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fgiga [73]
3 years ago
12

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

car can safely travel if the radius of the track is 84.0 m and the coefficient of friction is 0.42?
Physics
2 answers:
blagie [28]3 years ago
7 0

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force F_c. In this case, the only force that applies for that is the static frictional force f_sbetween the tires and the track. Then, we can write that:

f_s=F_c

And since f_s\leq \mu N and F_c=\frac{mv^{2}}{r}, we have:

\mu N\geq \frac{mv^{2}}{r}

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

N-mg=0\\\\\\implies N=mg

Substituting this expression for N and solving for v, we get:

\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

KIM [24]3 years ago
3 0

Answer:

v=18.6 m/s

Explanation:

We need to equal the friction force and the centripetal force:

F_{f}=F_{c}

\mu N=ma_{c}

  • N is the normal force (N = mg)
  • μ is the coefficient of friction (0.42)
  • a(c) is the centripetal force (a_{c}=v^{2}/R)

\mu mg=m*\frac{v^{2}}{R}

\mu g=\frac{v^{2}}{R}

v=\sqrt{R\mu g}

v=\sqrt{84*0.42*9.81}

v=18.6 m/s

I hope it helps you!

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Answer:

The answer is below

Explanation:

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In conclusion, their difference is the area in which they happen.

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Can we write names while writing conversation in board exam​
lora16 [44]

Answer:

ya we can write the imaginary character's name .

So that we  can identify these imaginary people, as we cannot simply write the conversation and leave it .

Or maybe sometimes the reader will get confused as there is no name for the two people .

So, i suggest that you should write the names

Explanation:

You can even ask to your class teacher for further clarification

5 0
3 years ago
A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric
andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

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A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
lesantik [10]
Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
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Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

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