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OleMash [197]
4 years ago
14

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

A = 23° above the x axis, and the angle θB = 53° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s acceleration when it is at point A
3) What is the y component of the car’s acceleration when it is at point A
4) What is the x component of the car’s acceleration when it is at point B
5) What is the y component of the car’s acceleration when it is at point B
Physics
1 answer:
Fantom [35]4 years ago
6 0

Centripetal acceleration of car is given by formula

a_c = \frac{v^2}{R}

now plug in the values in this

a_c = \frac{23^2}{210}

a_c = 2.52 m/s^2

Part b)

At position A we have

x component of acceleration is given as

a_x = -a_c cos23

a_x = -2.32 m/s^2

Part c)

At position A we have

y component of acceleration is given as

a_y = -a_c sin23

a_y = -0.98 m/s^2

Part d)

At position B we have

x component of acceleration is given as

a_x = -a_c cos53

a_x = -1.52 m/s^2

Part e)

At position B we have

Y component of acceleration is given as

a_y = a_c sin53

a_y = 2.01 m/s^2

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Firlakuza [10]

Answer:

Explanation:

Both these questions are based on the Universal Law of Gravitation, which is given by :

F = Gm1m2 / r²

2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5

   F = 6.67 x 10⁻⁵ x 8 / 1.5

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3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴

   F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29

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stiv31 [10]

Answer:

300 Newtons

Explanation:

Weight is the force of attraction between two bodies, one usually larger (like a planet), and one smaller (like a person). Force can be calculated using the formula: Force = mass × acceleration.

The mass here is 12kg, the acceleration, which in this case, is the acceleration due to gravity is 25m/s/s, by plugging in our values, we have

Force = 12 × 25 = 300 Newtons or 300 N for short.

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