First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
Answer:
Atomic mass is defined as the number of protons and neutrons in an atom
Answer:
10 m/s^2
Explanation:
Equation: F = ma.
a = acceleration
m = mass
F = force
Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.
F = 20
m = 2
a = ?
a = F/m
a = 20/2
a = 10 m/s^2
Average Velocity = Total Displacement / Total time
1st part of journey, 350 km at velocity 125 km/h
Time = 350 / 125 = 2.8 hours.
2nd part of journey, 220 km at velocity 115 km/h
Time = 220 / 115 = 1.9 hours
Average Velocity = Total Displacement / Total time
= (350 + 220) / (2.8 + 1.9)
= 570 / 4.7 ≈ 121.3 km/hr
Average Velocity ≈ 121 km/hr due south.
Option C.
Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards
Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m
