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GenaCL600 [577]

3 years ago

13

Which statement correctly describes how thermal energy tends to spontaneously flow

1 answer:

Nitella [24]3 years ago

3 0

#A for plato users, i took the test and got 100!!

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Which is the goal of technology

Tatiana [17]

TO bring joy and ease

8 0

4 years ago

Read 2 more answers

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

4 years ago

Which of the following best summarizes the quantum model of atoms?

Lelechka [254]

I believe it is A. Electrons reside in known positions in fixed orbits around the nucleus

6 0

3 years ago

Please help with this one

olasank [31]

Answer:

OPTION A is the correct answer

6 0

3 years ago

The acceleration due to gravity on or near the surface of Earth is 32 ft./s/s. Neglecting friction, from what height must a ston

svetoff [14.1K]

Given :

The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s

To Find :

From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.

Solution :

Initial velocity of stone, u = 0 ft/s.

Now, by equation of motion :

$2as = v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft$

Therefore, height from which stone is thrown is 289 ft.

3 0

3 years ago