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Ket [755]
3 years ago
7

(will give brainliest to whoever answers first and explains reasoning) A 10kg object is spun around in a circle with a centripet

al acceleration of 3.5m/s^2. What is the centripetal force acting on the object?
Physics
1 answer:
storchak [24]3 years ago
8 0

Answer:

35 N

Explanation:

F = ma

centripetal force = 10(3.5) = 35 N

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See this suggested solution.

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2 years ago
The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m
vovangra [49]
The equation of the car is given by the equation,

                          x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

                  dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

                 dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2. 

If we substitute these values to the equation for displacement,

(0)   , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2)    , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters. 
3 0
3 years ago
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