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Annette [7]
3 years ago
11

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

form electric field. What is the magnitude of the electric field if the field points 31.6° above the horizontal?
Physics
1 answer:
Tresset [83]3 years ago
4 0

Answer:231.16 N/C

Explanation:

Given

Electric Flux=147 N-m^2/C

Area(A)=0.824 m^2

Given Field point above 31.6 ^{\circ}

Therefore angle between Area vector Electric Field =90-31.6=58.4^{\circ}

We know that Flux is given by

\phi =\vec{E}\cdot \vec{A}

\phi =EAcos\theta

147=E\times 0.824\times cos(58.4)

E=231.16 N/C

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Lilit [14]
Electrons: negative charge
Protons: positive charge
Neutrons: negative charge

The atom would have to have more electrons than protons

Hope this helps :)
3 0
2 years ago
When did humans learn that the earth is not the center of the universe?
Vinvika [58]

Answer:

When did humans learn that the Earth is not the center of the universe?

Answer

1

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4 Answers

Asked in 3 Spaces





Science - Next Generation

Alexander Somm

, Consultant, Investor Relations at Novelpharm AG (2015-present)

Answered Oct 16

What, it isn’t?!

Sorry, I had to.

As far as I have read and understood, the Sumerians and later the Babylonians both had astronomical calendars that already differentiated planets and stars. Earth was not the center to them, the Sun likely was. That was around 2,200 - 1,600 BC.

After that, Greek philosopher Aristarchus of Samos (310 - 230 BC) was the first (recorded) to have believed the solar system was organized around the Sun, rather than the Earth. His heliocentric model was unpopular during Aristarchus’ lifetime, although it would inspire astronomers centuries later, such as Copernicus and Galileo.

Now, there are numerous archeological findings (cave paintings) and studies, that all suggest an understanding of complex astronomy in prehistoric times dating back as far as 40,000 years. This also explains how early, prehistoric migrants may have navigated the seas.

Explanation:

hope it helps

have a good day

4 0
2 years ago
The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an
Readme [11.4K]
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
F=8.99\cdot10^{9} Nm^2C^{-2}  \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
7 0
3 years ago
So u see what had happened was i need help again..
liubo4ka [24]

first off lemme just say this is really easy man, just look at the directions

Blank #1: -23

Blank #2: 23

8 0
2 years ago
A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when
ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, V_T = 60 \ m/s

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

F_D = kV_T^2

Where;

k is a constant

k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}

When the new drag force is half of the original drag force;

F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}=  \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}=  0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s

Therefore, the speed of the resistive force is 42.426 m/s

8 0
3 years ago
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