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2 years ago

10

A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect

on the kinetic energy of the ball? Explain.

2 answers:

otez555 [7]2 years ago

7 0

Answer:

Yes

Explanation:

doubling up would have a greater effect.

Arte-miy333 [17]2 years ago

5 0

Answer:

Yes

Explanation:

Kinetic energy is K.E1/2mv2 so that means it is directly proportional to mass and velocity.

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A 1-megabit computer memory chip contains many 27 fF capacitors. Each capacitor has a plate area of 3.09 × 10−11 m 2 . Determine

valentina_108 [34]

Answer:

Plate separation of each capacitor is 101.132 °A

Explanation:

The formula to calculate the capacitance in empty space as a function of distance (square parallel plates) is:

$C=\epsilon_{0}\frac{Area}{distance}$

clearing for distance:

$distance=\epsilon_0 \frac{Area}{Capacitance} \\\\\epsilon_0=8.8542(10)^{-12}C^2/Nm\\\\Area=3.09(10)^{-11}m^2\\Capacitance=27(10)^{-15}F\\\\Replacing\\\\distance=\frac{8.8542(10)^{-12}*3.09(10)^{-11}}{27(10)^{-15}} =1.0133(10)^{-8}m\\\\In A\\distance= 101.132 A$

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3 years ago

Which of the following forms matter?<br> A. Proteins<br> B.atoms<br> C.cells<br> D. DNA

Artemon [7]

Answer:b) atoms

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2 years ago

Read 2 more answers

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe

Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

$F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}$

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

$I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A$

Therefore, the bottom current is 12.8 A to the right.

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Answer:

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