A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
1 answer:
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Answer:
It took Kevin 26 minutes to run from markers 1 to 4
His average speed from mile markers 1 to 4 is 0.154 miles/minute
Kevin must run by average speed 0.1 miles/minute to finish the race
Explanation:
Lets explain how to solve the problem
A half marathon 13.1 miles
Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and
18 minutes to run from mile marker 2 to mile marker 4
→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4
→ The total time of the first 4 marker = 8 + 18 = 26 minutes
<em>It took Kevin 26 minutes to run from markers 1 to 4</em>
<em></em>
Average speed is total distance divided by total time
The average speed of Kevin as he ran from mile marker 1 to mile
marker 4 is the 4 miles divides by 26 minutes
→ Average speed = 4 ÷ 26 = = 0.154 miles/minute
<em>His average speed from mile markers 1 to 4 is 0.154 miles/minute</em>
<em></em>
It took Kevin 71 minutes to pass mile marker 9
Kevin need to complete the race in 112 minutes, then what must
Kevin's average speed be as he travels from mile marker 9 to the
finish line?
The total distance of the race is 13.1 miles, he ran 9 miles
→ The remaining distance = 13.1 - 9 = 4.1 miles
He must run 4.1 miles to complete the race
The total time is 112 minutes, he used 71 minutes to run the first 9 miles
→ The remaining time = 112 - 71 = 41 minutes
He must finish the 4.1 miles in 41 minutes
→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min
<em>Kevin must run by average speed 0.1 miles/minute to finish the race</em>
(a) See graph in attachment
The appropriate graph to draw in this part is a graph of velocity vs time.
In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.
Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at
t = 10 s
v = 15 m/s
(b)
The average acceleration of the horse can be calculated as:
where
v is the final velocity
u is the initial velocity
t is the time interval
In this problem,
u = 0
v = 15 m/s
t = 10 s
Substituting,
(c) 75 m
For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:
where
s is the distance travelled
u is the initial velocity
t is the time interval
a is the acceleration
In this problem,
u = 0
t = 10 s
Substituting,
(d) See attached graphs
In a uniformly accelerated motion:
- The distance travelled (x) follows the equation mentioned in part c,
So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.
- The acceleration is constant during the motion, and its value is
(calculated in part b)
therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.
