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FinnZ [79.3K]
3 years ago
14

Luke Autbeloe drops an approximately 5.0 kg object (weight = 50.0 N) off the roof of his house into the swimming pool below. Upo

n encountering the pool, the object encounters a 50.0 N upward resistance force (assumed to be constant). Use this description to answer the following questions. (Down is usually considered a negative direction) a. Which one of the velocity-time graphs best describes the motion of the object? Why?

Physics
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

Graph B represents correct graph for velocity time

Explanation:

Initially the object is dropped under the influence of constant force due to gravity

So here we can say its acceleration is constant due to gravity

hence here the velocity will increase downwards due to gravity and it is given as

v = 0 + (-10)t

now when it enters into the pool an upward force of same magnitude will act on it

so net force on it is

F = 50 - 50 = 0

so here we will have no acceleration and it will move with constant speed after that

so here correct graph is

Graph B

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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
An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?
Leya [2.2K]
41.2 = h-1/2g(t-1)^2 
<span> {-h = -1/2gt^2-1/2g+g*t-41.2 
</span><span> {h = 1/2gt^2 
</span><span> summing them up 
</span><span> 0 = -1/2g+g*t-41.2 
</span><span> 41.2 +4.9 = g*t 
</span><span> t = 46.1/9.8 = 4.70 sec 
</span><span> h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m </span>
3 0
3 years ago
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