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krok68 [10]
3 years ago
9

This graph indicates that the acceleration between 2.00 and 4.00 seconds is the same as that between 12.00 and 14.00 seconds.

Physics
1 answer:
adoni [48]3 years ago
5 0
I believe it’s false I’m probably wrong I ju at need points for my test
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Which of these ideas was part of the earliest model of the atom?
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The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

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3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
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Marysya12 [62]

D only kinetic energt because if it had heluim or any thing jt will float up

4 0
3 years ago
If the 3rd harmonic has a frequency of 600Hz, the frequency of the fundamental is ____________?
castortr0y [4]

Answer:

Fundamental frequency is 200 Hz.

Explanation:

It is given that,

Frequency of the 3rd harmonic is 600 Hz.

Let f is the fundamental frequency. We need to find the value of f. The frequency of third harmonic is given by :

3f=600\ Hz

So, fundamental frequency f is equal to :

f = 200 Hz

So, the fundamental frequency of the harmonics is 200 Hz. Hence, this is the required solution.

4 0
4 years ago
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