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3 years ago

15

If an aircraft slows down from 160 km/h to 10 km/h with a uniform of 5 m/s2

1 answer:

marusya05 [52]3 years ago

8 0

Explanation:

10km/h = 10×10/36=25/9m/sec

160km/h= 160×10/36=400/9

v=a×t

t=v/a

t=(400/9-25/9)/5

t=(125/3)/5=25/3=

$8 \ \frac{1}{3} sec$

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A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

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Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

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Force acting on the wrench, F = 80 N

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3 years ago

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o

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Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

$E_r = \frac{1}{2} mv^2$

Substitute,

14kg for m

9m/s for v

$E_r = \frac{1}{2} (14) (9)^2\\= 567J$

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

$E_R = \frac{1}{2} Iw^2$

The expression for the moment of inertia of the cylinder is,

$I = \frac{1}{2} mr^2$

The expression for angular velocity is,

$w = \frac{v}{r}$

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

$E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2$

$= \frac{mv^2}{4}$

$E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J$

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

$E = E_r + E_R\\\\$

substitute 567J for E(r) and 283.5J for E(R)

$E = 567J + 283.5\\= 850.5J$

The total energy of the cylinder is 850.5J

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