The distance between the two cities is 513.24 km.
<h3>Time of motion when the two trains meet</h3>
The time spent on the journey when the two trains meet is calculated as follows;
(Va - Vb)t = d
where;
- d is the distance between the trains before meeting
(76 - 65)t = 40
11t = 40
t = 40/11
t = 3.64 hr
<h3>Distance traveled by the fast train</h3>
d1 = 76 km/h x 3.64 h
d1 = 276.64 km
<h3>Distance traveled by the slow train</h3>
d2 = 65 km/h x 3.64 h
d2 = 236.6 km
The distance between the two cities = 276.64 km + 236.6 km
= 513.24 km
Learn more about relative velocity here: brainly.com/question/17228388
Answer:
The dart with the small mass will travel the farthest distance.
Explanation:
Acceleration is proportional to force times mass, and inertia is proportional to mass. Inertia is the reluctance of a moving body to stop, and a stationary body to start moving (inertia increses with mass). Assuming they both have the same aerodynamic design, and that they are both launched with the same force applied for the same time duration, the dart with less small mass will accelerate faster than the big mass dart. From this we can see that the small dart will have covered a longer distance before the effect of the force stops, when compared to the more massive dart.
Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
Simora [160]
Data D has the largest range.
Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23
Therefore, Data D has the largest range.
Answer:
t = 1.75
t = 0.04
Explanation:
a)
For part 1 we want to use a kenamatic equation with constant acceleration:
X = 1/2*a*t^2
isolate time
t = sqrt(2X / a)
Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2
t = sqrt(2*15m / 9.8m/s^2)
t = 1.75 s
b)
The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:
v = d / t
isolate time
t = d / v
plug in known variables
t = 15m / 340m/s
t = 0.04 s
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:
I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>
