On his way to class, a student on a skateboard is accelerating on a downhill stretch. Which of the following statements is true?
1 answer:
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Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s
so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull
Answer:
v=39.05 m/s
Explanation:
Given that
x= 56 cm
F= 158 N
m= 58 g = 0.058 kg
Lets take spring constant = k
At the initial position,before releasing the arrow
F= k x
By putting the values
F= k x
158= 0.56 k
k=282.14 N/m
Now from energy conservation
Lets take final speed of the arrow after releasing
k x²=mv²
282.14 x 0.56² = 0.058 v²
v=39.05 m/s