Answer:
The force has been reduced by 8018 N
Explanation:
The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

where:
F is the force exerted on the car
is the duration of the collision
m = 1400 kg is the mass of the car
is the change in velocity of the car
We can re-write the equation as

In the 1st collision, the time is 1.5 seconds, so the force is

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

Therefore, the force has been reduced by:

Answer:
times
Explanation:
First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.
Age of the universe:
(1 followed by 17 zeroes)
Lifetime of the top quark:
(we moved the decimal point 24 places to the right)
Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

1) The buoyant force acting on an object immersed in a fluid is:

where

is the density of the fluid,

is the volume of displaced fluid, and

is the gravitational acceleration.
2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is

. Using the relationship between density, volume and mass, we find

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is

, so we have

4) The weight of 1 Kg of titanium is instead:

So, the buoyant force is negligible compared to the weight.
Answer:
a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm
b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm
Explanation:
Given data:
L = 20 mm
F = 250 N
r₁ = 10 mm
r₂ = 15 mm
v = 0.3
E = 2.07x10⁵ MPa

a) The maximum contact pressure is:

The width of contact is:

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is
T = 0.3*P = 0.3 * 274.58 = 82.37 MPa
At a distance of
0.8*b = 0.8*0.029 = 0.023 mm