BACHELOR OF OCCUPATIONAL HEALTH AND SAFETY MANAGEMENT WITH
HONOURS (BOSHM)
SEMESTER MAY 2022
SBPH1103
PHYSICS
MATRICULATION NO
:
910420105443001
IDENTITY CARD NO.
:
910420105443
TELEPHONE NO.
:
017-3479477
E-MAIL
:
luqman202122@oum.edu.my
LEARNING CENTRE
:
PULAU PINANG LEARNING
CENTRE
QUESTION 1
a)
Calculate the density of the moon by assuming it to be a sphere of diameter 3475
km and having a mass of 7.35 × 1022 kg. Express your answer in g/cm3.
(3 marks)
b)
A car accelerates from zero to a speed of 36 km/h in 15 s.
i.
Calculate the acceleration of the car in m/s2.
ii.
If the acceleration is assumed to be constant, how far will the car travel
in 1 minute ?
iii.
Calculate the speed of the car after 1 minute.
(5 marks)
c)
Su Bingtian, Asia’s fastest man, is running along a straight line. Assume that he
starts from rest from point A and accelerates uniformly for T s, before reaching a
speed of 3 m/s. He is able to maintain this speed for 5 s. After that, it takes him 6
s to decelerate uniformly to come to a stop at point B.
i.
ii.
iii.
Sketch a speed versus time graph based on the information given above.
Find the value of T if the distance between A and B is 100 m.
Determine the deceleration.
(7 marks)
Question 1
a)
Given,
Diameter = 3475 km
Mass, 𝑚 = 7.35 × 1022 𝑘𝑔
Mass, 𝑚 = 7.35 × 1025 𝑔
Radius, 𝑟 = 1737.5 km
Radius, 𝑟 = 1.7375 × 108 𝑐𝑚
Density, 𝑝 =
𝓂
𝑉
4 3
𝜋𝑟
3
4
𝑉 = (3.142)(1.7375 × 108 )3
3
𝑉=
𝑉 = 2.197 × 1025 𝑐𝑚3
𝑝=
𝓂
𝑉
𝑝=
7.35 × 1025 𝑔
2.197 × 1025 𝑐𝑚3
∴ 𝑝 = 3.3455 g/cm3
b)
i.
Acceleration, 𝑎 =
𝑣−𝑢
𝑡
Given,
Initial velocity, 𝑢 = 0 m/s
Final velocity, 𝑣 = 36 km/h
convert to 𝑚/𝑠,
5
𝑣 = 36 km/h ×
18
𝑣 = 10 m/s
Time, 𝑡 = 15𝑠
𝑣−𝑢
𝑡
10 − 0
𝑎=
15
2
𝑎=
3
∴ 𝑎 = 0.67 𝑚/𝑠 2
𝑎=
ii.
1
Distance, 𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2
Given,
Initial velocity, 𝑢 = 0 m/s
Acceleration, 𝑎 = 0.67 𝑚/𝑠 2
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1
𝑠 = (0)(60) + (0.67)(60)2
2
∴ 𝑠 = 1206𝑚
iii.
Speed =
1206
60
∴ Speed = 20.1 𝑚/𝑠
Speed =
Distance
Time
Time, 𝑡 = 60𝑠
c)
i.
Speed − Time graph
𝑣 (𝑚𝑠 −1 )
3
5
T
6
ii.
Area of Trapezium, 𝑠 =
1
(𝑎 + 𝑏)ℎ
2
Given distance, 𝑠 = 100m ,
3𝑇 + 48
= 100
2
1
𝑠 = (5 + (𝑇 + 5 + 6))(3)
2
3𝑇 + 48
𝑠=
2
3𝑇 + 48 = 200
3𝑇 = 200 – 48
3𝑇 = 152
𝑇=
152
3
∴ 𝑇 = 50.67
iii.
