<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.
<h3>What is a momentum?</h3>
- In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
- It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
- The kilogram metre per second (kg m/s), or newton-second in the International System of Units (SI), is the unit used to measure momentum.
- The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion.
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The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2


The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
The answer would be B, George Darwin :)
80000 Joule is the change in the internal energy of the gas.
<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>
ΔW = -pdV
ΔW = -pd (V₂ -V₁)
ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)
= - 0.35×10⁵ pa.m³
= - 35000 (N/m³)(m³)
= -35000 Nm
ΔW = -35000 Joule
Therefore, work done by the system = -35000 Joule
<h3>Change in the internal energy of the gas,</h3>
ΔV = ΔQ + ΔW
Given:
ΔQ = 1.15×10⁵ Joule
ΔW = -35000 Joule
ΔU = 1.15×10⁵ Joule - 35000 Joule
= 80000 Joule.
Therefore, the change in the internal energy of the gas= 80000 Joule.
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