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3 years ago

9

Dani says, "This classroom is 11 meters long. A meter is longer than a yard, so if I measure the length of this classroom in yar

ds, I will get less than 11 yards."
Do you agree with Dani ?

1 answer:

WITCHER [35]3 years ago

6 0

Answer:

im not sure i just need points....

Explanation:

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3 years ago

An automobile whose speed is increasing at a rate of 0.800 m/s2 travels along a circular road of radius 10.0 m.

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(a) $a_t = 0.800 m/s^2$

The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:

$a_t = \frac{\Delta v}{\Delta t}$

This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is

$a_t = 0.800 m/s^2$

(b) $a_c = 0.9 m/s^2$

The centripetal acceleration component is given by

$a_c = \frac{v^2}{r}$

where

v is the tangential speed

r is the radius of the trajectory

When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):

$a_c = \frac{(3.00 m/s)^2}{10.0 m}=0.9 m/s^2$

(c) $1.2 m/s^2, 48.4^{\circ}$

The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:

$a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.8 m/s^2)^2+(0.9 m/s^2)^2}=1.2 m/s^2$

and the direction is given by:

$tan \theta =\frac{a_c}{a_t}=\frac{0.9 m/s^2}{0.8 m/s^2}=1.125\\\theta=tan^{-1}(1.125)=48.4^{\circ}$

where the angle is measured with respect to the direction of the tangential acceleration.

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3 years ago