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I am Lyosha [343]

2 years ago

12

Sasha is ordered Ampicillin 50mg/kg/day x 48 hours, to be given every 6 hours in 100mls of N/S run over 30 minutes. The tubing h

as a drop factor of 10 gtts/ml. Sasha weighs 64 lbs. What is the correct dose and infusion rate

1 answer:

Anna007 [38]2 years ago

5 0

Answer:

The correct dose = 1454.54 mg

and The jnfusion rate = 41.67 gitt/hr

Explanation: the correct dose will be 50mg/kg × kg/2.2 × 64lb

= 1454.54 mg

infusion rate will be

10 gtts/ml × 50mg/6 × 30/60

Infusion rate = 15000/360

= 41.67 gitt/hr

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The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

Fynjy0 [20]

Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

$E_z = \frac{-dV}{dz} = - nCzn^n^-^1$

Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

$=> D = 10$

$V(1, 0, 0) = 4V => A + 10 = 4$

Solving for A, we have:

$A = 4 - 10$

$A = -6$

$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

$B = -4$

$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

From above, A = -6

$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

$m = 4$

$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

From above, C = -2

$n = \frac{-16}{-2}$

$n = 8$

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1 year ago

A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

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All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

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What type of roadway has the highest number of hazards per mile?

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The roadway with the highest number of hazards is <span>city streets</span>

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In ocean waves, water particles move ________ and energy moves ________.

morpeh [17]

In ocean waves, water particles move with mechanical energy and energy moves with gravity

Not sure but hope it helps!

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Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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