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3 years ago

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45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d

isplacement at t = 5 s? (b) What is its velocity at this same time?

2 answers:

omeli [17]3 years ago

7 0

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

Brut [27]3 years ago

7 0

Answer:

Velocity at 5 seconds: 180 m/s

Ditance: 525 meters

Explanation:

In order to calculate displacement in this option we need to use the formula for distance when an object is experimenting aceleration:

$s=VoT+1/2at^2$

So now we just have to insert the data that we know:

$s=VoT+1/2at^2\\s= 30*5+ 1/2*30*25\\s=150+15*25\\s= 150+375\\s=525 m$

The distance after 5 seconds with an initial velocity of 30 m/s and an acceleration of 30 m/s will be 525 meters.

And the final velocity is calculated by the next formula:

$Vf= Vo+ A*t\\Vf= 30 m/s+ 30m/s*5\\Vf=180m/s$

So the final velocity at 5s would be 180 m/s

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

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choli [55]

Answer:

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Explanation:

The positive charge feels a force in the same direction as the electric field

F=Eq

F and E are vectors, q is a scalar

(if it were a negative charge the force would be in the opposite direction)

that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.

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You drive a car east on the highway at 26 m/s. Another car passes you moving east traveling at 32 m/s. How fast do you view the

tamaranim1 [39]

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<h3>Relative velocity of the car</h3>

The speed of the car passing you is determined by applying relative velocity principle as shown below;

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Vr = 26 m/s - 32 m/s

Vr = -6 m/s

Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

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Answer:

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