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1 answer:
Answer:
Explanation:
Initially the packet was ascending up with the balloon.
Taking upward as positive direction;
initial velocity, u = 4.9 m/s
final velocity = v m/s
initial height, h₁ = 100 m
final height, h₂ = 0
a = -9.8 m/s²
time taken = t seconds
Considering the second equation of motion.
S = ut + 1/2at^2
s = ut + 0.5at²
(h₂-h₁) = ut + 0.5at²
0- 100 = 4.9t + 0.5×(-9.8)×t²
-100 = 4.9t - 4.9t²
4.9t² -4.9t - 100 =0
t² -t - 20.41 = 0
Using : b±√(b²-4ac))/(2a).... 1
Where a =1, b = -1 c =-20.408
Substituting the values into equation 1
Solving it, we get t = 5.045s
v = u + at = 4.9 -9.8×5.045 = 4.9 - 49.44 = -44.54 m/s
The final velocity is -44.54 m/s, and it took 5.045s to reach the ground.
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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 ×
friction factor = 0. 52 ×
friction factor = 0. 52 × 0. 55
friction factor
b. When V = 3mls
Friction factor = 0. 52 ×
Friction factor = 0. 52 ×
Friction factor = 0. 52 × 0. 185
Friction factor
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × ×
×
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = ×
×
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
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Answer:
False
Explanation:
The given statement "Two objects must be in contact for them to exert a force on each other" is not true as there are many types of forces that doesn't require being in contact for exerting a force.
One such example is the gravitational force acting between two bodies. Gravitational force is the force of pull with which a body pulls another body without being in contact.
For two bodies of masses 'M' and 'm' separated at a distance of 'R', the gravitational force is given as:
The gravitational force acts always act between bodies that have mass. The bodies are not in contact yet experience force.
Therefore, the given statement is false.