You could heat the syrup, grease the spoon with shortening, because if you do, then the molasses won't stick when you pour it out of the spoon into the batter.
Answer:
distance between school and home is 21 miles
Explanation:
given data
in rush hour speed s1 = 28 mph
less traffic speed s2 = 42 mph
time t = 1 hr 15 min = 1.25 hr
to find out
distance d
solution
we consider here distance home to school is d and t1 time to reach at school
we get here distance equation when we go home to school that is
distance = 28 × t1 .......................1
and when we go school to home distance will be
distance = 42 × ( t - t1 )
distance = 42 × ( 1.25 - t1 ) ...................2
so from equation 1 and 2
28 × t1 = 42 × ( 1.25 - t1 )
t1 = 0.75
so
from equation 1
distance = 28 × t1
distance = 28 × 0.75
distance = 21 miles
Answer:
distance can describe the total distance moved and displacement shows how far something has moved from its starting position (in a straight line from point a to point b) the object doesn't have to move in a straight line, but that is how displacement is measured
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
Explanation:
Given that,
The mean kinetic energy of the emitted electron, 
(a) The relation between the kinetic energy and the De Broglie wavelength is given by :



(b) According to Bragg's law,

n = 1
For nickel, 



As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.