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Nataly [62]
3 years ago
14

A particle is moving in a straight line.at t second, it acceleration is (4-kt), where k is a constant. when t=6, the acceleratio

n of a particle is zero determine its velocity as a function of time​
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

v = 4t - t²/3

Explanation:

4 - k*6 = 0

k = 4/6 = 2/3

dv/dt = 4- 2/3 t

integral of dv = integral of (4-2/3 t) dt

v = 4t - t²/3

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A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d
pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}

Now, plug in the given values and solve for 'a'. This gives,

a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}

Now, plug in the given values and solve for 'S'. This gives,

S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

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what type of current measurements should never be taken on circuits that may contain high currents or unknown currents?
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Explanation:

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Explanation:

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Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was
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This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:


Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².


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