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3 years ago

Energy stored between charged particles is called

Physics

2 answers:

olga nikolaevna [1]3 years ago

7 0

Electrical potential energy is the energy stored between charged particles.

inna [77]3 years ago

5 0

It's either static energy or kinetic energy, but I'm more for static energy

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_____ variables are manipulated by the experiments

nikitadnepr [17]

Answer:

dependent variables

Explanation:

dependent varibeles are the thing you're measuring and independent variables are the thing you change in the exeriment to get a different dependent variable.

may I get brainliest please? :)

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4 years ago

What sport is best for someone who is 5'10, 220 lbs?

tamaranim1 [39]

Football is because 5'10 is pretty tall and 220 would be hard to tackle

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3 years ago

Jill leaves home and rides a distance of 70 km. It took her 2.5 hours. What is her speed?

Paraphin [41]

The answer is ...
28 km per hour

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3 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.44 kg and rotate with

Free_Kalibri [48]

Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

$I= \frac{1}{2}mr^{2}$

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

$K= \frac{1}{2}I\omega ^{2}$

Here, K is the rotational kinetic energy and $\omega$ is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.356 m.

$I= \frac{1}{2}(3.44)(0.356)^{2}$

$I= 0.218 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 0.218 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(0.218)(430) ^{2}$

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.775 m.

$I= \frac{1}{2}(3.44)(0.775)^{2}$

$I= 1.03 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 1.03 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(1.03)(430) ^{2}$

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

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4 years ago

A 50kg box is pulled up a distance of 5.0m up a 30 degree incline by a 300. N force. If there is friction between the box and th

Finger [1]

Answer:

Explanation:

5 0

3 years ago