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zepelin [54]
3 years ago
15

An investigation was done with an electromagnetic system made from a battery and wire wrapped around a nail. Different sizes of

nails were used. The number of paper clips the electromagnet could pick up was measured. What is the dependent variable in this experiment?
A) the size of nail used
B) the amount of paperclips picked up
C) the size of battery used
D) the type of wire used
Physics
1 answer:
alex41 [277]3 years ago
5 0
Dependant variable is something which you MEASURE during an experiment

So your answer would be : B
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Kipish [7]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

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One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. ope
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3 years ago
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.

4 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

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Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

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b). $T=2 \pi \sqrt{\frac{m}{k}}$

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c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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