Answer:
The total distance will be 400 m.
Explanation:
For portion AB:
Acceleration = 2
t= 10 s
Car start from rest , u=0 m/s
We know that
S= 100 m.
For portion BC:
V= u + at
V=0 + 2 x 10
V= 20 m/s
In this portion car moves with constant velocity 20 m/s for 10 s.
So distance S= V x t
S=20 x 10 =200 m.
For portion CD:
The velocity at point C will be 20 m/s
In this portion the final speed of car will be zero because given that at final car come to rest.
So the acceleration will be in the negative direction to stop the car.
We know that
S=100 m
The total distance AD=AB + BC+ CD
AD=100 +200 + 100 m
AD=400 m.
The total distance will be 400 m.
Speed with which initially car is moving is 21 m/s
Reaction time = 0.50 s
distance traveled in the reaction time d = v t
d = 21 * 0.50 = 10.5 m
deceleration after this time = -10 m/s^2
now the distance traveled by the car after applying bakes
so total distance moved before it stop
d = 22.05 + 10.5 = 32.55 m
so the distance from deer is 35 - 32.55 = 2.45 m
now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop
so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m
again by kinematics
so maximum speed would be 22.1 m/s
Answer:
Explanation:
radius of circle r = 0.9 m.
(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .
( b )
Tension in string T = m ω²r
Putting the values
60 = .072 x ω² x 0.9
ω² = 926
ω = 30.4 rad /s
angle made in 20 revolutions θ = 20 x 2π = 126.6 rad
time taken = θ / ω
= 126.6 / 30.4
= 4.16 s .
The correct answer among the choices given is option B. The energy transformation that occurs in the core of a nuclear reactor is from nuclear energy to thermal energy. In a power plant nuclear fission which involves nuclear energy to heat up water around it. This part is the core of the process.
