It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
Magnitude of displacement = 
Adding the squares gives displacement = 
Displacement = ≈ 724.7m
Answer:
Explanation:
Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .
Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A
friction force = .4 x 2.5 x 9.8
= 9.8 N
net force on block A = P - 9.8
acceleration = ( P - 9.8 ) / 2.5
force on block B = 9.8
acceleration = force / mass
= 9.8 / 6
for common acceleration
( P - 9.8 ) / 2.5 = 9.8 / 6
( P - 9.8 ) / 2.5 = 1.63333
P = 13.88 N .
Answer:
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Explanation: took the test
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