Answer: 7.5 rev/s
Explanation:
We are given the angular velocity
a helicopter's main rotor blades:
![\omega=450 rpm=450 \frac{rev}{min}](https://tex.z-dn.net/?f=%5Comega%3D450%20rpm%3D450%20%5Cfrac%7Brev%7D%7Bmin%7D)
However, we are asked to express this
in the International Systrm (SI) units. In this sense, the SI unit for time is second (
):
![\omega=450 \frac{rev}{min} \frac{1 min}{60 s}](https://tex.z-dn.net/?f=%5Comega%3D450%20%5Cfrac%7Brev%7D%7Bmin%7D%20%5Cfrac%7B1%20min%7D%7B60%20s%7D)
![\omega=7.5 \frac{rev}{s}](https://tex.z-dn.net/?f=%5Comega%3D7.5%20%5Cfrac%7Brev%7D%7Bs%7D)
Not knowing how to convert the type of measurement according to your way of learning from where ever you come from.
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer:
1224km/hr
Explanation:
To convert from m/s to km/hr
1000m = 1km
Divide both sides by 1000
1m = 1/1000 km................. (1)
60×60 seconds = 1 hr
3600s = 1hr
Divide both sides by 3600
1s = 1/3600 .............(2)
Divide (2) by (1)
1m/s = 1/1000 ÷ 1/3600 km/hr
1m/s = 1/1000 × 3600/1 km/hr
1m/s = 3600/1000 km/hr
1m/s = 3.6 km/hr .............(3)
To convert 340m/s to km/hr
Multiply (3) by 340
1× 340m/s = 3.6 × 340 km/hr
340m/s = 1224km/hr
I hope this was helpful, please mark as brainliest