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3 years ago

An object has a positive velocity to the right and a negative acceleration. what is the motion of the object?

1 answer:

5 0

The object's motion toward the right is slowing down ... the magnitude of his velocity is decreasing. If this goes on long enough, his velocity will dwindle to zero, and then become negative ... he'll start moving to the left.

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10 poin<br> What is the momentum of a 25kg cart traveling left with a velocity of 10<br> m/s?

Mashutka [201]

250 kg*m/s
Momentum is found by multiplying mass by velocity
25*10=250

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3 years ago

6. What is the charge of an object that lost 1.20x1024 electrons?

Montano1993 [528]

Answer:

Hydrogen ions contains +1 charge while sulphate ions have 2 charge. ... No .of ions in one formula unit of MgC12 =3 Total no of ions in MgC12 =1.20x1024 x3=3.6x1024 ions.

Explanation:

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3 years ago

How long does the president serve a term in office

Ilya [14]

Answer:

United States Electoral College to a four-year term, with a term limit of two terms (totaling eight years) or a maximum of ten years if the president acted as president for two years or less in a term

Explanation:

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3 years ago

Read 2 more answers

An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces a

marshall27 [118]

Refer to the free body diagram shown melow.

F = applied force
R = frictional force
m = 15 kg, the mass of the object

The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N

The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.

Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)

Answer:
The most reasonable answer is R = 5.5 N

8 0

3 years ago

Read 2 more answers

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