Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula
Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then
Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Answer:
as its mass and velocity will less so its momentum will be less than that of baseball
Weight changes due to the gravitational pull in mars,basically gravity
I think the first answer is hypocenter
the second answer is epicenter
Have a blessed day!
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