Answer:
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Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2
Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:
ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.
w=2*π*f
We know that the turntable is set to 33 1/3 rev/m so
the frequency 33.33/60=0.55 Hz
then w=2*π*0.55=3.45 rad/s
Finally the centripetal acceleration at differents radii results equal:
r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2
r=0.1 ac=3.45^2*0.1=1.20 m/s^2
r=0.14 ac=3.45^2*0.14=1.66 m/s^2
Well let's convert all these values out of standard form first:
2x10^-2 = 2x0.01 = 0.02m = 2cm
2x10^0m = 2x1 = 2m
2x10^-1m = 2x0.1 = 0.2m = 20cm
2x10^1m = 2x10 = 20m
Based on that, we know that 20cm is roughly equivalent to a basketball (at least it's closer than all the other values), so the answer is therefore 2 - 2x10^-1m
Answer: Option A: The number of trees sampled.
The accuracy can be understood as how close is the measured value to the true value. The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.