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GenaCL600 [577]
3 years ago
10

What is M=35g, V=17cm³​

Physics
1 answer:
cestrela7 [59]3 years ago
6 0

Answer: 25

Explanation:

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Which of the following is a circuit component whose function includes changing an open circuit to a closed circuit?
blondinia [14]

Answer:

<h2>B. Switch</h2>

Explanation:

<u>A device designed to open or close a circuit under controlled conditions is called a switch. The terms “open” and “closed” refer to switches as well as entire circuits. An open switch is one without continuity: current cannot flow through it.</u>

<h2><u>Hope this helps! Please consider marking brainliest!! </u></h2>
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3 years ago
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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
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<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>

<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>

<span>t = √(2y/g) </span>

<span>in the ft - lb - s system </span>

<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>

<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
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Derive the unit of intensity???
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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON
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