Answer:
300 m
Explanation:
The train accelerate from the rest so u = 0 m/sec
Final speed that is v = 80 m/sec
Time t = 30 sec
The distance traveled by first plane = 1200 m
We know the equation of motion 
where s is distance a is acceleration and u is initial velocity
Using this equation for first plane 
As the acceleration is same for both the plane so a for second plane will be 2.67 
The another equation of motion is 
using this equation for second plane 
s = 300 m
<span>First sum applied the Newton's second law motion: F = ma
Force = mass* acceleration
This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time,
force=(mass*velocity)/time
such that, (mass*velocity)/time=momentum/time
Therefore we get mass*velocity=momentum
Momentum=mass*velocity
Elephant mass=6300 kg; velocity=0.11 m/s
Momentum=6300*0.11
P=693 kg (m/s)
Dolphin mass=50 kg; velocity=10.4 m/s
Momentum=50*10.4
P=520 kg (m/s)
The elephant has more momentum(P) because it is large.</span>
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
Answer:
i. + 22.5 m ii. 4.0 m
Explanation:
i. Image distance
Using the lens formula
1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown
So, we find v
1/v = 1/f - 1/u
= 1/+18 - 1/+90
= (5 - 1)/90
= 4/90
v = 90/4
= + 22.5 m
So the image is real and formed 22.5 m away on the other side of the lens.
ii Length of Shark
Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.
m = image distance/object distance
= v/u
= +22.5/+90
= 0.25
0.25 = image height/object height
So,
object height = image height/0.25
= 1.0 m/0.25
= 4.0 m
So, the length of the shark is 4.0 m
