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LenaWriter [7]
3 years ago
12

Is it possible for an object’s instantaneous velocity and instantaneous acceleration to be of opposite sign at some instant of t

ime?
Physics
2 answers:
UkoKoshka [18]3 years ago
8 0
 <span>Of course. 

Throw a ball into the air. Right after you release it, the velocity is upward, but the acceleration is downward. 

Later, the velocity falls to zero; the acceleration is still downward. 

Still later, the velocity goes downward; the acceleration is still downward.</span>
telo118 [61]3 years ago
3 0

Answer:

Yes

Explanation:

First Example:

- Throwing the ball up in the air with an upward velocity e.g 5 m/s. The ball moves in the direction of velocity. While the acceleration due to gravity is g = -9.81 m/s^2 is pointing downwards throughout the journey.

Second Example:

- The car at a certain speed decelerates down the road. It has certain positive velocity but after the driver applies or car slows down the acceleration is in the opposite direction to the motion or negative.  

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A uniform crate with a massof 30 kg must be moved up along the 15° incline without tipping. Knowing that force P is horizontal,
lakkis [162]

Answer:

coefficient of friction =0.268

magnitude of force P=289.78N

Explanation:

The coefficient of friction is obtained by mgsinФ/mgcosФ=tanФ=tan15=0.268

force P is horizontal as stated in the question, horizontal component of P=mgcosФ=30*10*cos15=289.78

8 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}.

7 0
3 years ago
If the volume of one drop is 0.031 mL according to Stu Dent’s measurement, approximately what volume would 22 drops be? Answer w
olasank [31]

Answer:

Volume of 22 drop will be 0.68 ml

Explanation:

We have given volume of one drop = 0.031 ml

We know that 1 liter = 1000 ml

So 1ml=10^{-3}L

So 0.031 ml will be equal to 0.031\times 10^{-3}L

We have to find the volume of 22 drop

For finding volume of 22 drop we have to multiply volume of one drop by 22

So volume of 22 drop will be =22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL

So volume of 22 drop will be 0.68 ml

7 0
4 years ago
Ozone, O3 methane, CH4 oxygen gas, O2 Are classified as
erma4kov [3.2K]
I would think compounds.
5 0
3 years ago
Read 2 more answers
A wire as a length of 1.50m, diameter 0.60mm and resistance of 2ohms. calculate the resistance R of a wire of the same materials
Otrada [13]
  1. R=ρ×L/A

ρ=R×A/L

A=pie(r²)=pie(0.6²)=pie(0.36)=1.13

ρ=2ohm×1.13mm²/1500mm

ρ=0.0015ohm.mm

2. we were told that the wires were made of the same substance so the resistivity(ρ)is the same for both wires so:

R=ρ×L/A

R=0.0015ohm.mm×500mm/0.09mm²

R=8.3'ohm

so our resistance for the second wire is :

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>

3 0
3 years ago
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