The work done to accelerate the acrobat is given by
where F is the force applied and d the distance of application of the force.
If the barrel is 3.05 m long, then d=3.05 m. Therefore we can find the force:
Answer:
charge, q = ± 1.1 mC
Given:
Capacitance, 
Voltage, V = 110 V
Solution:
The charge on the capacitor plates can be calculated by using the definition of capacitance as :
q ∝ V
where
q = charge
V = potential difference or Voltage
Therefore,
q = CV
Now, charge, q :
q = 
Therefore, the charge on the positive plate is:
q = + 1.1 mC
the charge on the negative plate is:
q = - 1.1 mC
Answer:
36 Ω
Explanation:
Since the 3 resistors are connected in parallel.
The combined resistance of the resistor is
1/Rc = 1/R + 1/R + 1/R ...................... Equation 1
Where Rc = combined resistance of the three resistor, R1 = Resistance of each of the resistor
Rc = R/3 ....................... Equation 2
The formula of power is given as
P = V²/Rc
Rc = V²/P ................ Equation 3
Where V = Voltage, R = Combined Resistance, P = power.
Given: V = 48 V, 192 W.
Substitute into equation 3
Rc = 48²/192
Rc = 12 Ω
From equation 2
Rc = R/3
R = 3Rc
Where Rc = 12 Ω
R = 3×12
R = 36 Ω
Hence the resistance on each resistor = 36 Ω
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46
Answer:The line plot shows the number of cans a class of students at Parkway School collected for a canned food drive.
Part A: How many students collected cans of food?
Part B: What is the median number of cans of food collected?
Explanation:
