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chubhunter [2.5K]
2 years ago
15

Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma

of freshly baked bread during the cooking process. Is the mass of the baked loaf greater than, less than, or the same as the one pound of original dough?
Physics
1 answer:
Morgarella [4.7K]2 years ago
7 0

Answer:

The mass of the baked loaf will be less than the dough.

Explanation: When heat is applied to food substance or products like the one pound the substance or material gains a higher temperature, the increase in temperature causes moisture inherent or added to the product in this case the one pound dough to be lost, the one pound dough prepared at room temperature, once it is placed inside the oven at 350 degrees Fahrenheit it will lose moisture  in the form of vapor to the environment as noticed in the aroma, the moisture lost will eventually reduce it mass/weight (kilograms or grams) by some percentage or quantities(kilograms or grams)

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A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

3 0
2 years ago
Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind
Dimas [21]

Answer:

acceleration = -15.3g

Explanation:

given data

speed = 6.00 m/s.

thickness = 12

moves the entire = 12.0 cm

solution

we will use here equation that is

v² - u²  = 2 × a × s    ........................1

here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration

so we put here value and get acceleration

a = \frac{v^2-u^2}{2s}

a = \frac{0^2-6^2}{2\times 0.12}

a = -150 m/s² ( negative sign means it is a deceleration )

and

acceleration in units of g  

a = \frac{-150}{9.8}

a = -15.3 g

6 0
2 years ago
A 25-mm-diameter uniform steel shaft is 600 mm long between bearings. (7-33) (9 Pts) A. Find the lowest critical speed of the sh
Arte-miy333 [17]

Answer:

a = the lowest critical speed of the shaft 882.81 rad/s

b = new diameter 0.05m or 50mm

c = critical speed 1765.62rad/s

Explanation:

see the attached file

8 0
2 years ago
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P
KIM [24]

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

E=\frac{F}{Q}

But the force F

F= \frac{kQ1Q2}{r^2}

But the electric field intensity due to a point charge Q at a distance r meters away is given by

E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }

<em>From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases</em>

6 0
2 years ago
You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative char
kkurt [141]

Answer:attract each other

Explanation:

When two-sphere, one with a negative charge and another neutral is brought close together but do not touch then they try to attract each other.

This because of the polarization of the neutral sphere as it is placed in the vicinity of a negatively charged sphere. The negatively charged sphere will induce the positive charge in the neutral sphere and they will attract each other according to Columb law.

8 0
2 years ago
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