Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.
Answer:
D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Explanation:
Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.
An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.
Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Answer:the rate changes during the position of the object
Explanation:so there is no object that has the same rate but unless it is a specific one like a care but it changes during the position of the object
Answer:2.55 rad/s
Explanation:
Given
Diameter of ride=5 m
radius(r)=2.5 m
Static friction coefficient range=0.60-1
Here Frictional force will balance weight
And limiting frictional force is provided by Centripetal force
![f=\mu N=\mu m\omega ^2\cdot r](https://tex.z-dn.net/?f=f%3D%5Cmu%20N%3D%5Cmu%20m%5Comega%20%5E2%5Ccdot%20r)
weight of object=mg
Equating two
f=mg
![\mu m\omega ^2\cdot r=mg](https://tex.z-dn.net/?f=%5Cmu%20m%5Comega%20%5E2%5Ccdot%20r%3Dmg)
![\omega ^2=\frac{g}{\mu r}](https://tex.z-dn.net/?f=%5Comega%20%5E2%3D%5Cfrac%7Bg%7D%7B%5Cmu%20r%7D)
![\omega =\sqrt{\frac{g}{\mu r}}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Csqrt%7B%5Cfrac%7Bg%7D%7B%5Cmu%20r%7D%7D)
![\omega =2.55 rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D2.55%20rad%2Fsec)