Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
28384 *x soít cos estematema
Answer:
Program that removes all spaces from the given input
Explanation:
// An efficient Java program to remove all spaces
// from a string
class GFG
{
// Function to remove all spaces
// from a given string
static int removeSpaces(char []str)
{
// To keep track of non-space character count
int count = 0;
// Traverse the given string.
// If current character
// is not space, then place
// it at index 'count++'
for (int i = 0; i<str.length; i++)
if (str[i] != ' ')
str[count++] = str[i]; // here count is
// incremented
return count;
}
// Driver code
public static void main(String[] args)
{
char str[] = "g eeks for ge eeks ".toCharArray();
int i = removeSpaces(str);
System.out.println(String.valueOf(str).subSequence(0, i));
}
}
Answer:
the car to the right
Explanation:
its in the name the RIGHT of way hope it helps good luck
Answer:
Impulse =14937.9 N
tangential force =14937.9 N
Explanation:
Given that
Mass of car m= 800 kg
initial velocity u=0
Final velocity v=390 km/hr
Final velocity v=108.3 m/s
So change in linear momentum P= m x v
P= 800 x 108.3
P=86640 kg.m/s
We know that impulse force F= P/t
So F= 86640/5.8 N
F=14937.9 N
Impulse force F= 14937.9 N
We know that
v=u + at
108.3 = 0 + a x 5.8
So tangential force F= m x a
F=18.66 x 800
F=14937.9 N
