Answer:
The accurate statements are
2. The train is not an inertial frame of reference.
5. The train could be moving at a constant velocity in a circular path.
8. The train must be undergoing acceleration.
Explanation:
As we observe that the string forms an angle with the horizontal
we can formulate the force equation relevant to the given ball
similarly in the Y direction
Thus we conclude
Answer:
20.353125 V
Explanation:
m = Mass of proton =
q = Charge of proton =
= Velocity of proton at point A = 50 km/s
= Velocity of proton at point B = 80 km/s
The relationship derived from energy conservation is as follows:
The determined potential difference is 20.353125 V
Answer:
the time it takes after impact for the puck is 2.18 seconds
Explanation:
initially given information
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thickness = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
temperature of air = 15°C
to determine
time needed for the puck to reduce its speed by 10%
solution
we note that velocity changes from 0 to v
assuming initial velocity = v
therefore final velocity = 0.9v
implying a change in velocity is du = v
and clearance dy = h
shear stress acting on the surface is expressed as
= µ
therefore
= µ ............1
substituting the values
= 1.75 × ×
= 0.175 v
the area between the air and puck is given by
Area =
area =
area = 7.85 × m²
thus, the force on the puck can be represented as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
now applying Newton's second law
force = mass × acceleration
- force =
- 1.374 × v =
solving for t =
the time needed after impact for the puck is 2.18 seconds