Chapter 7
\text{Chapter 7}
Chapter 7
\text{Chapter 7}
Chapter 7
\text{Chapter 7}
Chapter 7
\text{Chapter 7}
Dr. Adel Abbout
\textit{Dr. Adel Abbout}
A) 0.44 mB) 0.39 mC) 0.23 mD) 0.13 mE) 0.56 m
\text{A) $0.44 \mathrm{~m}$}\\
\text{B) $0.39 \mathrm{~m}$}\\
\text{C) $0.23 \mathrm{~m}$}\\
\text{D) $0.13 \mathrm{~m}$}\\
\text{E) $0.56 \mathrm{~m}$}
A 3.0 kg block is released from a compressed spring (k=120 N/m). It travels over a horizontal
\text{A $3.0 \mathrm{~kg}$ block is released from a compressed spring $(\mathrm{k}=120 \mathrm{~N} / \mathrm{m})$. It travels over a horizontal}
surface (μk=0.20) for a distance of 2.0 m before coming to rest, Fig 1.
\text{ surface $(\mathrm{\mu_k}=0.20)$ for a distance of $2.0 \mathrm{~m}$ before coming to rest, Fig 1. }
How far was the spring compressed before being released?
\text{How far was the spring compressed before being released?}
A 3.0 kg block is released from a compressed spring (k=120 N/m). It travels over a horizontal
\text{A $3.0 \mathrm{~kg}$ block is released from a compressed spring $(\mathrm{k}=120 \mathrm{~N} / \mathrm{m})$. It travels over a horizontal}
surface (μk=0.20) for a distance of 2.0 m before coming to rest, Fig 1.
\text{ surface $(\mathrm{\mu_k}=0.20)$ for a distance of $2.0 \mathrm{~m}$ before coming to rest, Fig 1. }
How far was the spring compressed before being released?
\text{How far was the spring compressed before being released?}
ΔK=W(Fnet)=W(mg)+W(FN)+W(f)+W(Fs)
\Delta K=W(\vec{F}_\text{net})=W(m\vec{g})+W(\vec{F}_N)+W(\vec{f})+W(\vec{F}_s)
Kf=Ki−μkmgd+21kxi2
K_\text{f}=K_\text{i}-\mu_k m g d+\frac{1}{2}k x_\text{i}^2
W(mg)=W(FN)=0
W(m\vec{g})=W(\vec{F}_N)=0
W(f)=f.d=−fd=−μkmgd
W(\vec{f})=\vec{f}.\vec{d}=-f d=-\mu_k mg d
W(Fs)=21kxi2−21kxf2=21kxi2
\displaystyle W(\vec{F}_s)=\frac{1}{2}k x_\text{i}^2-\frac{1}{2}k x_\text{f}^2=\frac{1}{2}k x_\text{i}^2
(xf=0)
(x_\text{f}=0)
The block was released without speed (Ki=0) and stops at the end (Kf=0)
\text{The block was released without speed $(K_\text{i}=0)$ and stops at the end $(K_\text{f}=0)$}
xi=k2μkmgd
\displaystyle x_\text{i}=\sqrt{\frac{2\mu_k m g d}{k}}
xi=1202×0.2×3.0×9.8×2.0=0.44 m
\displaystyle x_\text{i}=\sqrt{\frac{2 \times0.2\times 3.0 \times9.8\times 2.0}{120}}=0.44 \text{ m}
Answer: A) 0.44 m
\text{Answer: A) 0.44 \text{m}}
Fs
\vec{F}_s
f
\vec{f}
mg
m\vec{g}
(FN=mg)
(F_N=mg)
A 3.00 kg block is dropped from a height of 40 cm onto a spring of spring constant k (see Fig 2).
\text{A $3.00 \mathrm{~kg}$ block is dropped from a height of $40 \mathrm{~cm}$ onto a spring of spring constant $\mathrm{k}$ (see Fig 2). }
If the maximum distance the spring is compressed =0.130 m, find k.
