Answer:
40 s
Explanation:
After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start
Let t be the number of seconds after the second skater starts will the second skater overtake the first skater
The distance traveled by the first skater after t seconds is
Similarly the distance traveled by the 2nd skater after t seconds is
Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater
We can substitute 
The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.
<h3 /><h3>What is mass?</h3>
Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)
To calculate the mass the engine burns each seconds, we use the formula below.
Formual:
- M = T/v............. Equation
Where:
- M = Mass per seconds of the rocket
- T = Thrust
- v = Velocity
From the question,
Given:
- T = 7.66×10⁵ N
- v = 3.05×10³ m/s
Substitute these values into equation 1
- M = (7.66×10⁵)/(3.05×10³)
- M = 2.5×10² kg/s
Hence, the mass of fuel burned in each second is 2.5×10² kg/s.
Learn more about mass here: brainly.com/question/25121535
#SPJ1
You can't change time to a measurement of distance
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values
So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
Answer:
V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2
V = 73.8 m/s
tan theta = Vy / Vx = 62/40 = 1.55
theta = 57.2 deg
