A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
<span>The given values are
F1 = 200 N</span>
F2 =?
Total = 400 N
Solution:
F1 + F2 = T
200 N + F2 = 400N
F2 = 400 - 200
F2 = 200
N
Answer
Radius of the wheel r = 2.1 m
Moment of inertia I = 2500 Kg m²
Tangential force applied F = 18 N
Time interval t = 16 s
Initial angular speed ω1 = 0
Final angular speed ω2 = ?
Let α be the angular acceleration.
Torque applied τ = Iα
F r = Iα
Angular acceleration α = F r/I
= 
= 0.015 rad/s²
(a)From rotational kinematic relation
Final angular speed ω₂ = ω₁ + αt
= 0 + (0.015 rad/s^2 * 16 s)
= 0.24 rad/s
(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²
= 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0
= 72 J
(c) Average power supplied by the child P = W/t = 
= 4.5 watt
Answer:
I think its a fuse because everything else makes sense and is used in electrical circuits and the fuse is the only one that stands out to me ¯\_(ツ)_/¯
Answer:
D. 21 ml
Explanation:
Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.
D. <u>21 ml</u>
