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Elena-2011 [213]
3 years ago
5

A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for

7.9 seconds. How far did the truck travel during that time interval?
Physics
1 answer:
Sonja [21]3 years ago
8 0

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

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JulijaS [17]

Answer:

<h2>          v= 21.47m/s      </h2>

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

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Assuming g= 9.81 m/s^2

U*m*g=\frac{mv^2}{r}

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substituting our given data in to expression we can solve for the speed V

0.5*9.81=\frac{v^2}{94}

making v the subject of formula we have

0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

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3 years ago
A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t
Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

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