Acceleration, 𝑎
Given,
Initial velocity, 𝑢 = 3 m/s
Final velocity, 𝑣 = 0 m/s
Time, 𝑡 = 6 𝑠
=
𝑣−𝑢
𝑡
𝑣−𝑢
𝑡
0−3
𝑎=
6
𝑎=
∴ 𝑎 = −0.5 𝑚/𝑠2
𝑡 (𝑠)
QUESTION 2
a)
An object of mass 2 kg is launched at an angle of 30o above the ground with an
initial speed of 40 m/s. Neglecting air resistance , calculate:
i.
the kinetic energy of the object when it is launched from the the
ground.
ii.
the maximum height attained by the object .
iii.
the speed of the object when it is 12 m above the ground.
(8 marks)
b)
According to a local scientist, a typical rain cloud at an altitude of 2 m will contain,
on average, 3×107 kg of water vapour. Determine how many hours it would take
a 2.5 kW pump to raise the same amount of water from the Earth’s surface to the
cloud’s position.
(3 marks)
c)
In Figure 1, two forces F1 and F2 act on a 5 kg object that is initially at rest. If the
magnitude of each force is 10 N, calculate the acceleration produced.
Figure 1
(4 marks)
Question 2
a)
i.
1
Kinetic energy = 2 𝑚𝑣 2
Given,
Mass, 𝑚 = 2 kg
Initial speed, 𝑣 = 40 m/s
1
𝑚𝑣 2
2
1
𝐾. 𝐸 = (2)(40)2
2
∴ K. E = 1600 Joule
𝐾. 𝐸 =
ii.
Given,
Mass, 𝑚 = 2 kg
Initial speed, 𝑣 = 40 m/s
Angle, 𝜃 = 30
Gravity, 𝑔 = −10 m/𝑠 2
Initial velocity in y − direction, 𝜈𝑦
𝑣𝑦 = 𝑣 sin 𝜃
𝑣𝑦 = (40) (sin 30)
𝑣𝑦 = 20 m/s
and y is,
𝑣𝑦 2
𝑦=
2𝑔
202
𝑦=
2(10)
∴ 𝑦 = 20 m in height
iii.
Given,
Mass, 𝑚 = 2 kg
Height, ℎ = 12 m
Gravity, 𝑔 = −10 m/𝑠 2
𝐸 = 𝑚𝑔ℎ
1
𝐸 = 𝑚𝑣 2
2
1
𝑚𝑔ℎ = 𝑚𝑣 2
2
2𝑔ℎ = 𝑣 2
𝑣 = √2𝑔ℎ
𝑣 = √2𝑔ℎ
𝑣 = √2(10)(12)
∴ 𝑣 = 15.5 𝑚/𝑠
b)
Given,
ℎ = 2𝑚
𝑚 = 3 × 107 𝑘𝑔
𝑃 = 2.5 𝑘𝑊
Gravity, 𝑔 = −10 m/𝑠 2
Power, 𝑃 =
Work done, 𝑊
Time (seconds), 𝑠
𝑊
𝑃
(3 × 107 )(10)(2)
𝑠=
2.5 × 103
𝑠 = 240000 seconds
𝑠=
Convert seconds to hour,
𝑠 = 240000 seconds
240000
𝑠=
3600
∴ 𝑠 = 66 hours 40 min
Work done by pump, 𝑊 = 𝑚𝑔ℎ
c)
Given,
𝐹1 = 10 𝑁
𝐹2 = 10 𝑁
𝑚 = 5 𝑘𝑔
Angle, 𝜃 = 60
𝑎 =
∑𝐹
𝑚
𝑎 =
10𝑖̂ + 5𝑖̂ + 8.7𝑗̂
5
15𝑖̂ + 8.7𝑗̂
5
𝑎 = 3𝑖̂ + 1.74𝑗̂
𝑎 =
Magnitude,
𝑎 = √32 + 1.742
∴ 𝑎 = 3.47 𝑚/𝑠 2
⃗⃗⃗
𝐹1 = 10𝑖̂
⃗⃗⃗
⃗⃗⃗2 | × (𝑐𝑜𝑠 60°𝑖̂ + 𝑠𝑖𝑛 60°𝑗̂)
𝐹2 = |𝐹
⃗⃗⃗
𝐹2 = 10(0.5𝑖̂ + 0.87𝑗̂)
⃗⃗⃗
𝐹2 = 5𝑖̂ + 8.7𝑗̂
QUESTION 3
a)
You would like to heat 10 litres of tap water initially at room temperature
using an old 2 kW heater that has an efficieny of 70%. Estimate the
temperature of the water after 20 minutes stating any assumptions made.
