Air, water and salt solution are given in the table.
1 answer:
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Answer:
The shortest possible stopping distance of the car is 175.319 meters.
Explanation:
In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:
(Eq. 1)
(Eq. 2)
After quick handling, we get that deceleration experimented by the car is equal to:
(Eq. 3)
Where:
- Deceleration of the car, measured in meters per square second.
- Kinetic coefficient of friction, dimensionless.
- Gravitational acceleration, measured in meters per square second.
If we know that and
, then deceleration of the car is:
The stopping distance of the car (), measured in meters, is determined from the following kinematic expression:
(Eq. 4)
Where:
- Initial speed of the car, measured in meters per second.
- Final speed of the car, measured in meters per second.
If we know that ,
and
, stopping distance of the car is:
The shortest possible stopping distance of the car is 175.319 meters.
Answer:
The angle is 18.3 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?
surface charge density, σ = 4 x 10^-5 C/m^2
Charge, q = 10^-8 C
mass, m = 0.008 kg
Let the angle is A and the tension in the string is T.
The electric field due to a plane is
Now equate the forces,