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VLD [36.1K]
3 years ago
9

When converting from a smaller to a larger unit of measurement, you move the _______ to the right. A. second number B. ratio C.

exponent D. decimal point
Physics
2 answers:
viktelen [127]3 years ago
7 0
The answer to your question is D
Ghella [55]3 years ago
5 0

Answer:

The correct answer is D. decimal point.

Explanation:

The question is wrong. It should be :

When converting from a smaller to a larger unit of measurement, you move the _______ to the left.

In order to convert from a smaller to a larger unit of measurement what we need to move is the decimal point to the left. Let's see an example :

We have the two following units

''g'' which is grams and ''kg'' which is kilograms.  

We know that the grams are a smaller unit of measurement that kilograms.

We also know the relationship :

1kg=1000g

For example if we want to convert 1500g to kilograms :

(1500g).(\frac{1kg}{1000g})=1.5kg

What we have done is that we divided the number ''1500'' by ''1000''.

This is equal to move the decimal point three places to the left ⇒

1500g=1.5kg

The correct answer is D. decimal point

You might be interested in
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
SSSSS [86.1K]

Answer:

{F_{tot} = -1.092*10^{-2}N

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

F =k\dfrac{q_1 q_2}{d^2}.

where d is the separation between the charges q_1 and q_2.

Now, in our case

q_1=-13.5nC=-13.5*10^{-9}C

q_2=-1.735nC=-1.735*10^{-9}C

q_3= 47*10^{-9}C

The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

Therefore, the force between them is

F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

and it is directed in the negative x-direction.

The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

8 0
3 years ago
A bicycle, a motorcycle, a sports car, and a pickup truck are traveling at the same velocity. Which vehicle requires the greates
trapecia [35]
D. Pickup truck because it's heavy and it gained more momentum
3 0
3 years ago
Read 2 more answers
​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
A 76g ball is attached to a string that is 1.5m long. It is spun so that it completes two full rotations every second. What is t
Marat540 [252]

The centripetal acceleration = 236.63 m/s²

The force = 17.98 N

<h3>Further explanation</h3>

Given

mass = 76 g = 0.076 kg

r = 1.5 m

f = 2 rps = 2 rotation per second

Required

The centripetal acceleration

The Force tension

Solution

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle  

\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}

F = centripetal force, N  

m = mass, Kg  

v = linear velocity, m / s  

r = radius, m  

The speed that is in the direction of the circle is called linear velocity  

Can be formulated:  

\tt \displaysyle v = 2 \pi.r.f

r = circle radius  

f = rotation per second (RPS)  

The linear velocity : 2 x 3.14 x 1.5 x 2 =18.84 m/s

The centripetal acceleration : ac = v²/R = 236.63 m/s²

The force : F = m x ac = 0.076 x 236.63 = 17.98 N

4 0
3 years ago
Gold has a density of 19.32 g/cm3. what is the volume of a sample of gold with a mass of 27.63 grams?
schepotkina [342]
Answer:  The volume is:  " 1.430 cm³ " .
_____________________________
Explanation:  
________________________
Volume, " V  = ? ; (unknown, we need to solve for this).
Density,  "D  = 19.32 g / cm³ ; 
mass , "m" = 27.63 g ; 
_____________________________________________
The formula for density is:  
____________________________________________
  D = m / V ;   Divide each side of the equation by:  "(1/m)" ;
                    to isolate " V" on one side of the equation; and to solve for "V" ; 
____________________________________________
1/m)*D = (1/m) * (m / V) ;
_______________________________
    to get:   
_______________________________
       D/m = ; ↔  1 / V ;  Take the reciprocal of EACH SIDE; to isolate "V" on each side of the equation:    
_____________________________
      m / D = V/1 ↔ V = m / D ;
_____________________________
    V = m / D ; 
_______________________________
         Now, plug in our given values for mass, "m" ; and Density, D";              to solve for "Volume, V " ;
_____________________________________
    V  =  m / D  =  (27.63 g)  ÷  (<span>19.32 g / cm</span>³)  ;                
 
                       =  (27.63 g) * (1 cm³ / 19.32 g) ;
 
                       =   (27.63 ÷ 19.32)  cm³  ;
  
                       =   1.4301242236024845 cm³ ;

                                              → Round to "4 significant figures" ; 
                       =   1.430 cm³ .
____________________________________________
   The volume is:  " 1.430 cm³ " .
______________________________________________
4 0
3 years ago
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