Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
Answer:
FATS
Explanation:
Fats are made up of carbon and hydrogen elements joined together in long groups called hydrocarbons. The simplest unit of fat is the fatty acid, of which there are two types: saturated and unsaturated.
Kinetic energy =1/2 mv^2
<span>m=2ke/v^2 </span>
<span>m=2(34)/3.6^2 </span>
<span>m=5.24 </span>
<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N
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Answer:
Explanation:
Near point = 56 cm .
near point of healthy person = 25 cm
person suffers from long sightedness
convex lens will be required .
object distance u = 25 cm
image distance v = 56 cm
both will be negative as both are in front of the lens.
lens formula
I/v - 1 / u = 1/f
- 1/56 +1/25 = 1/f
- .01785 + .04 = 1/f
1/f = .02215
f = 45.15 cm .
