Answer:
m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Explanation:
D = V0t + 0.5at^2
Where d is the distance
V0 is the initial velocity
A is the acceleration
T is time
From the graph a = 4/3 m/s2
D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2
D = 6 m
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
