Answer:
total work is 99.138 kJ
Explanation:
given data
diameter = 5 cm
depth = 75 m
density = 1830 kg/m³
to find out
the total work
solution
we know mass of volume is
volume =
volume =
so
work required to rise the mass to the height of x m
dw = gx dx
so total work is integrate it with 0 to 75
w =
w = × 0.05² × 1830× 9.81×
w = 99138.53 J
so total work is 99.138 kJ
C because the the smallest thing then the other ones because it never said what kind of size of it
Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.
Explanation:
Answer:
Smallest drop: Water
Largest drop: Dirt
Explanation:
The heat needed to change the temperature of a sample is:
(1)
with Q the heat (added(+) or removed(-)), c specific heat, m the mass and the change in temperature of the sample. So, if we solve (1) for
Sample A:
Sample B:
Sample C:
Note that the numbers 16744, 5400, 9450 are in the denominator of the expression that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.
So, the smallest drop is Sample A and the largest is Sample C.
(Important: The minus sign of implies the temperature is dropping)
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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