Answer:
Explanation:
To find the expression in terms of time t you take into account the following equation for the angular distance traveled by an object with angular acceleration w and initial angular position θo:
( 1 )
α is the angular acceleration, but in this case you have a circular motion with constant angular speed, then α = 0 rad/s^2. θo is the initial angular position, the information of the question establishes that Enrique is at 3-o'clock. This position can be taken, in radian, as π/4 (for 12-o'clock = 0 rads).
The angular speed is:
You replace the values of θo, α and w in the equation ( 1 ):
Furthermore, the arc length is:
Answer:
Explanation:
Given that,
Mass of a brick, m = 3.1 kg
The dimensions of the brick 225 m x 112 m x 75 m
We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :
F = mg
A = area with minimum dimensions i.e. 112 m x 75 m
Pressure is maximum when the area is least.
So,
So, the maximum pressure created by brick is .
The answer is A. 1000 times smaller
The metric prefix mili- is used to signify something 1000 times smaller than a measure. For example in 1 meter there are 1000 milimeters. Therefore; A is correct.
Hope this helped. Have a great night!
Answer:
v = 91.8 Km / h
Explanation:
We must start this exercise at the end, let's look for the lighter car acceleration (B), for this we use Newton's second law
fr = m a
a = fr / m
fr = μ N
N-W = 0
Let's replace
a = μ m g / m
a = μ g
a = 0.8 9.8
a = 7.84 m / s²
As the car B after the crash reached an initial velocity vo₀₂ and at the end of the fine speed zero, let us use kinematics
v² = v₀₂² - 2 a x
0 = v₀₂² - 2 a x
v₀₂ = 2 a x₂
v₀₂ = √ (2 7.84 26)
v₀₂ = 20.19 m / s
Let's perform the same procedure for car A, the acceleration is the same as it does not depend on the mass of the vehicles
v₀₁ = √ 2 a x₁
v₀₁ = √ (2 7.84 19)
v₀₂ = 17.36 m / s
Now let's use moment conservation, where the system is the two vehicles
Initial before crash.
p₀ = M v₁ + 0
After the crash
= M v₀₁ + m v₀₂
p₀ =
M v₁ = M v₀₁ + m v₀₂
v₁ = v₀₁ + m / M v₀₂
v₁ = 17.36 + 1841/3000 20.19
v₁ = 20.75 m / s
This is the speed of car 1 (A) just before the crash, now let's look for the speed when I apply the brakes the initial speed (v)
v₁² = v² - 2 a x₁
v = √ (v₁² + 2 a x₁)
v = √ (20.75² + 2 7.84 14)
v = 25.50 m / s
v = 25.50 m / s (1km / 1000m) (3600s / 1h)
v = 91.8 Km / h