CONSIDER A TRAIN WHICH CAN ACCELERATION OF 20CM/SEC AND SLOW DOWN WITH ACCELERATION OF 100 CM/SEC FIND THE MINIMUM TIME FOR TRAIN TO TRAVEL BETWEEN THE STATIONS 2.7 KM PER A PART
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time to put on the brakes, decelerate at the rate of 100, and come to a
screeching stop after a total distance of exactly 2.7 km.
The speed it reaches while accelerating is exactly the speed it starts decelerating from.
Speed reached while accelerating = (acceleration-1) (Time-1) = .2 time-1
Speed started from to slow down = (acceleration-2) (Time-2) = 1 time-2
The speeds are equal.
.2 time-1 = 1 time-2
time-1 = 5 x time-2
It spends 5 times as long speeding up as it spends slowing down.
The distance it covers speeding up = 1/2 A (5T)-squared
= 0.1 x 25 T-squared = 2.5 T-squared.
The distance it covers slowing down = 1/2 A (T-squared)
= 0.5 T-squared.
Total distance = 2,700 meters.
(2.5 + .5) T-squared = 2,700
T-squared = 2700/3 = 900
T = 30 seconds
The train speeds up for 150 seconds, reaching a speed of 30 meters per sec
and covering 2,250 meters.
It then slows down for 30 seconds, covering 450 meters.
Total time = 180 sec = 3 minutes, minimum.
Observation:
This solution is worth more than 5 points.
Considering train accelerate for 13000s and deaccelerate for 1000s
so total distance covered is = 20cm/s * 13000s + 100cm/s*100s
= (260000 + 10000 )cm
= 270000cm
= 2.7km
So total time taken is - 13100 seconds
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(output work) = (output force) x (distance)
450 N-m = (output force) x (3 meters)
Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)
= 150 newtons .
With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.
It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________
Well, my goodness !
I didn't even need to go through all of that.
Given:
-- The input force to the machine is 50 newtons.
-- The mechanical advantage of the machine is 3 .
That right there tells us that
-- The output force of the machine is 150 newtons.
We don't need any of the other given information.
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A dwarf planet looks like a planet, but is missing things to become a planet. For example, Pluto is now a dwarf planet.
I hope this helps;)
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Answer:
Explanation:
Let the area of one circular side be given by the formula :
However, the wire is a solid cylinder, then it means that the total area is 2 × =
However, there is the surface area to consider. This is the curved area of the wire. This is given as:
The length is x.
The breadth is calculated as follows - the length of the circle =
Then the area = lb
=
Therefore, the total area is given as
=
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Answer:
magnitude of force is 1725 N
direction is opposite to the direction of car moving i.e west
Explanation:
Given data
car mass = 1380 kg
speed = 27.0 m/s
time = 8 s
speed = 17 m/s
to find out
magnitude and direction of the net force
solution
we find here first acceleration thats formula is
a = v(initial) - v(final) / time
a =27 - 17 / 8
acceleration = 1.25 m/s
so force is = ma
force is = 1380 ( 1.25)
force = 1725 N
so magnitude of force is 1725 N
and this is here direction is opposite to the direction of car moving i.e west
Final answer:
The net force that leads to the deceleration of the 1380 kg car from 27.0 m/s to 17.0 m/s over 8.00 seconds has a magnitude of 1725 N and is directed towards the west.
Explanation:
To solve for the magnitude and direction of the net force that produces the deceleration in the car, you first need to calculate the acceleration and then use it to find the net force using the formula f = m * a. The acceleration here represents a deceleration because the speed is decreasing.
Acceleration can be found using the formula a = (v_f - v_i) / t. Here, v_i is the initial speed (27.0 m/s), v_f is the final speed (17.0 m/s), and t is the time interval (8.00 s). This gives you a = (17.0 m/s - 27.0 m/s) / 8.00 s = -1.25 m/s^2. The negative sign represents deceleration.
Now, apply this deceleration to the formula for force, f = m * a. Here, m is the mass of the car (1380 kg) and a is the value we computed (-1.25 m/s^2). So, f = 1380 kg * -1.25 m/s^2 = -1725 N. The negative sign indicates that the force is in the direction opposite to the initial motion of the car. Therefore, the magnitude of the net force is 1725 N, and the direction is towards the west or opposite to the car's original path.
Learn more about Net Force Calculation here:
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