sold is; 24 tubs each

The number of days it will take for both of them to sell same amount of tubs is; 4 days

Number of cookies that Nicole had already sold = 4 tubs

Number of cookies sold by Josie before counting = 0 cookies

Nicole now sells 5 tubs per day and

Josie now sells 6 tubs per day.

Let the number of days it will take for them to have sold the same amount of cookies be d.

Thus;

5d + 4 = 6d + 0

6d - 5d = 4

d = 4 days

Thus, total number of cookies for both are;

Total for Nicole = 4 + 5(4) = 24 cookies

Total for Josie = 6(4) = 24 cookies

Read more about proportion at; brainly.com/question/870035

You might be interested in

A Carnot machine operates with 25% efficiency, whose heat rejection reservoir temperature is 300K. Determine the temperature at

alisha [4.7K]

Answer:

The temperature at which observed heat is 400 K

Explanation:

Given data:

rejection reservoir temperature at exit $T_{L}$ is 300 k

the efficiency of a engine is η = 25%

we know that efficiency of Carnot is given as

$\eta = (1-\frac{T_{L}}{T_{H}})*100$

Putting all value to obtained temperature at which observed heat

$0.25 = (1-\frac{300}{T_{H}})$

$T_{H}$ = 400 K

5 0

3 years ago

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and 200 kPa while losing heat at a rat

jeyben [28]

Answer:606 m/s

Explanation:

Given

Steam Inlet temperature $400^{\circ} C$ and pressure 800 KPa

$h_1=3267.7 kJ/kg$

initial Velocity 10 m/s

Steam Outlet Velocity is $300^{\circ} C$ and pressure is 200 KPa

$h_2=3072.1 kJ/kg$

$\nu _2=1.31632$

From steam table

Heat loss 25 KW

inlet area 800 cm^2

applying Steady Flow Energy Equation

$h_1+\frac{1}{2}v_1^2++Q=h_2+\frac{1}{2} v_2^2+W$

$3267.7+\frac{1}{2000}10^2-25=3072.1+\frac{1}{2000}v_2^2$

$3267.7-3072.1+0.05-25=\frac{1}{2}v_2^2$

$v_2=\sqrt{195.65\times 2}=606 m/s$

and volume flow rate is $m=\dot{m}\mu _2=2.082\times 1.31623=2.74 m^3/s$

5 0

3 years ago

Read 2 more answers

A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. What is

coldgirl [10]

Answer:

d. 41.4

Explanation:

The initial diameter di = 100mm

The initial height hi {✓59m

Final height = 25 m

Final diameter = ?

Initial volume = after forging volume

D*(di)²*hi = D *(df)²*hf

D will cancel out from either sides of the equation

100² x 50 = df²x25

10000x2 = df²

20000 = df²

df = √20000

df = 141.42mm

Change in diameter = 141.42-100

= 41.42

The percentage change = 41.42/100*100

= 41.4%

The last option is the answer

4 0

3 years ago

A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v

In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0

3 years ago

Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow

Black_prince [1.1K]

Answer:

$\omega_y,\omega_x,\omega_Z$ all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function $\phi$ given as follows

$u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}$

Rotationality of fluid is given by $\omega$

$\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z$

$\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x$

$\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y$

So now putting value in the above equations ,we will find

$\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},$

$\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}$

So $\omega_y$ =0

Like this all $\omega_y,\omega_x,\omega_Z$ all are zero.

That is why velocity potential flow is irroational flow.

5 0

4 years ago