Answer:
Distance: 4.6 km Displacement= -0.2 km
Explanation:
The overall distance covered: 1.5 + 2.4 + 0.7 = 4.6 km
Displacement calculation: 1.5 - 2.4 + 0.7 = -0.2 km
The displacement could also simply be stated as 0.2 km depending on whether negative value is preferred.
Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.
At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa
At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m
Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa
Explanation:
The term 'collision' refers to the interaction between two objects. There are two distinct types of collisions: elastic and inelastic.
In this scenario, two identical carts are heading towards each other at the same speed, resulting in a collision. In an inelastic collision, the momentum is conserved before and after the incident, but kinetic energy is lost.
After the event, both objects combine and move together at a single velocity.
The graph representing a perfectly inelastic collision is attached, illustrating that both carts move together at the same speed afterward.
Answer:
A = 4.76 x 10⁻⁴ m²
Explanation:
Given data:
Person's weight = 625 N
Bike's weight = 98 N
Pressure per tire = 7.60 x 10⁵ Pa
Find: Contact area per tire
Total system weight = 625 + 98 = 723 N
Let F represent the force supported by each tire
2F = 723 N
Therefore, F = 361.5 N
Using the formula F = P × A
Contact area, A = 4.76 x 10⁻⁴ m²
I created the illustration found in the accompanying file.
There are two images included.
The upper one illustrates the impacts of:
- scaling vector A by a factor of 1.5, depicted in red with a dashed line.
- scaling vector B by -3, shown in purple with a dashed line.
The lower image displays the resultant vector: C = 1.5A - 3B.
The approach involves relocating the tail of vector -3B to the tip of vector 1.5A while maintaining the angles.
Next, an arrow is drawn from the tail of 1.5A to the position of -3B after this shift.
The arrow representing the result is vector C, marked with a black dashed line.
