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11Alexandr11 [23.1K]
3 years ago
14

Explain why the baking instructions on a box of cake mix are different for high and low elevations. Would you expect to have a l

onger or shorter cooking time at a high elevation?
Physics
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

All this is due to <u>atmospheric pressure</u> (which decreases with height) and affects the boiling point of water. That is why for recipes that carry some liquid (cake or bread, for example) it is necessary to modify both the ingredients and the baking time.

To better understand this, let's begin by explaining that the boiling point of water at an atmospheric pressure at sea level is 100\°C and as the height increases and the atmospheric pressure decreases, this boiling point  will diminish, as well.

This means the water will not boil at the same temperature at different pressures, because at lower pressure the boiling temperature will be lower.

In other words, as at this point the water boils at a lower temperature, it <u>heats less</u>, so that the <u>food takes longer to cook </u>(because it is necessary to increase the cooking time in order to have a well done food).

This is because as the height increases, the atmosphere becomes drier, the air has less oxygen and moisture evaporates quickly, therefore, it is sometimes necessary to add more water to compensate this evaporation during cooking.

Hence, at higher altitudes, the cooking time will be longer.

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An ocean wave traveling in one direction has a wavelength of 1.0 m and a frequency of 1.25 Hz. Take the direction of wave propag
aliina [53]

Answer:

a) v=1*1.25=1.25\: m/s

b) y(x,t)=2sin(2\pi( x-1.25 t)  

c) y(3,10)=1.73 m    

Explanation:

a) The speed of a wave is given by the following equation:

v=\lambda f

Where:

λ is the wavelength

f is the frequency

v=1*1.25=1.25\: m/s

b) The harmonic wave has the following equation:

y=Asin(kx-\omega t)

A is the amplitude (2 m)

k is the wavenumber (2π/λ)  

ω is the angular frequency (2πf)

y(x,t)=2sin(2\pi x-2\pi*1.25 t)  

y(x,t)=2sin(2\pi( x-1.25 t)  

c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).

y(3,10)=2sin(2\pi(3-1.25*10)

y(3,10)=2sin(2\pi(3-1.25*10)              

y(3,10)=1.73 m              

I hope it helps you!

 

6 0
3 years ago
The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

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(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

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