Friday, May 2, 2014
Sunday, September 22, 2013
Answers to Worksheet on Unit 11
Apologies for not giving the answers earlier.
1a 2.5°C
1b 1040 J
2a 900 J / K
2b 7200 J
2c 1800 J / K
2d No. More mass, more heat capacity (more ability to store heat before changing temperature)
3 4000 J / K
4a 900 J / (kg K)
4bi 3600 J
4bii 4000 J
5a 55°C
5b 12 min
5c Not all energy supplied to the water is absorbed by water, some energy is lost to the surroundings as heat.
6 K. Temperature change for K is higher, meaning that K is unable to hold that much heat before rising in temperature, thus specific heat capacity of K is lower.
7a 0.25 kg
7b 130 000 J
7c 35 J / kg
8a 2 100 000 J
8b 22 600 000 J
8c 400 000 J
9 1.21 kg
10a 36 000 J
10b 15.6°C
If there are any questions relating to the above answers, please do message me. Thanks.
1a 2.5°C
1b 1040 J
2a 900 J / K
2b 7200 J
2c 1800 J / K
2d No. More mass, more heat capacity (more ability to store heat before changing temperature)
3 4000 J / K
4a 900 J / (kg K)
4bi 3600 J
4bii 4000 J
5a 55°C
5b 12 min
5c Not all energy supplied to the water is absorbed by water, some energy is lost to the surroundings as heat.
6 K. Temperature change for K is higher, meaning that K is unable to hold that much heat before rising in temperature, thus specific heat capacity of K is lower.
7a 0.25 kg
7b 130 000 J
7c 35 J / kg
8a 2 100 000 J
8b 22 600 000 J
8c 400 000 J
9 1.21 kg
10a 36 000 J
10b 15.6°C
If there are any questions relating to the above answers, please do message me. Thanks.
Thursday, September 12, 2013
Answers to Worksheet 7
pg 55-56
1. A [gas supply on right pressing more ==> gas pressure has to be higher]
2. C [pressure of liquid = hpg]
3. D
Pressure on left = Pressure on right [pressure in hydraulic fluid same throughout]
F ÷ A = F ÷ A
10 N ÷ 10 cm² = (1500 x 10) ÷ A
A = 15000 cm²
4. A
Let weight of small cylinder be W and area of small cylinder be A
Therefore weight of big cylinder = 6W and area of big cylinder = 4A
Pressure of big cylinder = 6W ÷ 4A = 18 Pa
therefore 1.5 (W ÷ A) = 18 Pa
(W ÷ A) = 18 ÷ 1.5 = 12 Pa
Pressure of small cylinder = W ÷ A = 12 Pa
Section B
1b.
i. Weight on ground by each foot = 20000 N ÷ 4 = 5000 N
[since 4 feet, so weight is equally distributed among the 4]
Pressure = F ÷ A = 5000 N ÷ 200 cm² = 5000 N ÷ 0.02 m² = 250000 Pa
Alternatively, since pressure on one foot = pressure on all feet = 20000 N ÷ (4 x 200) cm² = 250000 Pa
ii. A = F ÷ Pressure = 600 N ÷ 250000 Pa = 0.0024 m²
2a.
i. Pressure at surface = atmospheric pressure = 101000 Pa
ii Pressure at 5m below = liquid pressure + Patm = hpg + 101000 = 5 x 1025 x 5 + 101000 = 152250 Pa
2b. 152250 - 101000 = 51250 Pa
3.
a. Pressure on small piston = F ÷ A = 120 N ÷ 20 cm² = 120 N ÷ 0.0020 m² = 60000 Pa
b. Pressure on large piston = pressure on small piston = 60000 Pa
c. Weight = Pressure x A = 60000 Pa x (300 cm² ÷ 10000) m² = 1800 N
4
a. Atmospheric pressure is equivalent to the pressure exerted by a mercury column of vertical height 76.0 cm
b. Pressure = hpg = 0.760 m x 13600 x 10 = 103360 Pa = 103000 Pa (3sf)
5a.
Pressure at point O = 0 Pa
Pressure at point P = hpg = (3h ÷ 4 ÷ 100) m x 13600 x 10 = 1020h Pa
Pressure at point Q = hpg = (h ÷ 100) m x 13600 x 10 = 1360h Pa
5b.
Shorter length of barometer as mercury higher density than water
Easier to observe as mercury is opaque, water is colourless and transparent [can add colouring though]
More accurate reading as mercury does not stick to glass, whereas water does
More accurate reading as mercury has a lower evaporation rate than water
6a.
Mercury in left arm will go down, in right arm will go up.
Air pressure in balloon higher than atmospheric pressure, pushes the mercury in left arm down.
[think about releasing a balloon filled with air, the balloon will sort of force the air out right? This sort of exert higher pressure. Also, the balloon will shrink after it is attached to the manometer.
