A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi
2 answers:
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Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:
where
is the acceleration due to gravity
is the mass of the block
is the speed of the block A just before touching block B
Solving for the speed,
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
where:
e = 0.7 is the coefficient of restitution in this case
is the final velocity of block B
is the final velocity of block A
is the initial velocity of block B
Solving,
Re-arranging it,
(1)
Also, the total momentum must be conserved, so we can write:
where
And substituting (1) and all the other values,
This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC
s = 68.94 m
3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.