On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m^3 , length 92.4 cm and diameter 2.15 cm f
1 answer:
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Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
Answer:
The correct answer is:
(A) to the left
(B) at speed -0.8725 m/s
Explanation:
The given values are:
Plate 1:
Mass,
m₁ = 201 g
Velocity,
v₁ = +1.79 m/s
Plate 2:
Mass,
m₁ = 335 g
Velocity,
v₁ = -2.47 m/s
According to the conservation of momentum, we get
⇒
then,
⇒
On substituting the values, we get
⇒
⇒
⇒
⇒ (to the left)