Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Answer:
a = 2m/s^2
Explanation:
Force (F) = 100 N
Mass (m) = 50 kg
Here,
F = m×a
100 = 50 × a
a = 100÷50
a = 2m/s^2
Thus, the acceleration on the cart is a = 2m/s^2
-TheUnknownScientist
Answer:
Explanation:
When the spring is compressed by .80 m , restoring force by spring on block
= 130 x .80
= 104 N , acting away from wall
External force = 82 N , acting towards wall
Force of friction acting towards wall = μmg
= .4 x 4 x 9.8
= 15.68 N
Net force away from wall
= 104 -15.68 - 82
= 6.32 N
Acceleration
= 6.32 / 4
= 1.58 m / s²
It will be away from wall
Energy released by compressed spring = 1/2 k x²
= .5 x 130 x .8²
= 41.6 J
Energy lost in friction
= μmg x .8
= .4 x 4 x 9.8 x .8
= 12.544 J
Energy available to block
= 41.6 - 12.544 J
= 29 J
Kinetic energy of block = 29
1/2 x 4 x v² = 29
v = 3.8 m / s
This will b speed of block as soon as spring relaxes. (x = 0 )
