In the alpha decay of Francium-221, what daughter element is produced? (You may need to refer to the periodic table in the lesso
2 answers:
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Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:
where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
Answer:
(i) 12 seconds
(ii) 216 meters from the initial position
(iii) 132 meters from the initial position
(iv) No
Explanation:
Speed of express train =36 m/s
Speed of local train =11 m/s
The initial distance between the local train and passenger train =100 m.
Due to the application of breaks, the express train slows at the rare of
So, the acceleration of the express train, .
(i) Let t be the time the express train takes to stop.
From the equation of motion,
v=u+at
where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.
In this case, v=0, u=36 m/s,
So, 0=36+(-3)t
seconds.
(ii) To compute the distance traveled, s, till the express train stops, using
meters.
(iii) The local train is moving at a speed of 11 m/s
So, in 12 seconds, the distance, d, traveled by the local train
d= 11x12=132 meters [as distance= speed x time]
(iv) Let 0 be the reference position which is the initial position of the express train.
So, at the initial time, the position of the local train is at 100m.
After 12 seconds:
The position of the express train is at 216 m [using part (ii)]
and the position of the local train is at 100+132=232m [using part (iii)].
So, the local train is still ahead of the express train, hence the trains didn't collide.