Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
(a)
According to Newton's second law, the force experienced by each balloon is given by:
F = ma
where
m = 0.021 kg is the mass
a = 1.1 m/s^2 is the acceleration
Substituting, we found:
The electrostatic force between the two balloons can be also written as
where
k is the Coulomb's constant
Q is the charge on each balloon
r = 16 m is their separation
Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:
(b) electrons
The magnitude of the charge of one electron is
While the magnitude of the charge on one balloon is
This charge can be written as
where N is the number of electrons that are responsible for this charge. Solving for N, we find:
Answer:
patient receiving drug 25 MCG/minute
Explanation:
given data
infusing = 15 ml/hr
drug = 50 mg
D5W = 500 ml
to find out
How many MCG/minute
solution
we know infusing rate is 15 ml/hr = 0.25 ml/min
so 0.25 ml drug content = 50 /500 × 0.25
0.25 ml drug content = 0.025 mg
so here
rate of drug will be 0.025 mg
rate of drug = 0.025 mg = 25 × gm/min
rate of drug = 25 MCG/minute
so patient receiving drug 25 MCG/minute
Answer:
<em>3.15 N towards the positive x-axis</em>
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Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F =
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>