Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.
The centripetal acceleration of any object moving in a circle is
(speed-squared) / (radius of the circle) .
Notice that we won't need to use the mass of the train.
We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.
Speed = (distance traveled) / (time to travel the distance).
Distance = 10 laps of the track. Well how far is that ? ? ?
1 lap = circumference of the track = (2π) x (radius) = 2.4π meters
10 laps = 24π meters.
Time = 1 minute 20 seconds = 80 seconds
The trains speed is (distance) / (time)
= (24π meters) / (80 seconds)
= 0.3 π meters/second .
NOW ... finally, we're ready to find the centripetal acceleration.
<span> (speed)² / (radius)
= (0.3π m/s)² / (1.2 meters)
= (0.09π m²/s²) / (1.2 meters)
= (0.09π / 1.2) m/s²
= 0.236 m/s² . (rounded)
If there's another part of the problem that wants you to find
the centripetal FORCE ...
Well, Force = (mass) · (acceleration) .
We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>
If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),
(0,r) + (s cos297,s sin297) = (6,0)
now just solve for r and s.
Pls mark me as brainliest
39.2 J
Explanation:
Step 1:
To find the potential energy the following formula is used.
Potential Energy = m × g × h
Where,
m = Mass
g = Acceleration due to gravity
h = Height
Step 2:
Here m = 4 kg, g = 9.8 m/s², h = 1 m
Potential Energy = ( 4 × 9.8 × 1)
= 39.2 J
Answer:
v = 24 cm and inverted image
Explanation:
Given that,
The focal length of the object, f = +8 cm
Object distance, u = -12 cm
We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :
Put all the values,
So, the image distance from the lens is 24 cm.
Magnification,
The negative sign of magnification shows that the formed image is inverted.
