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Fittoniya [83]
3 years ago
15

Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f

rom point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
Physics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put  12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

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E) The centripetal force is always perpendicular to the velocity.

Explanation:

Due to gravity and inertia, the satellite follows a uniform circular motion. In this movement, the velocity is always tangent to the orbit and the centripetal force is directed towards the center. Therefore, there is no net acceleration in the same direction of velocity, which implies that it remains constant.

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What are some of the characteristics scientists are looking for when they search for other Earth-like planets?
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Which is the BEST description of how eyeglasses work?
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C I think.

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3 years ago
A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet
Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

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3 0
3 years ago
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

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