Answer:
answer is 0.1428
Explanation:
Data:- vf=5.0 , vi=0.0 , t=35 , a=? so appling first eq of motion vf=vi+at we have to find a=vf-vi/t , a=5.0-0.0/35 , a=5/35 ,a=0.1428m/sec²
Answer:
<h2>b) 4230 J
</h2>
Explanation:
Step one:
given data
extension= 40cm
Spring constant K= 52.9N/cm
Step two:
Required
the Kinetic Energy KE
the expression to find the kinetic energy is
KE= 1/2ke^2
substituting our data we have
KE= 1/2*52.9*40^2
KE=0.5*52.9*1600
KE= 42320Joules
<u>The answer is b) 4230 J
</u>
Answer:
The image distance is 20.0 cm.
Explanation:
Given that,
Power = 1.55 dp
Distance between book to eye = 26.0+3.00=29.0 cm
We need to calculate the focal length
Using formula of focal length

Put the value into the formula



We need to calculate the image distance
Using lens formula


Put the value into the formula



Hence, The image distance is 20.0 cm.
Answer:
the answer is c. producers, detrivores