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Valentin [98]
3 years ago
11

A child is prescribed a drug X; the recommended manufacturer’s dose is 2.50 mg/m2. The body surface area (BSA) of the child is 1

.50 m2. What dose should be administered to the child? Record your answer using two decimal places.
Physics
1 answer:
guapka [62]3 years ago
3 0

Answer: 3.75 mg

Explanation:

We need a dose of 2.50 mg per square meter of BSA, and we also know that the BSA of the child is 1.50 m^2, then if the dosage should be 2.50mg for one m^2, for 1,5 m^2 we should administrate an amount of:

D = 1.5 m^2*(2.50mg/m^2) = 1.5*2.5 mg = 3.75 mg

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A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
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Answer:

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Explanation: you're welcome

8 0
2 years ago
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Which electrons have the greatest amount of potential energy?
Andrei [34K]
The atoms furthest from the nucleus

7 0
3 years ago
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
murzikaleks [220]
The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
__________________________________________    
                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
    "cm³ " or "mL"; 
_____________________________________________
           {Note the exact equivalent:  1 cm³ = 1 mL }.
____________________________________________
         →  The formula is:  " D = m / V "  ; 
___________________________________________
   in which:

     "D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given); 

     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
_________________________________________________
                 We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________ 
              D  =  m / V ; 
_________________________________________________________
             And we rearrange; to isolate "m" (mass) on ONE side of the    equation; and then we plug in our known/given values;
 to solve for "m" (mass);  in units of "g" (grams) ;
___________________________________________________
    Multiply each side of the equation by "V" ; 
____________________________________________________
             V * { D  =  m / V } ;  to get:
____________________________________________________
      V * D = m ;   ↔   m = V * D ;
___________________________________________________
           Now, we plug in the given values for "V" (volume) and "D" (density) ;     to solve for the mass, "m" ;
______________________________________________________
           m  =  V * D ;
 
           m  =  (23.4 cm³) * (8.9 g / 1 cm³)  = (23.4 * 8.9) g = 208.26 g ;
  
 →  Round to "208 g" (3 significant figures);  
____________________________________
The answer is:  " 208 g " .
_____________________________________________________
7 0
3 years ago
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