What is the force applied to an object that has a mass of 200 kg and is accelerated to 40 m/s^2?
2 answers:
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Answer:
Approximately (assuming that this spring is ideal.)
Explanation:
The displacement of a spring is the new length of the spring relative to the original length.
For example:
- When the
-spring in this question is stretched to
, the displacement is
.
- Likewise, if this spring is stretched to
, the displacement would be
.
If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of displaces this spring by
, while a force of
displaces this spring by
, then:
.
In this question, it is given that a force of would stretch this spring by
. Thus, the force
required to stretch this spring by
would satisfy:
.
Rearrange and solve for :
.
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = ( ) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth