Before coming into conclusion first we have to understand both scalar and vector .
A scalar quantity is a physical quantity which has only magnitude for it's complete specification.
A vector quantity is that physical quantity which not only requires magnitude but also possesses direction for it's complete specification.
So the most important factor that differentiate vector from scalar is the direction.
As per the question the student is doing an experiment where he is recording the data obtained during the process.
In order to arrange them in data table, he should ask about the direction of the quantity under consideration.
Hence the correct option is the third option(C)i.e does the measurement include direction?
Answer:
All of the above are true.
Explanation:
(a). true
whenever charge particle move back and froth from its mean position then it will produce oscillating electric and magnetic fields, . so an em wave can be obtain by accelerating charge
(b). true
the electric field and the magnetic field have vibrations in the perpendicular direction along the motion of the wave so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave
(c) true .
The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields and also both fields are perpendicular to the direction of propagation of the wave.
(d) true .
An electromagnetic wave carry energy through vacuum with a speed of 
so , all of the above are true.
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
<span>When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 86 m/s^2
t = 1.7 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m
vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m
Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
Bergeron–Findeisen Process.