*the reason why the balloon grows bigger is to reduce the pressure of air inside to be the same as atmospheric pressure, but given a chance, the balloon will want to go back to its original size]
6b
mercury in left arm will go up, in right arm will go down, until both same level
6c
use liquid of lower density (example water, alcohol, etc)
7a
For Figure 7.4, pressure in mmHg = Patm +h = 760 +100 = 860 mmHg
For Figure 7.5, pressure in mmHg = Patm - h = 760 - 50 = 710 mmHg
7b
compare densities of immiscible liquids
8a
mass of X = density x volume = 2.00 x (2 x 10 ) = 40 g
weight of X = mg = 0.040 kg x 10 = 0.4 N
8b
Pressure = F ÷ A = 0.4 N ÷ 2 cm² = 0.4 N ÷ 0.0002 m² = 2000 Pa
8c
mass of Y = density x volume = (5.00 x A x h) g
weight of Y = (5.00 x A x h ÷ 1000) x 10 = 0.05 Ah N
Pressure of Y = Weight of Y ÷ A
2000 = 0.05 Ah ÷ (A ÷10000) = 500 h
h of Y = 4 cm
mass of Z = density x volume = (10.00 x A x h) g
weight of Z = (10.00 x A x h ÷ 1000) x 10 = 0.10 Ah N
Pressure of Z = Weight of Z ÷ A
2000 = 0.10 Ah ÷ (A ÷10000) = 1000 h
h of Z = 2 cm
1. A [gas supply on right pressing more ==> gas pressure has to be higher]
2. C [pressure of liquid = hpg]
3. D
Pressure on left = Pressure on right [pressure in hydraulic fluid same throughout]
F ÷ A = F ÷ A
10 N ÷ 10 cm² = (1500 x 10) ÷ A
A = 15000 cm²
4. A
Let weight of small cylinder be W and area of small cylinder be A
Therefore weight of big cylinder = 6W and area of big cylinder = 4A
Pressure of big cylinder = 6W ÷ 4A = 18 Pa
therefore 1.5 (W ÷ A) = 18 Pa
(W ÷ A) = 18 ÷ 1.5 = 12 Pa
Pressure of small cylinder = W ÷ A = 12 Pa
Section B
1b.
i. Weight on ground by each foot = 20000 N ÷ 4 = 5000 N
[since 4 feet, so weight is equally distributed among the 4]
Pressure = F ÷ A = 5000 N ÷ 200 cm² = 5000 N ÷ 0.02 m² = 250000 Pa
Alternatively, since pressure on one foot = pressure on all feet = 20000 N ÷ (4 x 200) cm² = 250000 Pa
ii. A = F ÷ Pressure = 600 N ÷ 250000 Pa = 0.0024 m²
2a.
i. Pressure at surface = atmospheric pressure = 101000 Pa
ii Pressure at 5m below = liquid pressure + Patm = hpg + 101000 = 5 x 1025 x 5 + 101000 = 152250 Pa
2b. 152250 - 101000 = 51250 Pa
3.
a. Pressure on small piston = F ÷ A = 120 N ÷ 20 cm² = 120 N ÷ 0.0020 m² = 60000 Pa
b. Pressure on large piston = pressure on small piston = 60000 Pa
c. Weight = Pressure x A = 60000 Pa x (300 cm² ÷ 10000) m² = 1800 N
4
a. Atmospheric pressure is equivalent to the pressure exerted by a mercury column of vertical height 76.0 cm
b. Pressure = hpg = 0.760 m x 13600 x 10 = 103360 Pa = 103000 Pa (3sf)
5a.
Pressure at point O = 0 Pa
Pressure at point P = hpg = (3h ÷ 4 ÷ 100) m x 13600 x 10 = 1020h Pa
Pressure at point Q = hpg = (h ÷ 100) m x 13600 x 10 = 1360h Pa
5b.
Shorter length of barometer as mercury higher density than water
Easier to observe as mercury is opaque, water is colourless and transparent [can add colouring though]
More accurate reading as mercury does not stick to glass, whereas water does
More accurate reading as mercury has a lower evaporation rate than water
6a.
Mercury in left arm will go down, in right arm will go up.
Air pressure in balloon higher than atmospheric pressure, pushes the mercury in left arm down.
[think about releasing a balloon filled with air, the balloon will sort of force the air out right? This sort of exert higher pressure. Also, the balloon will shrink after it is attached to the manometer.
*the reason why the balloon grows bigger is to reduce the pressure of air inside to be the same as atmospheric pressure, but given a chance, the balloon will want to go back to its original size]
6b
mercury in left arm will go up, in right arm will go down, until both same level
6c
use liquid of lower density (example water, alcohol, etc)
7a
For Figure 7.4, pressure in mmHg = Patm +h = 760 +100 = 860 mmHg
For Figure 7.5, pressure in mmHg = Patm - h = 760 - 50 = 710 mmHg
7b
compare densities of immiscible liquids
8a
mass of X = density x volume = 2.00 x (2 x 10 ) = 40 g
weight of X = mg = 0.040 kg x 10 = 0.4 N
8b
Pressure = F ÷ A = 0.4 N ÷ 2 cm² = 0.4 N ÷ 0.0002 m² = 2000 Pa
8c
mass of Y = density x volume = (5.00 x A x h) g
weight of Y = (5.00 x A x h ÷ 1000) x 10 = 0.05 Ah N
Pressure of Y = Weight of Y ÷ A
2000 = 0.05 Ah ÷ (A ÷10000) = 500 h
h of Y = 4 cm
mass of Z = density x volume = (10.00 x A x h) g
weight of Z = (10.00 x A x h ÷ 1000) x 10 = 0.10 Ah N
Pressure of Z = Weight of Z ÷ A
2000 = 0.10 Ah ÷ (A ÷10000) = 1000 h
h of Z = 2 cm
Answers to Worksheet 6
pg 46-47
4.
