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3 years ago

A ring, seen from above, is pulled on by three forces. the ring is not moving. how big is the force f

1 answer:

7 0

All we know is that when the three force vectors are added up, their sum is zero. We don't know anything about them individually.

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Coal and other fossil fuels are considered to be "nonrenewable" sources of energy. How does the use of nonrenewable fuels affect

nignag [31]

Answer: D Although the total energy remains constant, nonrenewable fuels convert chemical energy into forms that are difficult or impossible to use again.

Explanation:

The first law of thermodynamics says that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.

7 0

3 years ago

At what speed, as a fraction of c, does a moving clock tick at four fifth the rate of an identical clock at rest?

ololo11 [35]

Answer:

The seed as a fraction of the speed of light is $\frac{3}{5}c$

Solution:

As per the question:

Suppose, $t_{i}$ be the rate of an identical clock between two time intervals.

For a moving clock, moving with velocity 'v', at the clock tick of four-fifth:

t = $\frac{5}{4}t_{i}$

Now,

Using the relation of time dilation, from Einstein's relation:

$t = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$\frac{5}{4}t_{i} = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

Squaring both sides:

$(\frac{5}{4})^{2} = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}})^{2}$

$\frac{25}{16} = \frac{1}{{1 - \frac{v^{2}}{c^{2}}}}$

$1 - \frac{16}{25} = \frac{v^{2}}{c^{2}}$

$\frac{v}{c} = \sqrt{\frac{9}{25}}$

$\frac{v}{c} = \frac{3}{5}$

$v = \frac{3}{5}c$

6 0

3 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 $\alpha$ :

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