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.
Which of the following are characteristics of noble gases?
✅
- An inert gas is one that does not undergo chemical reactions. The noble gases have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why they are said to be inert.
✅
- Noble gases are the least reactive of all elements. This is because they already have the desired eight total 's' and 'p' electrons in their outermost (highest) energy level.
Given,
the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
Answer:
acceleration of the rocket is given as
Explanation:
As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min
so we have
so we have
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
