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3 years ago

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A particle moves along a line so that its position at any time is t>0 or t=0 is given by the function s(t)=(t^3)-(6t^2)+(8t)+

2 where s is measured in meters and t is measured in seconds. a) find instantaneous velocity at any time t. b) find acceleration of the particle at any time t. c) when is the particle at rest?. d) describe motion of the particle. at what values of t does the particle change directions

1 answer:

Butoxors [25]3 years ago

6 0

A. The instantaneous velocity at any time is 3t^2 -12t +8.
b. The acceleration of the particle at any time is 6t - 12.
c. The acceleration when particle is at rest is 3t^2 -12t+8 = 0.
d. The particle travels like a cubic graph.

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The momentum of an object is determined to be 7.2 x 10-3 cm kg x m/s. Express this as provided or use any equivalent unit. How i

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Complete Question:

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g).

Answer:

7.2 gm/s.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

Given the following data;

Momentum = 7.2 * 10^-3 kgm/s

1 kg = 1000 g

Substituting the unit in kilograms with grams, we have;

Momentum = 7.2 * 10^-3 * 1000 gm/s

<em>Momentum = 7.2 gm/s. </em>

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0.4 sec the frequency equals

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An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (cons

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A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m

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Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

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After collision,

Velocity = 2.0 m/s

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}$

$-3i+2j=3.0\times v_{2}$

$v_{2}=-i+0.66j$

The direction of the momentum

$tan\theta=\dfrac{0.66}{-1}$

$\theta=tan^{-1}\dfrac{0.66}{-1}$

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The direction of the momentum with respect to east

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A person pushes two boxes with a horizontal force F of magnitude of 100 N.

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The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".

According to the question,

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$m_A = 9\ kg$

Mass of box B,

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$F_{app} = 100 \ N$

From the Newton's law,

→ $F_{app} = (\frac{F_{app}}{m_A+m_B} )a$

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Bu substituting the values, we get

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We can see that between the two boxes, the action-reaction pair exist.

then,

→ $F_{action-reaction} = m_b \ a$

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→ $= 18.2 \ N$ (magnitude)

Thus the above solution is appropriate.

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