Answer:
Explanation:
Given
average speed of train
Maximum acceleration=0.05g
Now centripetal acceleration is
r=7346.93 m
(b)Radius of curvature=900 m
therefore
70 kg to mg i need to show the work of how i did it
1 answer:
You might be interested in
<h2>Answer: x=125m, y=48.308m</h2>
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:
x-component:
(1)
Where:
is the projectile's initial speed
is the angle
is the time since the projectile is launched until it strikes the target
is the final horizontal position of the projectile (the value we want to find)
y-component:
(2)
Where:
is the initial height of the projectile (we are told it was launched at ground level)
is the final height of the projectile (the value we want to find)
is the acceleration due gravity
Having this clear, let's begin with x (1):
(3)
(4) This is the horizontal final position of the projectile
For y (2):
(5)
(6) This is the vertical final position of the projectile
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²