Answer:
resultant is a single force that can replace the of a number of forces , equililbrant is a force that is exactly opposite to resultant
Explanation:
Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is

where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE




We know that capacitrnce and charge is related as


v = 9.523 V
Electric field is given as

= 
E = 5291.00 V/m
E = 5291.00 N/C
Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A
Answer:
It is easier to stop the bicycle moving at a lower velocity because it will require a <em>smaller force</em> to stop it when compared to a bicycle with a higher velocity that needs a<em> bigger force.</em>
Explanation:
The question above is related to "Newton's Law of Motion." According to the <em>Third Law of Motion</em>, whenever an object exerts a force on another object <em>(action force)</em>, an equal force is exerted against it. This force is of the same magnitude but opposite direction.
When it comes to moving bicycles, the force that stops their movement is called "friction." Applying the law of motion, the higher the speed, the higher the force<em> </em>that is needed to stop it while the lower the speed, the lower the force<em> </em>that is needed to stop it.
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km