\text{If the maximum distance the spring is compressed $=0.130 \mathrm{~m}$, find $\mathrm{k}$.}
A) 490 N/mB) 980 N/mC) 1840 N/mD) 1250 N/mE) 2800 N/m
\text{A) $490 \mathrm{~N} / \mathrm{m}$}\\
\text{B) $980 \mathrm{~N} / \mathrm{m}$}\\
\text{C) $1840 \mathrm{~N} / \mathrm{m}$}\\
\text{D) $1250 \mathrm{~N} / \mathrm{m}$}\\
\text{E) $2800 \mathrm{~N} / \mathrm{m}$}
A 3.00 kg block is dropped from a height of 40 cm onto a spring of spring constant k (see Fig 2).
\text{A $3.00 \mathrm{~kg}$ block is dropped from a height of $40 \mathrm{~cm}$ onto a spring of spring constant $\mathrm{k}$ (see Fig 2). }
If the maximum distance the spring is compressed =0.130 m, find k.
\text{If the maximum distance the spring is compressed $=0.130 \mathrm{~m}$, find $\mathrm{k}$.}
d
d
x
x
ΔK=W(mg)+W(Fs)
\Delta K=W(m\vec{g})+W(\vec{F_s})
Be careful, Fg works even when the spring is being compressed
\text{Be careful, $\vec{F}_g$ works even when the spring is being compressed}
ΔK=mg(d+x)+(21kxi2−21kx2)
\displaystyle \Delta K=m g (d+x)+(\frac{1}{2}kx^2_\text{i}-\frac{1}{2}kx^2)
=
=
0
0
Block was at rest Ki=0.
\text{Block was at rest \textcolor{red}{$K_\text{i}=0$.} }
At maximum compressin, the block stops momentarilty. Kf=0
\text{At maximum compressin, the block stops momentarilty. \textcolor{red}{$K_\text{f}=0$}}
0=mg(d+x)−21kx2
\displaystyle 0=m g (d+x)-\frac{1}{2}kx^2
k=x22mg(d+x)=0.13022×3.0×9.8×(0.40+.130)=1844 N/m
\displaystyle k=\frac{2m g (d+x)}{x^2}=\frac{2\times3.0 \times 9.8\times (0.40+.130)}{0.130^2}=1844\text{ N/m}
Answer C)
\text{Answer C)}
(A more difficult version of this problem is in the next slide)
\text{(A more difficult version of this problem is in the next slide)}
Warning
\text{Warning}
The relation
\text{The relation}
W(Fs)=21kxi2−21kxf2
\displaystyle \text{$W(\vec{F}_s)=\frac{1}{2}k x_\text{i}^2-\frac{1}{2}k x_\text{f}^2$}
the origin of your axis is the position where the spring is neither stretched nor compressed
\text{ the origin of your axis is the position where the spring is neither stretched nor compressed}
is correct only if:
\text{is correct only if:}
x
x
xf
x_\text{f}
Correct
\text{Correct}
Wrong
\text{Wrong}
Wrong
\text{Wrong}
x
x
xf
x_\text{f}
l0
l_0
0
0
0
0
l0
l_0
x
x
xf
x_\text{f}
0
0
l0
l_0
A 16 kg crate falls from rest from a height of 1.0 m onto a spring scale with a spring constant
\text{A $16 \mathrm{~kg}$ crate falls from rest from a height of $1.0 \mathrm{~m}$ onto a spring scale with a spring constant }
of 2.74×103 N/m. Find the maximum distance the spring is compressed.
\text{of $2.74 \times 10^3 \mathrm{~N} / \mathrm{m}$. Find the maximum distance the spring is compressed.}
A) 40 cmB) 2.0 cmC) 60 cmD) 7.0 cmE) 5.0 cm
\text{A) $40 \mathrm{~cm}$}\\
\text{B) $2.0 \mathrm{~cm}$}\\
\text{C) $60 \mathrm{~cm}$}\\
\text{D) $7.0 \mathrm{~cm}$}\\
\text{E) $5.0 \mathrm{~cm}$}
1.0 m
1.0 \text{ m}
d
d
x
x
0=mg(d+x)−21kx2
\displaystyle 0=m g (d+x)-\frac{1}{2}kx^2
A 16 kg crate falls from rest from a height of 1.0 m onto a spring scale with a spring constant
\text{A $16 \mathrm{~kg}$ crate falls from rest from a height of $1.0 \mathrm{~m}$ onto a spring scale with a spring constant }
of 2.74×103 N/m. Find the maximum distance the spring is compressed.
