Answer:
Keeping the speed fixed and decreasing the radius by a factor of 4
Explanation:
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :
We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"
It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,
R' = R/4
New centripetal acceleration will be,
So, the centripetal acceleration of the ball can be increased by a factor of 4.
Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:
You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:
hence, the speed of the Hubble is approximately 384km/min
Answer:
Biomass-Total of mass of organisms in a given area/volume
Biofuel-A fuel directly from living matter.
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
