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1 year ago

calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.

Physics

1 answer:

WARRIOR [948]1 year ago

8 0

Answer:

216,890 J

Explanation:

We first define the Kinetic Energy Formula:

K.E =

$\frac{1}{2} m {v}^{2}$

where m is the mass of the object in kg, and v is the velocity in m/s.

Given from the information from the question, we substitute the values in to find K.E.

K.E of car =

$\frac{1}{2} (820) {(23)}^{2} \\ = 216890 \: joules$

Note that the SI unit for Energy is Joules or J.

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3 years ago

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1. What was the Michelson-Morley experiment designed to do?2. When was the Michelson-Morley experiment done?3. What was the ethe

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Answer:

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4 years ago

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Answer:

Part a)

$KE = 77.95 J$

Part b)

$L = 3.16 m$

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

$KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2$

$KE = \frac{7}{10} mv^2$

$KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)$

$KE = 77.95 J$

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

$mgLsin\theta = KE$

$\frac{42}{9.81}(9.81)L sin36 = 77.95$

$L = 3.16 m$

Part c)

by equation of energy conservation we know that

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3 years ago

A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5

podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity ( $\omega_o$ ) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

$\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\$

$\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s$

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

$\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)} =112.4\ rev$

θ = 323 rad

4 0

3 years ago

￼calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.

calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.