The answer is actually True, I just took the test and it was correct.
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity
will be expressed by the following equation:

Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
Answer:
The mass of Uranium present in a 1.2mg sample is 
Explanation:
The ration between Uranium mass and total sample mass is:
For a sample of mass 1.2 mg, the amount of uranium is:

Work done by a given force is given by

here on sled two forces will do work
1. Applied force by Max
2. Frictional force due to ground
Now by force diagram of sled we can see the angle of force and displacement
work done by Max = 

Now similarly work done by frictional force



Now total work done on sled


Answer:
The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Explanation:
Given that,
Mass of halfback = 98 kg
Speed of halfback= 4.2 m/s
Mass of corner back = 85 kg
Speed of corner back = 5.5 m/s
We need to calculate their mutual speed immediately after the touchdown-saving tackle
Using conservation of momentum

Where,
= mass of halfback
=mass of corner back
= velocity of halfback
= velocity of corner back
Put the value into the formula



Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s