F = kq1q2/r²
F = 9.0(10^9)(10)(50)/20²
F = 1.125 * 10^10 N
Use this formula for an object starting from rest and accelerating:
Distance = (1/2) (acceleration) (time)²
In this problem, the distance is 40 meters, and acceleration is gravity.
So . . .
40 m = (1/2) (9.8 m/s²) (time)²
Divide each side by (4.9 m/s²) :
time² = (40 m) / (4.9 m/s²)
time² = 8.16 sec²
Take the square root of each side :
√(time²) = √(8.16 sec²)
time = 2.86 seconds
Round it to 2.9 sec <em>(choice B)</em>
When the engine is near the governed RPMand the transmission is in the lower gears, the braking effect of the engine is at its greatest .
It is specified that the legal requirements for the braking system that:
A reasonable distance should be practiced in stopping your vehicle.
Adjournment from the jar or shudder should be smooth and free.
A good braking system’s other necessities are:
Unaffected by heat, water, road grit, dust etc.
Constant continued application while moving downwards from a hill should not decrease its effectiveness and it must have good anti fade characteristics
Adequate durability with economical maintenance and adjustment Its durability must be enough and must have proper adjustment and maintenance.
The variables that show a direct relationship are :
- The speed of a car and the distance traveled
- Number of students in a cafeteria and the amount of food consumed
- The distance a planet is from the sun and that planet's temperature
- The mass of a space shuttle and its acceleration through space
In direct relationship, when one factor is increased/decreased , it will directly cause the other factor to be increased/decreased
Any photos so i can help you with that?