Answer: The time required for the impluse passing through each other is approximately 0.18seconds
Explanation:
Given:
Length,L = 50m
M/L = 0.020kg/m
FA = 5.7×10^2N
FB = 2.5×10^2N
The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.
Ca(t) + CB(t) = 50
Where CA and CB are the velocities of the wire A and B
t = 50/ (CA + CB)
But C = Sqrt(FL/M)
Substituting gives:
t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))
t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))
t = 50 / (168.62 + 111.83)
t = 50/280.15
t = 0.18 seconds
Answer:
This slide shows the three forces that act on a baseball in flight. The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion
Answer:
Her speed is 1.1 m/s, and her velocity is 0 m/s
Explanation:
Speed = Distance covered/Time
Given
Distance = 400m
Time = 6minutes = 6*60 = 360 secs
Substitute the given parameter into the formula;
Speed = 400/360
Speed = 1.1m/s
Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)
Velocity = Displacement/Time
Velocity = 0/3600
Velocity = 0m/s
Hence her speed is 1.1 m/s, and her velocity is 0 m/s
Answer: Q=5.46 L/s
COP=2.58
Explanation:
Given that
Cp = 4.18 kJ/(kg.C
density = 1 kg/L
Heat rejected Qr= 570 kJ/min
Power in put W= 2.65 KW
From first law of thermodynamics
U = W+ q
q = Heat absorbed
U = internal energy
W = workdone
U = 570 kJ/min = 9.5 KW
9.5 = 2.65 + q
q = 6.85 KW
COP = q/W
COP = 6.58 / 2.65
COP=2.58
Lets take volume flow rate is Q
So mass flow rate of water m = ρ Q
q = m Cp ΔT
6.85 = 1 x Q x 4.18 ( 23-5)
Q=0.091 L/min
Q=5.46 L/s
P (gravitational force) = m (mass) x g
<=> P = 0.05 x 10
<=> P = 0.5N