\text{of $2.74 \times 10^3 \mathrm{~N} / \mathrm{m}$. Find the maximum distance the spring is compressed.}
kx2−2mgx−2mgd=0
kx^2-2mgx -2m g d=0
ΔK=W(mg)+W(Fs)
\Delta K=W(m\vec{g})+W(\vec{F_s})
Be careful, Fg works even when the spring is being compressed
\text{Be careful, $\vec{F}_g$ works even when the spring is being compressed}
ΔK=mg(d+x)+(21kxi2−21kx2)
\displaystyle \Delta K=m g (d+x)+(\frac{1}{2}kx^2_\text{i}-\frac{1}{2}kx^2)
Block was at rest Ki=0.
\text{Block was at rest \textcolor{red}{$K_\text{i}=0$.} }
At maximum compressin, the block stops momentarilty. Kf=0
\text{At maximum compressin, the block stops momentarilty. \textcolor{red}{$K_\text{f}=0$}}
During the exam, you can solve it numerically using your calculator.
\text{During the exam, you can solve it numerically using your calculator.}
2740x2−313.6x−313.6=0
2740 x^2-313.6x-313.6=0
x=0.40 m or x=−0.29 m
x=0.40 \text{ m}\text{ or } x=-0.29 \text{ m}
⇒
\Rightarrow
Answer A)
\text{Answer A)}
1.0 m
1.0 \text{ m}
the solusion is x=40 cm
\text{the solusion is }x=40 \text{ cm}
d
d
x
x
A 16 kg crate falls from rest from a height of 1.0 m onto a spring scale with a spring constant
\text{A $16 \mathrm{~kg}$ crate falls from rest from a height of $1.0 \mathrm{~m}$ onto a spring scale with a spring constant }
of 2.74×103 N/m. Find the maximum distance the spring is compressed.
\text{of $2.74 \times 10^3 \mathrm{~N} / \mathrm{m}$. Find the maximum distance the spring is compressed.}
kx2−2mgx−2mgd=0
kx^2-2mgx -2m g d=0
The expression of the solution can be found this way:
\text{The expression of the solution can be found this way:}
x=2k2mg±(2mg)2+8kmgd
\displaystyle x=\frac{2mg\pm\sqrt{(2mg)^2+8k mgd}}{2k}
x=kmg±(mg)2+2kmgd
\displaystyle x=\frac{mg\pm\sqrt{(mg)^2+2k mgd}}{k}
Δ=(2mg)2+8kmgd
\Delta=(2mg)^2+8k mgd
1.0 m
1.0 \text{ m}
At time t=0 a single force F acts on a 2.0 kg particle and changes its velocity from
\text{At time $t=0$ a single force $\mathbf{F}$ acts on a $2.0 \mathrm{~kg}$ particle and changes its velocity from }
vi=(4.0i−3.0j)m/s at t=0 to vf=(4.0i+3.0j)m/s at t= 3.0 s.
\text{$\mathbf{v}_{\mathbf{i}}=(4.0 \mathbf{i}-3.0 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at $\mathrm{t}=0$ to $\mathbf{v}_{\mathrm{f}}=(4.0 \mathbf{i}+3.0 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at $\mathrm{t}=$ $3.0 \mathrm{~s}$.}
During this time the work done by F on the particle is:
\text{During this time the work done by $\mathbf{F}$ on the particle is:}
A) 2.0 JB) 6.0 JC) 0D) 50 JE) 10 J
\text{A) $2.0 \mathrm{~J}$}\\
\text{B) $6.0 \mathrm{~J}$}\\
\hspace{-5mm}\text{C) 0}\\
\text{D) $50 \mathrm{~J}$}\\
\text{E) $10 \mathrm{~J}$}
At time t=0 a single force F acts on a 2.0 kg particle and changes its velocity from
\text{At time $t=0$ a single force $\mathbf{F}$ acts on a $2.0 \mathrm{~kg}$ particle and changes its velocity from }
vi=(4.0i−3.0j)m/s at t=0 to vf=(4.0i+3.0j)m/s at t= 3.0 s.