Total energy at top = total energy at bottom + energy lost to surroundings
GPE + KE = GPE + KE + Energy lost
mgh + 0 = 0 + ½ x mv² + energy lost [not moving at top=> no ke; no gpe at bottom]
2 kg x 10 x 6 m = ½ x 2 kg x 10² + energy lost
Energy lost = 20 J
5.
Since walking up a stairs, thus work is done against gravity.
Work done = F x d = Weight x height
= (80 kg x 10) x 4 m= 3200 J
Power = Work Done t = 3200 J ÷ 16 s = 200 W
6.
Total Work Done at start = F x d = 8 N x 10 m = 80 J [input]
Work done against friction = F x d = 6 N x 10 m = 60 J [output]
Work done to increase KE = 80 - 60 = 20 J [output]
*energy is conserved as total energy input = total energy output
pg 50-51
5.
a. Power = Work Done ÷ t = 1200 J ÷ 10 s = 120 W
b. GPE = mgh = 50 kg x 10 x 2 m = 1000 J
c. Some energy lost as / converted to heat due to friction between student and the stairs.
6.
a. GPE = mgh = 20000 kg x 10 x 10 m = 2000000 J
b. distance = Work done ÷ Force = 345000 J ÷ 50000 N = 6.9 m = 7 m (1sf)
7.
a. Work done = F x d = 50 N x 3 m = 150 J
b. Work done = F x d = Weight x height = (10 kg x 10) x (3 x 9) [no of floors between 1 and 10 is 9]
= 2700 J
8.
a. GPE is converted to electrical energy
Power = energy converted ÷ t
= mgh ÷ t
= (m ÷ t) x g x h = 4000 kg/s x 10 x 600 m = 24000000 W
b. if 90% efficient, electrical energy = 90% x answer in (a) = 21600000W
pg 52
1a.
i. Work done = F x d = 26 N x 4 m = 104 J [4 m is in direction of force]
ii. Gain in KE = ½ x mv² = ½ x 2 kg x 8² = 64 J
iii. Gain in GPE = mgh = 2 kg x 10 x (4 sin 30°) = 40 J
[sin 30° = opp ÷ hypo = height ÷ 4 => height = 4 sin 30°]
1b. i = ii + iii. Energy is conserved. Work done [input] is converted to KE and GPE [output]
4.
Total energy at top = total energy at bottom + energy lost to surroundings
GPE + KE = GPE + KE + Energy lost
mgh + 0 = 0 + ½ x mv² + energy lost [not moving at top=> no ke; no gpe at bottom]
2 kg x 10 x 6 m = ½ x 2 kg x 10² + energy lost
Energy lost = 20 J
5.
Since walking up a stairs, thus work is done against gravity.
Work done = F x d = Weight x height
= (80 kg x 10) x 4 m= 3200 J
Power = Work Done t = 3200 J ÷ 16 s = 200 W
6.
Total Work Done at start = F x d = 8 N x 10 m = 80 J [input]
Work done against friction = F x d = 6 N x 10 m = 60 J [output]
Work done to increase KE = 80 - 60 = 20 J [output]
*energy is conserved as total energy input = total energy output
pg 50-51
5.
a. Power = Work Done ÷ t = 1200 J ÷ 10 s = 120 W
b. GPE = mgh = 50 kg x 10 x 2 m = 1000 J
c. Some energy lost as / converted to heat due to friction between student and the stairs.
6.
a. GPE = mgh = 20000 kg x 10 x 10 m = 2000000 J
b. distance = Work done ÷ Force = 345000 J ÷ 50000 N = 6.9 m = 7 m (1sf)
7.
a. Work done = F x d = 50 N x 3 m = 150 J
b. Work done = F x d = Weight x height = (10 kg x 10) x (3 x 9) [no of floors between 1 and 10 is 9]
= 2700 J
8.
a. GPE is converted to electrical energy
Power = energy converted ÷ t
= mgh ÷ t
= (m ÷ t) x g x h = 4000 kg/s x 10 x 600 m = 24000000 W
b. if 90% efficient, electrical energy = 90% x answer in (a) = 21600000W
pg 52
1a.
i. Work done = F x d = 26 N x 4 m = 104 J [4 m is in direction of force]
ii. Gain in KE = ½ x mv² = ½ x 2 kg x 8² = 64 J
iii. Gain in GPE = mgh = 2 kg x 10 x (4 sin 30°) = 40 J
[sin 30° = opp ÷ hypo = height ÷ 4 => height = 4 sin 30°]
1b. i = ii + iii. Energy is conserved. Work done [input] is converted to KE and GPE [output]
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