kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration
Complete question:
A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
Answer:
the total units of insulin admistered each morning
= 22 units of qam and humulin
Explanation:
given
44 units and Humnlin N
with concentration 50/100 = 1/2 = 0.5
∴ 44 × 0.5 ≈ 22 units in the morning
regular insulin administered each day
(22 + 35)units of qam and humulin
= 57units
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
In parallel
Solving the above quadratic equation
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Answer:
15,505 N
Explanation:
Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student
-ΔU = ΔK
-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²
-(0 - mgh) = 1/2mv² - 0
mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.
mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 1 m)
v = √(19.6 m²/s²)
v = 4.43 m/s
Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s
So Ft = m(v' - v)
F = m(v' - v)/t
substituting the values of the variables, we have
F = 70 kg(0 m/s - 4.43 m/s)/0.02 s
= 70 kg(- 4.43 m/s)/0.02 s
= -310.1 kgm/s ÷ 0.02 s
= -15,505 N
So, the force transmitted to her bones is 15,505 N
Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))
