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Leto [7]
3 years ago
8

The maximum centripetal acceleration a car can sustain without skidding out of a curved path is 9.40 m/s2 . If the car is travel

ing at a constant speed of 40.0 m/s on level ground, what is the radius R of the tightest unbanked curve it can negotiate?
Physics
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer:

The radius = 170.21 m

Explanation:

The given data are : -

The centripetal acceleration of a car = 9.40 m/s².

Speed of a car = 40.0 m/s .

We have to calculate the radius ( r ) of of curve.

The centripetal acceleration ( a ) is given by

a = \frac{v^{2} }{r}

r = \frac{v^{2} }{a}  = \frac{40^{2} }{9.40}  = \frac{1600}{9.40} = 170.21 m

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2 years ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
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F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

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m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

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FN : Normal force : perpendicular to the floor

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Ty = Tsin θ = T*sin 24.7°  (N)

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∑Fy = m*ay ay = 0  

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We apply the formula (2) :

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Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

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We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

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