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KonstantinChe [14]

3 years ago

13

You are out camping. As it gets dark, it starts to get cold so you light a fire. What are two energy transformations that explai

n the light and the heat given off by the fire?

1 answer:

I am Lyosha [343]3 years ago

4 0

Conduction and convection you gave off energy to start the fire and the fire gives off heat to you

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10. A force F S 1 of magnitude 6.00 units acts on an object at the ori- gin in a direction u 5 30.0° above the positive x axis (

MArishka [77]

Answer:

The x component of the x axis is: $F_x = F_1\cos(30)=5\cos(30)= 4.33 u$

The y component of the y axis is: $F_y =F_2+ F_1\sin(30)=6.00+5\sin(30)= 8.50 u$

The magnitude is given by: $F = \sqrt{F_x+F_y}= 9.54 u$

The direction is given by: $\theta = \tan^{-1}\frac{F_y}{F_x}= 63.0 ^\circ$

See the graphical magnitude and direction below.

Explanation:

The x component of the x axis is: $F_x = F_1\cos(30)=5\cos(30)= 4.33 u$

The y component of the y axis is: $F_y =F_2+ F_1\sin(30)=6.00+5\sin(30)= 8.50 u$

The magnitude is given by: $F = \sqrt{F_x+F_y}= 9.54 u$

The direction is given by: $\theta = \tan^{-1}\frac{F_y}{F_x}= 63.0 ^\circ$

8 0

2 years ago

What is the difference between condensation and conduction

aniked [119]

Answer:

Convection is the heat transfer due to the bulk movement of molecules within fluids such as gases and liquids, including molten rock. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. These clouds may produce precipitation, which is the primary route for water to return to the Earth's surface within the water cycle.

Explanation:

Their a difference....... A huge One

8 0

3 years ago

Which of the following is an example of technology influencing science?

ElenaW [278]

I would say your answer is A.

7 0

2 years ago

Read 2 more answers

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

<u>r < a:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

<u>r = a:</u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>a < r < b:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>r = b:</u>

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

<u>r < c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u>At r = b:</u>

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>At r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

PHYSICAL SCIENCE HELP PLEASE !!!!! Picture below is for question 4

Eva8 [605]

Question 1) excited and moving around a lot.

Question 2) heat

Question 3) specific heat

Question 4) As temperature increases, energy transferred increases.

Hope these answers help!

3 0

3 years ago