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grin007 [14]
2 years ago
11

Which qualifications are typical for a Manufacturing career? Check all that apply.

Physics
1 answer:
kicyunya [14]2 years ago
8 0
D should be the answer
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Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.
Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0
2 years ago
What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)
mariarad [96]

<u>Answer:</u>

<em>The correct equation for measuring the average microscopic weight  for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>

<u>Explanation:</u>

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

  • Take the correct weight of each isotope (that will be in decimal form)
  • Multiply the weight of each isotope by its abundance
  • Add each of the results together.

<em>This gives the required  average microscopic weight of the three isotopes.</em>

3 0
2 years ago
What is one latitude where there is no continental barriers?
AveGali [126]
The equator has no continental borders.
5 0
3 years ago
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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
Elements that easily transmit electricity and heat display the property know as ______
andre [41]
It is known as conductivity
7 0
3 years ago
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