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2 years ago

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A 5.0-n projectile leaves the ground with a kinetic energy of 220 j. at the highest point in its trajectory, its kinetic energy

is 120 j. to what vertical height, relative to its launch point, did it rise if there was no air resistance

1 answer:

NikAS [45]2 years ago

3 0

First, we get the difference between the kinetic energies such that,
difference = (220J - 120J)
difference = 100 J
The difference in kinetic energy is the equivalent of the potential energy which is calculated through the equation,
PE = mgh
To calculate for the height, we derive the equation in a form,
h = PE/mg
The product of the mass and acceleration due to gravity is the weight.
h = (100 J) / (5 N)
h = 20 m

<em>Hence, the answer is 20 m. </em>

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5.5 rad/sec

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Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

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Answer:

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Answer:

Part a)

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Part b)

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Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

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$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

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number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

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Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

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