It’s not flat because how can we be 3D when our planet is flat
D. Velocity because it describes a speed and direction
A) The answer is 11.53 m/s
The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi
Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²
Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h
So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h
We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m
1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s
b) The answer is 6.78 m
The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE
Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²
Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h
KE = PE
1/2 m * v² = m * a * h
Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?
1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
<h2>
Answer:</h2>
-310J
<h2>
Explanation:</h2>
The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e
ΔE = Q + W ----------------------(i)
If heat is released by the system, Q is negative. Else it is positive.
If work is done on the system, W is positive. Else it is negative.
<em>In this case, the system is the balloon and;</em>
Q = -0.659kJ = -695J [Q is negative because heat is removed from the system(balloon)]
W = +385J [W is positive because work is done on the system (balloon)]
<em>Substitute these values into equation (i) as follows;</em>
ΔE = -695 + 385
ΔE = -310J
Therefore, the change in internal energy is -310J
<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>
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