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3 years ago

7

Saki is ice-staking to the left with velocity of 5 m/s. A steady breeze starts blowing. Causing saki to accelerate leftward at a

constant rate of 1.5 m/s2. After 6s , what is saki velocity

1 answer:

Pavlova-9 [17]3 years ago

4 0

Answer:

Final velocity will be equal to 14 m/sec

Explanation:

We have given initial velocity u = 5 m/sec

Constant acceleration is given $a=1.5m/sec^2$

Time t = 6 sec

We have to find the final velocity

From first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity , a is acceleration and t is time

So $v=5+1.5\times 6=14m/sec$

So equal final velocity will be equal to 14 m/sec

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An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x

romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

$v(x)=y'(x)$

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

$a(x)=v'(x)=y''(x)$

We are given the function for y

$y(x)=3.0+4.8x +6.4x^2$

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

$v(x)=y'(x)=4.8+12.8x$

And the acceleration is $a(x)=v'(x)=12.8$

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3 years ago

Who represents the worst of the iron age in his attempt to feed jupiter once living flesh?

wolverine [178]

Lycaon represents the worst of the iron age in his attempt to feed jupiter once living flesh.

<h3>What is iron age?</h3>

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It was preceded by the Stone Age, and the Bronze Age.

Who is lycaon?

In Greek mythology, Lycaon was a king of Arcadia who, in the most popular version of the myth, tested Zeus' omniscience by serving him the roasted flesh of Lycaon's own son Nyctimus, in order to see whether Zeus was truly all-knowing.

Thus, Lycaon represents the worst of the iron age in his attempt to feed jupiter once living flesh.

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4 0

2 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\theta-\theta_{o} = 435\,rad$ , the angular deceleration is:

$\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}$

$\alpha = -8.780\,\frac{rad}{s^{2}}$

Now, the time interval of the Deceleration Phase is obtained from this formula:

$\omega = \omega_{o} + \alpha \cdot t$

$t = \frac{\omega - \omega_{o}}{\alpha}$

If $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\alpha = -8.780\,\frac{rad}{s^{2}}$ , the time interval is:

$t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }$

$t = 9.954\,s$

The total time needed for the grinding wheel before stopping is:

$t_{T} = 2.40\,s + 9.954\,s$

$t_{T} = 12.354\,s$

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

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3 years ago

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb, stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

$W =k\int_{6}^{10.1} x$

$W = \dfrac{k}{2}[x^2]_6^{10.1}$

$W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)$

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

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3 years ago

What are five atomic models that have been proposed over time ?

Art [367]

Air, water , atomic acid , and pocket rocket

5 0

3 years ago