Round 1
Tuesday, September 15, 2015
HighFour Physics
Category D: Grades 11 – 12
The use of calculator is required.
Answer #1:
Ohm’s
Answer #2:
Pascal’s / Pascal’s principle
Answer #3:
Isothermal
Answer #4:
Explanation:
Double Shear
It is considered as double shearing because the sections are sliding past off
each other, thus creating two sheared areas.
Answer #5:
Solution:
52
m
Use the formula ρ = and manipulate to arrive at m = Vρ.
V
One of the ways of solving the given problem is:
V = 12.5 ft x 15.5 ft x 8.0 ft = 1,550 ft 3
1m 3
3
V = 1,550 ft �
� = 43.92 m3
3.28ft
g
g
100cm 3
1L
ρ = 1.19 �
=
1,190
�
�
�
L 1000cm3
m3
1m
m = 1,190
g
(43.92 m3 ) = 52,264.8 g
m3
m = 52,264.8 g �
Answer #6:
Brinell
1kg
� = 52.2648 kg ≅ 52 kg
1000g
Round 1
Tuesday, September 15, 2015
HighFour Physics
Category D: Grades 11 – 12
The use of calculator is required.
Answer #7:
Solution:
503
Create a force diagram as shown below to better visualize the forces acting
in equilibrium.
T2
T1
50°
600 N
∑ 𝐹𝐹𝐹𝐹= 0:
∑ 𝐹𝐹𝐹𝐹= 0:
𝑇𝑇2 cos 50° − 𝑇𝑇1 = 0
𝑇𝑇2 sin 50° − 600 = 0
Substitue value of T2 in Equation 1:
;
;
cos(50°)T2 = T1 (Equation 1)
sin(50°)T2 = 600 (Equation 2)
T2 = 783.2444 N
(cos(50°))(783.2444 N) = T1
T1 = 503.4598 N ≈ 503 N
Answer #8:
Explanation:
Proportional Limit
The proportional limit is the point on the stress-strain curve below which,
the stress is proportional to strain.
Answer #9:
Corrosion
Answer #10:
Explanation:
Convection
Convection can also refer to the flow of heat associated with the
movement of a fluid (such as gas or liquid). Examples could be illustrated as
when hot air from a furnace enters a room, or the transfer of heat from a
hot surface to a flowing fluid.
Round 1
Tuesday, September 15, 2015
HighFour Physics
Category D: Grades 11 – 12
The use of calculator is required.
Answer #11:
184.7
Solution:
Use the formula
F
A
∆L
∆L
= Y � � and manipulate to achieve F = Y � � A
L
L
(0.002 m)2
0.001 m
F = (117.6 x 10 Pa) �
�π�
� = 184.7256 N
4
2m
9
≈ 184.7 N
Answer #12:
Explanation:
Isolated / Isolated System
This system is called isolated system because neither mass nor energy is
allowed to cross the boundary.
Answer #13:
Solution:
absorbed 1.012W
Use the formula P = v𝑖𝑖;
1V
P = (220 mV) �
� (4.6 A) = 1.012 W
1000mV
Since the current enters the element through its positive end, the
calculated power is being absorbed by the element.
Answer #14:
364
Solution:
Use the formula FC =
FC =
m 2
(80 kg) �10 �
s
(22 m)
mv2
r
= 363.6364 N ≈ 364 N
Round 1
Tuesday, September 15, 2015
HighFour Physics
Category D: Grades 11 – 12
The use of calculator is required.
Answer #15:
Explanation:
Friction
The friction is the force that resists an
action being acted upon on one body.
force
friction
Θ
Answer #16:
Explanation:
Zero / 0
In an adiabatic process, no heat transfer occurs. Hence, there is zero heat
energy flowing into the gas.
Answer #17:
Explanation:
Concurrent
Concurrent forces intersect at a single point. This is the most common force
system encountered in physics and engineering problems. Below is an
example of a concurrent force system.
F = 20 kN
P = 10 kN
A
T = 30 kN
Forces P, F, and T meet at a single point A.
Answer #18:
Explanation:
Compression
Since the forces are acting towards the object on both ends, the bar is in a
state of compression.
Round 1
Tuesday, September 15, 2015
HighFour Physics
Category D: Grades 11 – 12
The use of calculator is required.
Answer #19:
Solution:
22.95
-
Answer #20:
Solution:
Use the formula v = R𝑖𝑖
v = (500 Ω)(45.9 mA) �
1A
1000mA
500 Ω
+
45.9 mA
� = 22.95 V
6
1
Use the formula y = v0y t ∓ gt 2 to solve for time.
2
m
Solve for v0y: v0y = v0 sin50° = �40 � sin50° = 30.6418
s
m
y = 0 (since starting and end points are at ground level)
g = -9.8 m/s2 (assign upward direction as positive)
Solve for t:
0 = �30.6418
0 = �30.6418
1
m
𝑚𝑚
� t − �9.8 2 � t 2
2
s
𝑠𝑠
m
𝑚𝑚
� t − �4.9 2 � t 2
s
𝑠𝑠
Solving for t will yield:
t = 6.25 seconds ≈ 6 seconds
s