Uncertainty means that your result may be very random, so you can't trust the first or second or so observation, making several samples critical for accuracy. <span />
Answer:
Explanation:
Relative velocity is defined as the velocity of an object B in the rest frame of another object A.
Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.
Answer:
it relates to the light propensity to travel over one straight line without having any interference in its trajectory
Explanation:
Answer:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
<u>h' = 55.3 m</u>
