Answer:
distance between the ships changing 29.255 knot
Explanation:
Given data
time = noon to 5 pm = 5 hours
A = 50 nautical miles due west of B
A = sailing west at 22 knots
B = sailing north at 20 knots
to find out
How fast is the distance between the ships changing
solution
we consider a to b distance is z and at noon ship at 90 degree angle of AB at x and y distance from AB respectively
so here z² = x² + y² ..............1
and 2z z' = 2x x' + 2y y'
so z' = ( x x' + y y' ) / z
and after 5 hours
x = 50 + 22(t) = 50 + 22(5) = 160
y = 20(t) = 20 (5) = 100
so from equation 1
z² = x² + y²
z² = 160² + 100²
z =
z = 188.68
so z' = ( x x' + y y' ) / z
z' = ( 160 (22) + 100(20) ) / 188.68
z' = 29.255
distance between the ships changing 29.255 knot
Answer:
745.4K ~ 472.3 C
Explanation:
This is an Ideal Gas Law problem where we have to manipulate the equation a bit. Let's start with the basic:
PV = nRT will be used for both the initial and final, so we will rearrange this problem to state:
(V(initial))/(T(Initial)) = nR/P
Since we know that the pressure, number of moles of He, and ideal gas constant (R) remain the same from start to finish so we can write the problem as such:
(V(initial))/(T(Initial)) = nR/P = (V(final))/(T(final))
or
(V(initial))/(T(Initial)) = (V(final))/(T(final))
Now lets define some of these values:
T(initial) = 25degree (assuming degrees Celsius) ~ 298.15K
V(initial) = 2.0L
V(final) = 5.0L
T(final) = ?
Since we are solving for T(final) let's rearrange the problem once more to be solving for T(final):
T(final) = (V(final)T(Initial))/V(initial)
Now plug in your values:
T(final) = (5.0L*298.15K)/(2.0L) ~ 745.4K ~ 472.3degrees Celsius
Answer:
Explanation:
Given
Distance between home and school is 39 km
Total time taken
i.e.
Total Distance traveled =39+39+39=117 km
displacement=39 km
Thus Average speed
Average velocity
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