Answer:
d) 1/32 microgram
Explanation:
First half life is the time at which the concentration of the reactant reduced to half.
Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.
Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.
Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.
Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.
The initial mass of the sample = 1 microgram
After 5 half-lives, the mass should reduce to 1/32 of the original.
So the concentration left = 1/32 of 1 microgram = 1/32 microgram
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
The drag force acting on the rocket is 80N.
<h3>Give an explanation of drag force?</h3>
The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.
The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.
The drag force acting on the rocket F = ma
Given,
m = 4kg, a = 20ftm/s²
Substituting m and a values in the above formula,
The drag force acting on the rocket F = 4×20
The drag force acting on the rocket F = 80N.
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Answer:
25°C
Explanation:
Using the linear expansivity formula expressed as;
∝ = ΔL/lΔθ
∝ is coefficient of lineat expansion = 1.2 ∙ 10⁻⁵ °C⁻¹
ΔL is the change in length = 6.00036-6
ΔL = 0.00036m
l is the original length = 6m
Δθ is the change in temperature =θ₂-20
Substituting into the formula;
1.2 ∙ 10⁻⁵ °C⁻¹ = 0.00036/6(θ₂-20)
cross multiply
1.2 ∙ 10⁻⁵ * 6 = 0.00036/(θ₂-20)
7.2 ∙ 10⁻⁵= 0.00036/(θ₂-20)
0.00036 = 7.2 ∙ 10⁻⁵(θ₂-20)
0.00036 = 7.2 ∙ 10⁻⁵θ₂-144∙ 10⁻⁵
7.2 ∙ 10⁻⁵θ₂ = 0.00036+0.00144
7.2 ∙ 10⁻⁵θ₂ = 0.0018
θ₂ = 0.0018/0.000072
θ₂ = 25°C
Hence the temperature at which this bar must be acidic for its compression is 6,00036 m is 25°C
The distance increases per minute, levels out, then continued increasing.
