To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.
The angular velocity as a function of the period is described as
Where,
Angular velocity
T = Period
At the same time the relationship between Angular velocity and linear velocity is described by the equation.
Where,
r = Radius
Our values are given as,
We also know that the radius of the earth (r) is approximately
Usando la ecuación de la velocidad angular entonces tenemos que
Then the linear velocity would be,
x
The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is 463.96
GEL IT IS COMBINATION OF SOLID AND LIQUID
Answer:
Poke or kick the boxes. The sand box is the one that resists the most
Explanation:
The one that resists the most a change in motion (inertia) is the one with the greater mass—the one filled with sand. Since we cannot open the boxes or perform any weight measurement we can distinguish them by their inertia. Sand has greater mass and thus greater inertia
The power delivered by the elevator is 34496 W.
<h3>What is power?</h3>
Power is the rate at which energy is used up.
To calculate the power delivered by the elevator, we use the formula below.
Formula:
- P = mgh/t.............. Equation 1
Where:
- P = Power delivered by the elevator
- m = Total mass on the elevator
- g = acceleration due to gravity
- h = Height
From the question,
Given:
- m = 1.1×10³ kg
- h = 40 m
- t = 12.5 s
- g = 9.8 m/s²
Substitute these values into equation 1
- P = 1.1×10³×40×9.8/12.5
- P = 34496 W.
Hence, the power delivered by the elevator is 34496 W.
Learn more about power here: brainly.com/question/24858512
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Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g 
let's calculate
a = 0.65 9.8 
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
