Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
It does work or increases thermal energy
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-diameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. If the total irreversible head loss of the system is determined to be 1.5 m, determine the initial velocity of the water from the tank. Disregard the effect of the kinetic energy correction factors.
Answer:
Time - taken = 2.5 s
deceleration= -8 m/s²
Solution:
Given:
speed, v = 8 m/s
distance, d = 20m
To Find:
deacceleration = ?
As we know speed is defined as
v = d/t
plugging in the values
t = 20/ 8
t = 2.5s
Now from deceleration formula
a = - v/ t
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken and acceleration is 2.5 s and -8 m/s²
respectively.
Learn more about deceleration here:
brainly.com/question/13354629
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Answer:
2 m/s²
30 m/s
Explanation:
1) F = ma
1000 N = (500 kg) a
a = 2 m/s²
2) Given:
v₀ = 0 m/s
a = 2 m/s²
t = 15 s
Find: v
v = at + v₀
v = (2 m/s²) (15 s) + 0 m/s
v = 30 m/s
