The answer is <span>D: its acceleration is 1.25 meters/second^2.</span>
Answer:
Explanation:
In an experimental research, the control group is the group that serves as the neutral group that is not given any form of treatment and serves as the group in which the experimental groups are firstly compared to. Thus, <u>the control group in the question described is the Third group</u>.
While experimental groups are the groups that receive treatments required to make an inference from the experiment. From this description, <u>it can be deduced that the First and the Second group are the experimental groups.</u>
Answer:
0.239 T
Explanation:
Applying,
F = Bvqsin∅................ Equation 1
Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.
make B the subject of the equation
B = F/(vqsin∅)................. Equation 2
From the question,
Given: F = 1.15×10⁻¹³ N, v = 3.0×10⁶ m/s, ∅ = 90°(perpendicular)
Constant: q = 1.602 x 10⁻¹⁹ C
Substitute into equation 2
B = 1.15×10⁻¹³ /(3.0×10⁶×1.602 x 10⁻¹⁹×sin90°)
B = 1.15×10⁻¹³/(4.806×10⁻¹³)
B = 0.239 T.
Hence the magnetic field = 0.239 T
They all end with suffix "-ide"
In short, Your Answer would be Option C
Hope this helps!
Given that,
Puck slide total Deltax = 12m
Puck and board Mk = 0.10
Find the initial speed = ?
We know that,
Vf^2-Vi^2 = 2a Deltax
-Vi^2 = -2MkgDeltax ............(1)
then,
fk = -Mk mg = ma
a = -Mkg .........(2)
From equation (1),
Vi^2 = 2MkgDeltax
Vi = âš2MkgDeltax [g=9.8m/s^2]
Vi = âš2(0.10)(0.98)(12)
Vi = âš0.2(9.8)(12)
Vi = âš1.92(12)
Vi = âš23.52
Vi = 4.8 m/s
Therefore the initial speed of the puck Vi = 4.8 m/s
