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3 years ago

This tasty snack starts as a kernel and is “popped” then

1 answer:

6 0

Popcorn can be popped by either of the three forms of heat transfer:

Conduction in a pan with hot oil on a stove element.

Convection by an air popper and warm air rising over a heating element... no direct contact with heat source.

Radiation is what occurs in a microwave. Invisible radiant heat activates water molecules in the popcorn.

All the above heat the kernel over 100 Celsius. Water vaporizes/boils (latent heat) and erupts through the kernel.

Mmmm popcorn watch out for the lipids (fat in the oil and butter)

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1.How far is Object Z from the origin at t = 3 seconds?

GenaCL600 [577]

Answer:

Please find the answer in the explanation

Explanation:

1.) How far is Object Z from the origin at t = 3 seconds

The distance of the object Z from the origin will be the slope of the graph.

Slope = 4/2 = 2m

2.) Which object takes the least time to reach a position 4 meters from the origin ?

According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.

3.) Which object is farthest from the origin at t = 2 seconds?

The correct answer is still object Z because it has the highest slope.

4 0

3 years ago

What is most likely to happen to light that hits an opaque object?

ddd [48]

B. is the answer.

C is not correct because the light is actually reflected off of an opaque object.

4 0

3 years ago

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

-BARSIC- [3]

Answer:

4.2s

Explanation:

Given parameters:

Power = 2190W

Mass of box = 1.47 x 10⁴g

distance = 6.34 x 10⁴mm

Unknown:

Time = ?

Solution:

Power is the rate at which work is done;

Mathematically;

Power = $\frac{work done}{time}$

Time = $\frac{work done}{power}$

Work done = weight x height

convert mass to kg;

100g = 1kg;

1.47 x 10⁴g = 14.7kg

convert the height to m;

1000mm = 1m

6.34 x 10⁴mm gives 63.4m

Work done = 14.7 x 9.8 x 63.4 = 9133.4J

Time taken = $\frac{9133.4}{2190}$ = 4.2s

6 0

3 years ago

A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K

Zinaida [17]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 </span>J/g•K

7 0

3 years ago

Read 2 more answers

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago