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Ghella [55]
3 years ago
14

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

s the driver speeding?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Answer:

(a) 16.777mi

(b)Yes, he was speeding

Explanation:

(a)

Let's do the proper operations in order to convert km to mi:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

We can conclude that the trip length in miles was:

d=16.77706389mi

(b)

Let's calculate the speed of the man during the trip:

v=\frac{d}{t}

But first, let's do the proper operations in order to convert min to h:

16min*\frac{1h}{60min} =2.666666667h

Now, the speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

As we can see:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

So, we can conclude that the driver was speeding

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Explanation:

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The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :

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Here, a = g

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v = 9.07 m/s

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Three identical metal spheres are hung from a ceiling on rigid non-conducting rods. Sphere 1, sphere 2, and sphere 3 have charge
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Answer:

a) The net electric force on the middle sphere due to spheres 1 and 3 is 404N. b) The direction of the net electric force is to the left.

Explanation:

In order to solve this problem, we must first draw our free body diagram (See attached picture).

By definition, two opposite charges attract each other, so the force due to sphere 1 will go to the left (being the left direction negative) and the force due to sphere 3 will go to the right (being the right direction positive).

Knowing this we can use the electric force formula to calculate each of the forces:

F_{e}=k_{e}\frac{q_{1}q_{2} }{r^{2} }

where k_{e}=8.99x10^{9}N\frac{m^{2} }{C^{2} }

So knowing this, we can now calculate each force. Let's start with the force exerted by sphere 1 over sphere 2:

F_{12}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(71x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }

Which gives me a force of:

F_{12}=-1148.48N

In this case the force will be negative because it's directed towards sphere 1, this is to the left.

We can do the same with the force due to sphere 3:

F_{23}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(46x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }

Which gives me a force of:

F_{23}=744.09N

In this case the force is positive because it's directed towards sphere 3, this is it goes to the right.

With these two values I can now find the net force electric force on the middle sphere due to spheres 1 and 3.

F_{net}=∑F

F_{net}=F_{12}+F_{23}

F_{net}=-1148.48N+744.09N

F_{net}=-404.39N

a) Since part a of the problem only asks us for the magnitude, then the net electric force on the middle sphere due to spheres 1 and 3 is 404.39N

b) since the answer when solving this problem was negative, this means that the force will be directed towards sphere 1, this is to the left.

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