I believe the answer would be 7.5 m/s^2
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
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
Answer:
D. Strong nuclear forces hold the nucleus of an atom together. Weak nuclear forces are involved when certain types of atoms break down.
Explanation:
Answer:
m=417.24 kg
Explanation:
Given Data
Initial mass of rocket M = 3600 Kg
Initial velocity of rocket vi = 2900 m/s
velocity of gas vg = 4300 m/s
Θ = 11° angle in degrees
To find
m = mass of gas
Solution
Let m = mass of gas
first to find Initial speed with angle given
So
Vi=vi×tanΘ...............tan angle
Vi= 2900m/s × tan (11°)
Vi=563.7 m/s
Now to find mass
m = (M ×vi ×tanΘ)/( vg + vi tanΘ)
put the values as we have already solve vi ×tanΘ
m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)
m=417.24 kg
