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3 years ago

Which has more heat, a home oven at 500 degrees Celsius, or a one ton commercial oven at 500 degrees celsius?

Physics

1 answer:

alisha [4.7K]3 years ago

8 0

A commercial oven because it is more durable plus it’s more professional so they use more heat then we do at home

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Graph the following data tables on different graphs.

Anna [14]

Answer:

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3 years ago

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

4 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

4 years ago

Which answer correctly describes the grey lines in this Hering illusion? (Optical Illusions Lesson)

Ratling [72]

The answer that correctly describes the grey lines in this Hering illusion is "The grey lines are bent." Option A. This is further explained below

<h3>What is the Hering illusion?</h3>

The Hering Illusion is one of several illusions in which a major component of a basic line picture is obscured.

In conclusion, The statement for the grey lines in this Hering illusion is "The grey lines are twisted.".