Basically its just an air like fluid with chemicals.... just like air it just goes as it pleases and fills space
Answer:
a) F = 2250 Ib
b) F = 550 Ib
c) new max force ( F newmax ) = 2850 Ib
Explanation:
A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up
max capacity of elevator = 24000 Ibs
counterweight = 1000 Ibs
To calculate the force (F) :
we first calculate the Tension using this relationship
Counterweight (1000) - T = ( 1000 / g ) ( g/4 )
Hence T = 750 Ib
next determine F
750 + F - 2400 = 2400 / 4
hence F = 2250 Ib
B ) calculate Tension first
T - 1000 = ( 1000/g ) ( g/4)
T = 1250 Ib
F = 2400 -1250 - 2400/ 4
F = 550 Ib
C ) determine design limit
Max = 2400 * 1.2 = 2880 Ib
750 + new force - 2880 = 2880 / 4
new max force ( F newmax ) = 2850 Ib
Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
<span>Suppose you mixed two chemicals in the lab until you could not tell the two apart. After some time passed, a white powder formed which would not dissolve, and settled on the bottom. The mixture was first homogeneous then heterogeneous. </span>
