Final speed = initial speed + (acceleration x time)
(final speed - initial speed) = acceleration x time
Time = (final speed - initial speed) / acceleration
Answer:
v = 2.928 10³ m / s
Explanation:
For this exercise we use Newton's second law where the force is the gravitational pull force
F = ma
a = F / m
Acceleration is
a = dv / dt
a = dv / dr dr / dt
a = dv / dr v
v dv = a dr
We substitute
v dv = a dr
∫ v dv = 1 / m G m M ∫ 1 / r² dr
We integrate
½ v² = G M (-1 / r)
We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon
v² = 2G M (- 1 / R +2.73 10³+ 1 / R)
We calculate
v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))
v² = 14.6828 10⁷ (0.1783 -0.1199)
v = √8.5748 10⁶
v = 2.928 10³ m / s
<span>a) 1960 m
b) 960 m
Assumptions.
1. Ignore air resistance.
2. Gravity is 9.80 m/s^2
For the situation where the balloon was stationary, the equation for the distance the bottle fell is
d = 1/2 AT^2
d = 1/2 9.80 m/s^2 (20s)^2
d = 4.9 m/s^2 * 400 s^2
d = 4.9 * 400 m
d = 1960 m
For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes
d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T
d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s
d = 4.9 m/s^2 * 400 s^2 - 1000 m
d = 4.9 * 400 m - 1000 m
d = 1960 m - 1000 m
d = 960 m</span>
Answer:
C. 3.00 s
Explanation:
Given:
Δy = 1.80 m − 46.0 m = -44.2 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 3.00 s
Answer:
To describe motion accurately and completely, a frame of reference is necessary. frame of reference ( or reference frame) consists of an abstract coordinate system and the set of physical reference points that uniquely fix ( locate and orient ) the coordinate system and standardize measurements within that frame.
Explanation:
The different observations occur because the two observers are in different frames of reference. A frame of reference is a set of coordinates that can be used to determine positions and velocities of objects in that frame; different frames of reference move relative to one another.