\text{$\mathbf{v}_{\mathbf{i}}=(4.0 \mathbf{i}-3.0 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at $\mathrm{t}=0$ to $\mathbf{v}_{\mathrm{f}}=(4.0 \mathbf{i}+3.0 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at $\mathrm{t}=$ $3.0 \mathrm{~s}$.}
During this time the work done by F on the particle is:
\text{During this time the work done by $\mathbf{F}$ on the particle is:}
A) 2.0 JB) 6.0 JC) 0D) 50 JE) 10 J
\text{A) $2.0 \mathrm{~J}$}\\
\text{B) $6.0 \mathrm{~J}$}\\
\hspace{-5mm}\text{C) 0}\\
\text{D) $50 \mathrm{~J}$}\\
\text{E) $10 \mathrm{~J}$}
ΔK=W(F)
\Delta K=W(\bf{F})
ΔK=Kf−Ki=21mvf2−21mvi2
\displaystyle \Delta K=K_\text{f}-K_\text{i}=\frac{1}{2}m v_\text{f}^2-\frac{1}{2}m v_\text{i}^2
ΔK=Kf−Ki=21m(42+32)−21m(42+(−3)2)=0
\displaystyle \Delta K=K_\text{f}-K_\text{i}=\frac{1}{2}m (4^2+3^2)-\frac{1}{2}m (4^2+(-3)^2)=0
ΔK=W(F)=0
\Delta K=W(\bf{F})=0
Answer A)
\text{Answer A)}
A projectile is fired from the top of a 40 m high building with a speed of 20 m/s.
\text{A projectile is fired from the top of a $40 \mathrm{~m}$ high building with a speed of 20 $\mathrm{m} / \mathrm{s}$.}
What will be its speed when it strikes the ground?
\text{What will be its speed when it strikes the ground?}
A) 82 m/sB) 10 m/sC) 34 m/sD) 16 m/sE) 50 m/s
\text{A) $82 \mathrm{~m} / \mathrm{s}$}\\
\text{B) $10 \mathrm{~m} / \mathrm{s}$}\\
\text{C) $34 \mathrm{~m} / \mathrm{s}$}\\
\text{D) $16 \mathrm{~m} / \mathrm{s}$}\\
\text{E) $50 \mathrm{~m} /\mathrm{s}$}
Let us consider a general case where the angle θ is not known.
\text{Let us consider a general case where the angle $\theta$ is not known.}
A projectile is fired from the top of a 40 m high building with a speed of 20 m/s.
\text{A projectile is fired from the top of a $40 \mathrm{~m}$ high building with a speed of 20 $\mathrm{m} / \mathrm{s}$.}
What will be its speed when it strikes the ground?
\text{What will be its speed when it strikes the ground?}
Let us consider a general case where the angle θ is not known.
\text{Let us consider a general case where the angle $\theta$ is not known.}
The work of a conservative force, does not depend on the path we take.
\text{The work of a conservative force, does not depend on the path we take.}
mg
\text{$m\vec{g}$}
The work along the orange path is easiest to calculate
\text{The work along the \textcolor{orange}{orange} path is easiest to calculate}
W(mg)=mgy+mgdcos900=mgy
W(m\vec{g})=mgy+mgd\cos 90^0=mgy
ΔK=W⇒Kf=Ki+mgy
\Delta K=W \Rightarrow K_\text{f}=K_\text{i}+mgy
vf2=vi2+2gy⇒
v_\text{f}^2=v_\text{i}^2+2gy\Rightarrow
vf=vi2+2gy=202+2×9.8×40
v_\text{f}=\sqrt{v_\text{i}^2+2gy}=\sqrt{20^2+2\times 9.8\times 40}
vf≈34 m/s
v_\text{f}\approx 34 \text{ m/s}
Answer C)
\text{Answer C)}
Fig. 1 gives the only force Fx that can act on a particle. If the particle has a kinetic energy
\text{Fig. 1 gives the only force $F_x$ that can act on a particle. If the particle has a kinetic energy}
of 10 J at x=0, find the kinetic energy of the particle when it is at x=8.0 m.
