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2 years ago

What is Not considered as metric system ? a. Length b. Mass c. Time d. Volume e. Temperature

Engineering

2 answers:

vampirchik [111]2 years ago

7 0

Answer:

D.) Volume

Explanation:

borishaifa [10]2 years ago

5 0

Answer:

c. Time

Explanation:

The metric system measures mass in grams

The metric system measures mass in grams or kilograms, distance in meters or kilometers, and volume in liters. It measures temperature in Kelvin or Celsius degrees instead of the Fahrenheit degrees used in the imperial system.

Metric is based on increments of the number 10; english is a linear measurement based on inches, feet and yards.

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How can you fix a Nintendo switch if it is a blue screen

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2 years ago

Write the definition of a function printLarger, which has two int parameters and returns nothing. The function prints the larger

vredina [299]

Answer:

#include <iostream>
using namespace std;
void printLarger(int a, int b){
if(a > b){
cout<<a;
}else{
cout<<b;
}
}
int main()
{
printLarger(4, 5);
return 0;
}

Explanation:

The solution code is written in C++.

Firstly define a function printLarger that has two parameters, a and b with both of them are integer type (Line 5). In the function, create an if condition to check if a bigger than b, print a to terminal (Line 7-8). Otherwise print b (Line 9-10).

In the main program, test the function by passing 4 and 5 as arguments (Line 16) and we shall get 5 printed.

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3 years ago

In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala

Vilka [71]

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, $|F_{electrostatic}| = |F_{gravitational}|$

where

$F_{electrostatic}=q.E$

where q is the charge of the droplet and E is the electric field. On the other hand

$F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g$

where m,V ,d and r are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

$q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)$

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))

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3 years ago