Answer:
v = 21 m / s
Explanation:
We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data
v = v₀ + a t
the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s
a = (v-v₀) / t
a = (11 - 6) / 8
a = 0.625 m / s²
when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s
v = v₀ + a t
v = 11 + 0.625 16
v = 21 m / s
Answer:
Explanation:
1. Many buyers and sellers participate in the market.
2. Sellers offer identical products.
3. Buyers and sellers are well informed about products.
4. Sellers are able to enter and exit the market freely.
Answer: 477W
Explanation:
Given the following :
Mass (m) = 7.3kg
Initial Velocity (u) = 0
Final velocity (v) = 14m/s
time (t) = 1.5s
Power = workdone (W) / time (t)
The workdone can be calculated as the change in kinetic energy (KE) :
Recall ;
KE = 0.5mv^2
Therefore, change in KE is given by:
0.5mv^2 - 0.5mu^2
Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)
Change in KE = 715.4J
Therefore ;
Average power = Workdone / time
Workdone = change in KE = 715.4N
Average power = 715.4 / 1.5
Average power = 476.93333 W
= 477W
Answer:
Explanation:
The wheel and falling student will have common acceleration .
For rotational motion of wheel
Tx r = I α , T is tension in the crank , α is angular acceleration of wheel , I is moment of inertia , r is radius of the wheel.
= I a / r
T = I a / r²
For motion of student
Mg - T = Ma , M is mass of the wheel.
Mg - I a / r² = Ma
Mg = Ma +I a / r²
Mg = (M +I / r²)a
a = Mg / (M +I / r²)
= 51 x 9.8 / ( 51 + 9.6 / .3² )
499.8 / (51+ 106.67 )
= 499.8 / 157.67
= 3.17 m / s².
If time t is taken to fall by 12 m
12 = 1/2 a t²
24 / a = t²
24 / 3.17 =t²
t²= 7.57
t = 2.75 s
velocity to reach sidewalk
v = u + at
= 3.17 x 2.75
= 8.72 m / s