# data[].attributes.body
P1: 76 + 30 = 106cm Hg
P2: 76cm Hg
P3: 76 - 15 = 61cm Hg

# data[].attributes.body
a)
A: 76 cm Hg
B: 76 + 4 = 80 cm Hg
C: 76 + 4 + 6 = 86 cm Hg
D: since pressure at B is same as pressure at D,
80 cm Hg

b) by using P =hpg,
A: 1.03x10^5 Pa
B:  1.03x10^5 + (0.04 x 1.36 x 10^4 x 10) = 1.08x10^5 Pa
C: 1.03x10^5 + (0.1 x 1.36 x 10^4 x 10) = 1.17 x 10^5 Pa
D: 1.08x10^5 Pa

c)
1.08x10^5 Pa

# data[].attributes.body
P1= 76cmHg + 30cmHg
=106cmHg
P2=76cmHg
P3=76cmHg - 15cmHg
=61cmHg

# data[].attributes.body
P1:
76+30=106cmHg
P2:
76-20=56cmHg
P3:
76-15=61cmHg

# data[].attributes.body
Pressure at P1: 30+76=106cmHg
Pressure at P2: 76cmHg
Pressure at P3: 76-15=61cmHg

# data[].attributes.body
a) X1 = 106 cmHg
X2 = 76 cmHg
X3 = 61 cmHg

# data[].attributes.body
Pressure at P1: 46cmHg
P2: 56cmHg
P3: 61Hg

# data[].attributes.body
a) x1 = 76cmHg + 30cmHg = 106cmHg
x2 = 76cmHg
x3 - 76 -15= 61cmHg

# data[].attributes.body
a)
A = 76 cmHg
B = 80 cmHg
C = 86 cmHg
D = B = 80 cmHg

b)
A = 1.03 x 10^5
B = 1.08 x 10^5
C = 1.17 x 10^5
D = 1.08 x 10^5

C) 1.08 x 10^5

# data[].attributes.body
a) a=76cmHg
b= 80 cmHg
c= 90cmHg
D= 80cmHg

# data[].attributes.body
a)
A = 0 cmHg
B = 18 cmHg
C = 76 cmHg
D = 101 cmHg
E = 126 cmHg

b)
(13600)(10)(0.76) = 1.03 x 10^5 Pa

c) there is no change as only changes to density height and gravitational force affects liquid pressure, as shown in the equation P = hpg

# data[].attributes.body
Q1a)
A: 0cmHg
B:76-58=18cmHg
C:58+18=76cmHg
D:76+25=101cmHg
E:76+50=126cmHg
b)
1.03x10*5 Pa
c)There will be no change
Q2a)
76-55=21cmHg
bi)
Height of mercury drops
bii)
Pressure is constant
Q3a)
76+5=81cmHg
b)
P=0.81x(1.36x10*4)x10
=110.16x10*3
Q4

# data[].attributes.body
a) A=76cmHg
B=80cmHg
C=86cmHg
D=80cmHg
b)1.03x10^5

# data[].attributes.body
a) A:76cmHg
B: 76+4=80cmHg
C: 76+4+6=86cmHg
D: 80cmHg
b) A: 1.03x10^5Pa
B:

# data[].attributes.body
a)
A:76
B:80
C:86
D:80

# data[].attributes.body
(a)
A: 0cmHg
B: 76-58=18cmHg
C: 76cmHg
D: 76+25=101cmHg
E: 76+50=126cmHg

(b) 1.03x10^5 Pa
(c) The pressure will decrease because the height decreases

# data[].attributes.body
a)A=0cmHg
B=18cmHg
C=76cmHg
D= 101cmHg
E=126cmHg
b)1.36x10^4x0.76x10
=1.03x10^5Pa
c)There will be no change since the diameter doesn't make a difference in the barometer.

# data[].attributes.body
Q1a)
A: 0 cm Hg
B: 76 - 58 = 18 cm Hg
C: 76 cm Hg
D: 76 + 25 = 101 cm Hg
E: 76 + 50 = 126 cm Hg

b)
1.03x10^5 Pa

c)
There will be no change as pressure is dependent on depth, and not width. Hence, changing the diameter does not affect any values  obtained from the barometer.

Q2a)
76 - 55 = 21 cm Hg

b)
i)The height of the mercury column drops.
ii)The pressure remains the same.

