Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
Answer:
3658.24m
Explanation:
Hello!
the first thing that we must be clear about is that the train moves with constant acceleration
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf = final speed
=160km/h=44.4m/s
Vo = Initial speed
=42.9km/h=11.92m/s
A = acceleration
=0.25m/s^2
X = displacement
solving

the distance traveled by the train is 3658.24m
Hi there!

Use the equation:

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:
v2 = ?
m1 = 0.048 kg (converted)
m2 = 2.95
v1 = 391


They have to have a positive charge and negative so then thats what u get