Answer:
has units of distance
has units of distance over time
has units of distance over 
has units of distance over 
Explanation:
Since the expression for the distance is:
then:
has units of distance
has units of distance over time
has units of distance over
has units of distance over
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
See this suggested solution.
1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.
2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.
3. using the conditions described in 2. the answer is C.
Answer:
If the resistors are in parallel they must all experience the same voltage drop.
R * I = V
8 * 3 = 24 V (if i is 3 amps)
24V / 4 ohm = 6 amps
24V / 16 ohm = 1.5 amps
Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
Answer:
Thanks for the 50 points. Answer eill be always 29.0m
