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andrew11 [14]
3 years ago
10

you cover 10 meters in a time of 1 second .Is your speed the same if you cover 20 meters in 2 seconds?

Physics
1 answer:
IrinaVladis [17]3 years ago
8 0

Definition:
                 Speed = (distance covered) / (time to cover the distance).

Speed in the first example = (10 meters) / (1 second) = 10 m/s .

Speed in the second example = (20 meters) / (2 seconds) = 10 m/s.

Apparently, it is.  This little exercise demonstrates it.

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Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

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sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

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Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

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Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

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Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

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Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

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Answer:

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If the acceleration is negative, the net force is negative.

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