(5 marks)
b)
Determine the amount of heat needed to completely transform 1 g of water at
15°C to steam at 115°C.
(Obtain any relevant data that you need from the internet. Cite the source of that
data in your answer)
(5 marks)
Question 3
a)
Given,
𝑚𝑤𝑎𝑡𝑒𝑟 = 10 ℓ = 10 𝑘𝑔
Initial water temperature, 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 20℃
∆𝑇 = final temperature – initial temperature
𝑃𝑜𝑤𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑟 = 2 𝑘𝑊 = 2000 𝐽
𝑃𝑜𝑤𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑟 𝑤𝑖𝑡ℎ 70% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦,
𝑃𝑜𝑤𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑟 = 2000 𝐽 × 70%
𝑃𝑜𝑤𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑟 = 1400 𝐽
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟, 𝑐 = 4182 J/kg ℃
𝑄 = 𝑡𝑃
𝑄 = (1200)(1400)
𝑄 = 1680000 𝐽
𝑚𝑐∆𝑇 = 𝑡𝑃
1680000 𝐽 = 𝑚𝑐∆𝑇
1680000 = (10) (4182)( 𝑇𝑓𝑖𝑛𝑎𝑙 − 20℃)
1680000
41820
𝑇𝑓𝑖𝑛𝑎𝑙 − 20℃ = 40.17℃
𝑇𝑓𝑖𝑛𝑎𝑙 = 40. 17℃ + 20℃
∴ 𝑇𝑓𝑖𝑛𝑎𝑙 = 60.17℃
𝑇𝑓𝑖𝑛𝑎𝑙 − 20℃ =
Time, 𝑡 = 20 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Convert to seconds,
Time, 𝑡 = 20 × 60
Time, 𝑡 = 1200 𝑠
𝑄 = 𝑚𝑐∆𝑇
𝑄 = 𝑡𝑃
𝑚𝑐∆𝑇 = 𝑡𝑃
b)
Given,
𝑚𝑤𝑎𝑡𝑒𝑟 = 1 𝑔
Initial water temperature = 15℃
Final steam temperature = 115℃
∆𝑇 = final temperature – initial temperature
Heat of fusion vaporization of water, 𝛥𝐻𝑣 = 2257 𝐽/g
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟, 𝑐 = 4.18 J/g ℃
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚, 𝑐 = 2.09 J/g ℃
𝑇ℎ𝑒 ℎ𝑒𝑎𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑔𝑜 𝑓𝑟𝑜𝑚 15℃ 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 100℃ 𝑤𝑎𝑡𝑒𝑟,
𝑄1 = 𝑚𝑐∆𝑇
𝑄1 = (1)(4.18)(100℃ − 15℃)
𝑄1 = 355.3 𝐽
𝑇ℎ𝑒 ℎ𝑒𝑎𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 100℃ 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 100℃ 𝑣𝑎𝑝𝑜𝑟,
𝑄2 = 𝑚𝛥𝐻𝑣
𝑄2 = (1)(2257)
𝑄2 = 2257 𝐽
𝑇ℎ𝑒 ℎ𝑒𝑎𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑔𝑜 𝑓𝑟𝑜𝑚 100℃ 𝑣𝑎𝑝𝑜𝑟 𝑡𝑜 115℃ 𝑣𝑎𝑝𝑜𝑟,
𝑄3 = 𝑚𝑐∆𝑇
𝑄3 = (1)(2.09)(115℃ − 100℃)
𝑄3 = 31.35 𝐽
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑠,
𝑄𝑇𝑜𝑡𝑎𝑙 = 𝑄1 + 𝑄2 + 𝑄3
𝑄𝑇𝑜𝑡𝑎𝑙 = 355.3 + 2257 + 31.35
∴ 𝑄𝑇𝑜𝑡𝑎𝑙 = 2643.65 𝐽
# source from : https://socratic.org/questions/how-many-joules-of-heat-are-needed-to-change50-0-grams-of-ice-at-15-0-c-to-stea