\text{ of $10 \mathrm{~J}$ at $\mathrm{x}=0$, find the kinetic energy of the particle when it is at $x=8.0 \mathrm{~m}$.}
A) 0 JB) 20 JC) 30 JD) 60 JE) 10 J
\hspace{-2mm}\text{A) $0 \mathrm{~J}$}\\
\text{B) $20 \mathrm{~J}$}\\
\text{C) $30 \mathrm{~J}$}\\
\text{D) $60 \mathrm{~J}$}\\
\text{E) $10 \mathrm{~J}$}
Fig. 1 gives the only force Fx that can act on a particle. If the particle has a kinetic energy
\text{Fig. 1 gives the only force $F_x$ that can act on a particle. If the particle has a kinetic energy}
of 10 J at x=0, find the kinetic energy of the particle when it is at x=8.0 m.
\text{ of $10 \mathrm{~J}$ at $\mathrm{x}=0$, find the kinetic energy of the particle when it is at $x=8.0 \mathrm{~m}$.}
W(Fx)=Area under the curve
W(F_x)=\text{Area under the curve}
W(Fx)=21(20N×4m)−21(10N×4m)=20J
\displaystyle W(F_x)=\frac{1}{2}(20 N\times 4 m)-\frac{1}{2}(10 N\times 4 m)=20J
Answer C)
\text{Answer C)}
ΔK=W(Fs)⇒Kf=Ki+W(Fs)=10+20=30J
\Delta K=W(F_s)\Rightarrow K_\text{f}=K_\text{i}+W(F_s)=10+20=30J
A 200 kg box is pulled along a horizontal surface by an engine. The coefficient of friction
\text{A $200 \mathrm{~kg}$ box is pulled along a horizontal surface by an engine. The coefficient of friction}
between the box and the surface is 0.400.
\text{ between the box and the surface is 0.400. }
Answer C)
\text{Answer C)}
The power the engine delivers to move the box at constant speed of 5.00 m/s is:
\text{The power the engine delivers to move the box at constant speed of $5.00 \mathrm{~m} / \mathrm{s}$ is:}
A) 3920 WB) 1960 WC) 980 WD) 490 WE) 0 W
\text{A) $3920 \mathrm{~W}$}\\
\text{B) $1960 \mathrm{~W}$}\\
\hspace{-2mm}\text{C) $980 \mathrm{~W}$}\\
\hspace{-2.2mm}\text{D) $490 \mathrm{~W}$}\\
\hspace{-5mm}\text{E) $0 \mathrm{~W}$}
T
\vec{T}
f
\vec{f}
m
m
A 200 kg box is pulled along a horizontal surface by an engine. The coefficient of friction
\text{A $200 \mathrm{~kg}$ box is pulled along a horizontal surface by an engine. The coefficient of friction}
between the box and the surface is 0.400.
