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Julli [10]
3 years ago
11

(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of +93 m/s. What is the average acceleration

of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?
Physics
1 answer:
Doss [256]3 years ago
4 0
A. the average acceleration would be 0 since there is no change in velocity

b. Average acceleration = change in velocity/time

93- (-93)/12

= 186 /12

= 15.5 m

hope this helps

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A speaker generates a continuous tone of 440 Hz. In the drawing, sound travels into a tube that splits into two segments, one lo
chubhunter [2.5K]

Answer:

The minimum difference between the lengths of the two tubes should be 0.385 meters.

Explanation:

As we known that for any two waves to arrive in phase at any point the difference in the path traveled by the waves should be an integral multiple of the wavelength of the wave.

Mathematically we can write:

\Delta x=n\frac{\lambda }{2}

For the given wave we have

\lambda =\frac{v}{\nu }

Applying values we get

\lambda =\frac{339}{440 }=0.77m

Thus the minimum difference in the lengths of the tubes can be obtained by putting the value of n = 1

\therefore \Delta x=1\times \frac{0.77}{2}=0.385m

7 0
4 years ago
Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

B = B_o(-\hat k)

now the velocity is from bottom to top

so we have

v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

now we have

\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

\vec F = e v_o B \hat i

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7 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

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4 years ago
Which energy drives convection within the ocean
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4 years ago
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