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2 years ago

Under which circumstances would protesters’ right to assembly most likely be protected?

Physics

1 answer:

ICE Princess25 [194]2 years ago

7 0

Answer:

Explanation:

Mob violence is never protected. Organizers are always careful about not letting things get out of hand. Not D

That remark sort of applies to C as well. Property destruction is never protected.

B is something assemblies are allowed to do. So B.

A falls into the same category as C and D. You may not like the fact that he/she is doing something illegal, but you can't take away their privacy.

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Can someone tell me what my displacement is?

natulia [17]

60017 is or answer ok you got it

3 0

3 years ago

A spacecraft travels at 1.5 X 108 m/s relative to Earth. A process onboard the

kvasek [131]

Answer:

73.6 minutes

Explanation:

relative time = time interval / √(1 - observer velocity² / speed of light²)

we have relative time. we want time interval.

rearrange

time interval = relative time x √(1 - observer velocity² / speed of light²)

convert 85 mins into seconds

85 x 60 = 5100

1.5 x 10⁸ as a number is 150000000

for c = 299 792 458

time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)

for c = 3 x 10⁸

time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)

time interval = 5100 x 0.866

time interval = 4415.71

divide by 60 for back into minutes

time = 73.6 minutes

4 0

2 years ago

Graphing Motion

KIM [24]

The answer will be D

3 0

3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

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

Where:

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

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

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

Then,

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

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

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

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

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

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

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

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

A cylinder of compressed gas has a pressure of 488.2 kPa. The next day the cylinder of gas has a

Luda [366]

Answer:

20 °C

Explanation:

Ideal gas law:

PV = nRT

Rearranging:

P / T = nR / V

Since n, R, and V are constant:

P₁ / T₁ = P₂ / T₂

488.2 kPa / T = 468 kPa / 281.15 K

T = 293.29 K

T = 20.1 °C

Rounded, the temperature was 20 °C.

6 0

3 years ago