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sukhopar [10]
3 years ago
6

2. What is the speed of a car that travels 73.4 kilometers (km) in 5 hours?

Physics
2 answers:
maksim [4K]3 years ago
7 0
Speed =dist./time
=73.4/5
=14.68 km/hr
horrorfan [7]3 years ago
7 0

me when i find out there's ice cream in the fridge.

:)

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Using the formula:


a = (Vf - Vi) / t


Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:


a = (8.15 - 0.0) / 5

a = 1.63 m/s^2


If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

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What was done to test if Comets are responsible for 1/2 the Earth's water?
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✯Hello✯

↪  A satellite was crashed into a comet (on purpose of course)

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3 years ago
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Explain your problem of<br>Federalism<br>​
telo118 [61]

That is more of a History of English question.

6 0
3 years ago
You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend
Bogdan [553]

Answer:

a.\tau=200J b.\alpha=0.44 \frac{rad}{s^2} c. \alpha=0.33\frac{rad}{s^2} d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by \tau=rF\sin \theta, being r the radius, F the force aplied and \theta the angle between the vector force and the vector radius. Since \theta=90^{\circ}, \, \sin\theta=1 and so \tau=rF=2m100N=200Nm=200J.

b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}.

c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0
3 years ago
Students with disabilities who require assistance in class are generally mainstreamed into regular classes rather than putting t
Brilliant_brown [7]
HELLO!

THIS STATEMENT IS T) TRUEEE

DONT FORGET TO RATE
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3 years ago
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