<span>An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left.
(1) 22 N to the right
(2) 13 N to the left.
magnitude = 22 - 13
magnitude = 9
Direction would be to the right.
So magnitude is 9N direction to the right.</span>
Answer:
7.468 kN
Explanation:
Here the force is given in Newton
Some of the prefixes of the SI units are
kilo = 10³
Mega = 10⁶
Giga = 10⁹
Tera = 10¹²
The number is 7468.0
Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.
1 kilonewton = 1000 Newton
So, 7468 N = 7.468 kN
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
In that case, their momentum must be equal.
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10
In short, Your Answer would be: 10 m/s
Hope this helps!
