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.
1) d = V*t >>>as you double the av. speed the distance become doubled.
2) after you draw the victors you will find the total displacement = 1 meter to the left.
3) V =d/t =( 8km)/(1.25hr) = 6.4km/hr
There are problems with the first sentence, and it's not really needed when
working with this question. So let's just take the 20 (Hz ?) frequency from
the first sentence, and ignore the rest of it for right now.
Wavelength = (speed) / (frequency) =
(331 m/s) / (20 Hz) = <em>16.55 meters</em>.
It will be -1 because in valence shell there will be 7 electrons so it will take 1 from other atom to complete its outer shell