Answer:
The sled needed a distance of 92.22 m and a time of 1.40 s to stop.
Explanation:
The relationship between velocities and time is described by this equation: 
, where 
is the final velocity, 
is the initial velocity, 
the acceleration, and 
is the time during such acceleration is applied.
Solving the equation for the time, and applying to the case: 
, where 
because the sled is totally stopped, 
is the velocity of the sled before braking and, 
is negative because the deceleration applied by the brakes.
In the other hand, the equation that describes the distance in term of velocities and acceleration:
, where 
is the distance traveled, is the initial velocity, the time of the process and, is the acceleration of the process.
Then for this case the relationship becomes: 
.
<u>Note that the acceleration is negative because is a braking process.</u>
If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.
For a given wave in a given medium, if the frequency doubles,
the wavelength becomes 50% shorter.
That is, it becomes half as long as it was originally.
Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
