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3 years ago

What factors affect potential energy

Physics

2 answers:

jok3333 [9.3K]3 years ago

6 0

Mass ,gravity and height

Ugo [173]3 years ago

3 0

Gravitational potential energy is determined by three factors : mass, gravity and height . All three factors are directly proportional to energy

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30. The slope of a velocity-time graph will give

CaHeK987 [17]

Answer:

The acceleration

Explanation:

The slope of velocity time graph is the acceleration

5 0

3 years ago

According to Hooke's law

SOVA2 [1]

Answer:

According to Hook's law, we know,

strain/stress =Constant

Explanation: So, the ratio between stress and strain is always constant.

So, if stress is increased, then strain changes in that way so that this ratio always remains constant.

7 0

3 years ago

If Charge A is positive, then Charge B must be:

Eduardwww [97]

It must be positive also
Explanation The energy is coming out of A back into b and then going into A again and out

4 0

2 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

A metal sphere has a charge of +10C. What is the net charge after 8.0 x 10^13 electrons have been placed on it?

masya89 [10]

Answer: 9.9999872 C

Explanation: In to answer this question we have to use the charge of the electron, that is eqaul to -1.6*10^-19 C.

After that, we have to calculate the charge given by 8.0*10^13 electrons, then we an additional charge of: 8.0*10^13 * -1,6*10^-19 C=1.28*10^-5C

Finally the net charge of the metal sphere, initially charged by +10C is:

10C-1.28*10^-5C=9.9999872 C

3 0

4 years ago