Answer:
Conversion tables show:
1 m = 1.09361 yds
1 Lit = .26418 gal = 1.05672 qt or 1 qt = .944632 Lit
1 lb = .45359 kg = 2.2046 Lbs / Kg
So X yds = X m * 1.09361 yds / m = 1.09361 * X yds
Likewise X Lit = X qt / 1.05672 qt/ Lit = X / 1.05672 Lit = .94632 X Lit
So X Lbs = X kg * 2.2046 Lbs / Kg = 2.2046 Lbs
Velocity vs. time graph shows the acceleration as a slope whereas displacement vs. time graph shows the velocity as a slope. So, the given statement is false.
Answer: Option B
<u>Explanation:</u>
To understand the acceleration graphically, consider the x axis of the graph as the run and the y axis as the velocity rise. Now, as we all know that,

We can estimate this through the graph. let's draw the motion of an object with time if it's velocity is changing in every second by 4 m/s. Now if we draw this on graph, we will see that there is a slope between the two corresponding values of time and velocity. This slope defines the acceleration for the object with time.
Now, in the same way, if we draw a distance and time graph respective to the y and x axis; we'll get a slope which defines the velocity of the object i.e. change in distance with time.
Hence, with a velocity vs time graph, we get the slope for acceleration whereas with the distance and time graph, we get the slope for velocity. So both the cases, we see there is no velocity slope on an acceleration and time graph. Hence the statement is false.
Answer:
S = V × t
Explanation:
first convert 36km/h to m/s
36km/h=36*5/18=180/18=10m/s
d =s*t
=10*10=100m
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 ×
C
when force were attractive
Q1 = 1.07 ×
C
Q2 = -1.39 ×
C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F =
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 ×
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 ×
C²
and
we have Q2 = 93 ×
C - Q1
put here value
Q1² - 93 ×
Q1 + 1.49 ×
= 0
solve we get
Q1 = 2.06 ×
C
and
Q1Q2 = 1.49 ×
2.06 ×
Q2 = 1.49 ×
Q2 = 7.23 ×
C
and
if force is attractive we get here
Q1Q2 = - 1.49 ×
C²
then
Q1² - 93 ×
Q1 - 1.49 ×
= 0
we get here
Q1 = 1.07 ×
C
and
Q1Q2 = - 1.49 ×
2.06 ×
Q2 = - 1.49 × 
Q2 = -1.39 ×
C