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Harrizon [31]
2 years ago
8

A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?
Physics
1 answer:
kenny6666 [7]2 years ago
5 0

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, N - mg = F_c

N-mg=\frac{mv^2}{r}

Now substitute the given values, we get

N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}

N=15.68+46.08

N = 61.76  N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

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Answer:

tan \theta = \mu_s

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An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

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\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

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mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

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4 0
3 years ago
A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo
kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

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• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

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• at a point a distance d to the right of the pivot point,

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∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

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6 0
2 years ago
Why is an object's density expressed as a relationship between two units?
vfiekz [6]
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ElenaW [278]

Answer:

Explanation:

right

8 0
2 years ago
Calculate the critical angle for Zircon which has a refractive index of n = 2*
MA_775_DIABLO [31]

Answer:

Given that refractive index of the material is √2. i.e. n = √2. Hence, critical angle for the material is 45°

8 0
3 years ago
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