Q3a)
76 + 5 = 81 cm Hg

b)

Q4
by comparing total pressure,
density of x = 667 kgm^-3

# data[].attributes.body
a: 76cmHg
b:80cmHg
c:86cmHg
d:80cmHg

# data[].attributes.body
(a)  A: 0cmHg
b: 76-58=18cmHg
c: 58+18=76cmHg
d: 25+58+18= 101cmHg
e: 50+58+18=126cmHg

(b): 1.36x10^4x(0.76x10)=103360=
1.03x10^5Pa
(c) No change

# data[].attributes.body
a)A= 0cmHg
B= 18cmHg
C=76cmHg
D=101cmHg
E= 126cmHg
b)1.36x10^4x0.76x10=1.03x10^5Pa
c) There will be no changes as the diameter does not matter, only the density of the liquid determines the height.

# data[].attributes.body
a) 76 - 55 = 21 cmHg

b)
i) The height will decrease as the decrease in atmospheric pressure is lesser.
ii) It will not change, it is a vacuum and its pressure will remain at 0 Pa.

# data[].attributes.body
(a) 76-55=21cmHg
(b)(i): The height in the mercury column will decrease.
(ii): The pressure will remain the same.

# data[].attributes.body
a)76cmHg - 55cmHg
=21cmHg
b) (i)The height will decrease
(ii)The pressure will remain the same at 0 Pa
c)

# data[].attributes.body
a)76-55=21
b)
i)the height will decrease as the atmospheric pressure changes
ii)the pressure in the space above the mercury will remain the same as it is a vacuum.

# data[].attributes.body
3(a). 81cmHg
(b). 81cmHg
4. 20kg/m3?

# data[].attributes.body
3.(a) 76+5 =81cmHg
(b) 0.81×1.36×10⁴×10=1.10×10⁵ (3sf)
4. 667 kg/m³ (3sf)

# data[].attributes.body
a)76+5 = 81 cmHg

b) (13600)(0.81)(10) = 1.1 x 10^5

# data[].attributes.body
(1000)(10)(0.06) = 600
total pressure at a = total pressure at b
600 + atm = (0.09)(10)(p) + atm
600 = 0.9p
density is 6.67 x 10^2 = 667 Kg/m^3

# data[].attributes.body
a) 76 + 5 = 81 cm Hg
b) 6.8 times 10^5

# data[].attributes.body
a)76+5=81cmHg
b)hpg
=0.05 x 13600 x 10
=6800Pa

# data[].attributes.body
6cm H2O=9cm liquid X
pressure
0.06 x 1000 x 10=0.09 x density x 10
density=666.666666...
=6.67 x10^2 (3s.f.)
=667 kgm^-3

# data[].attributes.body
1. D
2. C x (D)
3. D
4. B
5. B
6. B
7. B
8. C
9. B
10. B

# data[].attributes.body
1) D
2) C
3) D
4) i don't know :(
5) B
6) B
7) B
8) B
9) B
10) B

# data[].attributes.body
1.D (correct)
2.C (wrong)
3.D (correct)
4.B (correct)
5.B (correct)
6.B (correct)
7.B (correct)
8.C (correct)
9.A (wrong)
10.B (correct)

# data[].attributes.body
1. D
2. dD
3. D
4.B
5. B
6. B
7. B
8. cC
9. B
10. B

# data[].attributes.body
1.D
2.C X(D)
3.A X(D)
4.D X(B)
5.B
6.B
7.B
8.B X(C)
9.B
10.B

# data[].attributes.body
1. D
2. C (D)
3. D
4. B
5. B
6. A (B)
7. B
8. C
9. B
10. B

# data[].attributes.body
Q1 D
Q2 C
Q3 B
Q4 B
Q5 C
Q6 B
Q7 B
Q8 B
Q9 D
Q10 B

# data[].attributes.body
Q1

# data[].attributes.body
(i)Since it is constant acceleration, the friction should equal to applied force, which is 20N
(ii)F=ma
F=1x4
=4N
Therefore,
20N-4N=16N

# data[].attributes.body
(i) 20N
(ii) f-friction=ma
20-f=1x4
20-f=4
f=16N

# data[].attributes.body
i) Constant velocity = no acceleration
Forward Force = Friction
Friction is 20N

ii) 20 N - Friction = 1 x 4
20N - 4N = Friction
Friction = 16N

# data[].attributes.body
(i) 20N
(ii) 16N

# data[].attributes.body
i)20N
ii)16N

# data[].attributes.body
I) 20N
II) 16N

# data[].attributes.body
i) 20N
ii) 16N

# data[].attributes.body
1800N

# data[].attributes.body
(60)(10) - resistance = (-20)(60)
600N - resistance = -1200N
resistance = 1800N
water resistance = 1800N

# data[].attributes.body
f-friction=ma
600-water resistance=60x(-20)
600-water resistance=-1200
water resistance=1800N

# data[].attributes.body
force when entering water=60x(-20)=-1200N
60x10=600N
60N+1200N=1800N