Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you. ...
Step 2: Make a List. ...
Step 3: Identifying the Atoms in Each Element. ...
Step 4: Multiplying the Number of Atoms. ...
Step 5: Placing Coefficients in Front of Molecules. ...
Step 6: Check Equation. ...
Step 7: Balanced Chemical Equation.
Distribute the 3 to 4m and 6. this makes 12m-18=12
now add 18 to both sides.
12m=30
now divide by 12
m=30/12
Answer:
acceleration=force/mass
Due to variation in the mass[ as same force as per magnitude acts on both of them] the acceleration developed in the bug will be more and the acceleration developed in the car is negligible.
impulse=change in momentum
The impulse of bug is more and that of car is small.
momentum changes=change in momentum
force=change in momentum/time of contact
the force exerted by car on the bug will be more than the converse
Total. momentum is conserved , so magnitude of change in momemtum or impulse is same. By Newtons third law, since the bug have smaller mass, its change in magnitude of velocity is larger.
Step-by-step explanation:
I think it would be B ?? Because it’s the shortest one ♀️
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
