Given :
Initial velocity, u = -15 m/s.
Acceleration , a = 2 m/s².
Time taken to applied brake, t = 2.5 s.
To Find :
The velocity of the car at the end of the braking period.
How far has the car moved during the braking period.
Solution :
By equation :
Now, distance covered by car is :
Hence, this is the required solution.
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The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant .
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
= 0
=
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
or, T =
=
=
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
=
=
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
Answer:
671.76 kg or 6590 N
Explanation:
So the buoyant force generated by the floating ice is equals to the mass of water displaced by the submerged ice. We also need to account for gravity of ice. The resulting additional mass that the ice sheet can support is the difference between the mass of water displaced by ice and the mass of ice submerged totally in water.
So the ice piece can support an additional 671.76 kg of bear, or 671.76 * 9.81 = 6590 N
Answer:
Explanation:
Initial velocity in air, Vo = 0 m/s
Final velocity in air, Vi = 16 m/s
Initial velocity in water, Vf = 3 m/s
Total distance, S = 127 m
Total time, T = 12 s
Using the equation of motion,
(V - U)t = s
S = s1 + s2
Let T = t1 + t2
127 = (16 × t1) + 3 × (12 - t1)
127 = 16t1 + 36 - 3t1
91 = 13t1
t1 = 91/13
= 7 seconds
Time taken in air, t1 = 7 seconds
t2 = 12 - 7
= 5 seconds.