\text{ between the box and the surface is 0.400. }
Answer A)
\text{Answer A)}
The power the engine delivers to move the box at constant speed of 5.00 m/s is:
\text{The power the engine delivers to move the box at constant speed of $5.00 \mathrm{~m} / \mathrm{s}$ is:}
T
\vec{T}
f
\vec{f}
m
m
constant speed means that a=0
\text{constant speed means that $\vec{a}=0$}
we deduce that T=f
\text{we deduce that }T=f
P(T)=T.v=Tv=μkmgv
P(T)=\vec{T}.\vec{v}=T v=\mu_k mg v
=0.400×200×9.8×5.00
=0.400 \times200 \times 9.8\times 5.00
=0.400×200×9.8×5.00
=0.400 \times200 \times 9.8\times 5.00
=3920 W
=3920\text{ W}
mg
{m\vec{g}}
N
\vec{N}
A helicopter lifts an 80 kg man vertically from the ground by means of a cable. The upward
\text{A helicopter lifts an $80 \mathrm{~kg}$ man vertically from the ground by means of a cable. The upward }
acceleration of the man is 2.0 m/s2. Find the rate at which the work is being done on the man
\text{ acceleration of the man is $2.0 \mathrm{~m} / \mathrm{s}^2$. Find the rate at which the work is being done on the man}
by the tension of the cable when the speed of the man is 1.5 m/s.
\text{by the tension of the cable when the speed of the man is $1.5 \mathrm{~m} / \mathrm{s}$.}
A) 1.8×103 WB) 1.1×103 WC) 1.2×104 WD) 1.4×103 WE) 2.5×104 W
\text{A) $1.8 \times 10^3 \mathrm{~W}$}\\
\text{B) $1.1 \times 10^3 \mathrm{~W}$}\\
\text{C) $1.2 \times 10^4 \mathrm{~W}$}\\
\text{D) $1.4 \times 10^3 \mathrm{~W}$}\\
\text{E) $2.5 \times 10^4 \mathrm{~W}$}
A helicopter lifts an 80 kg man vertically from the ground by means of a cable. The upward
\text{A helicopter lifts an $80 \mathrm{~kg}$ man vertically from the ground by means of a cable. The upward }
acceleration of the man is 2.0 m/s2. Find the rate at which the work is being done on the man
\text{ acceleration of the man is $2.0 \mathrm{~m} / \mathrm{s}^2$. Find the rate at which the work is being done on the man}
Answer D)
\text{Answer D)}
by the tension of the cable when the speed of the man is 1.5 m/s.
\text{by the tension of the cable when the speed of the man is $1.5 \mathrm{~m} / \mathrm{s}$.}
T
\vec{T}
mg
m\vec{g}
a
\vec{a}
mg+T=ma
m\vec{g}+\vec{T}=m\vec{a}
projection on the y-axis:
\text{projection on the y-axis:}
−mg+T=ma⇒
-m{g}+{T}=m{a} \Rightarrow
T=ma+mg
{T}=m{a}+m{g}
T=80×(2+9.8)=944 N
{T}=80 \times (2+9.8)=944 \text{ N}
P=dtdW(T)=T.v=Tv=944×1.5=1416 W
\displaystyle P=\frac{dW(\vec{T})}{dt}=\vec{T}.\vec{v}=T v=944\times 1.5=1416 \text{ W}
A force F=(3.00i+7.00j)N acts on a 2.00 kg object that moves from an initial position
\text{A force $\mathbf{F}=(3.00 \mathbf{i}+7.00 \mathbf{j}) \mathrm{N}$ acts on a $2.00 \mathrm{~kg}$ object that moves from an initial position}
r1=(3.00i−2.00j)m to a final position r2=(5.00i+4.00j)m in 4.00 s.
\text{$\mathbf{r}_1=(3.00 \mathbf{i}-2.00 \mathbf{j}) \mathrm{m}$ to a final position $\mathbf{r}_2=(5.00 \mathbf{i}+4.00 \mathbf{j}) \mathrm{m}$ in $4.00 \mathrm{~s}$.}
What is the average power due to the force during that time interval?
\text{What is the average power due to the force during that time interval?}
A) 8.0 WB) 7.00 WC) 12.00 WD) 6.00 WE) 16.0 W
\hspace{-2mm}\text{A) $8.0 \mathrm{~W}$}\\
\text{B) $7.00 \mathrm{~W}$}\\
\hspace{2 mm}\text{C) $12.00 \mathrm{~W}$}\\
\text{D) $6.00 \mathrm{~W}$}\\
\text{E) $16.0 \mathrm{~W}$}
A force F=(3.00i+7.00j)N acts on a 2.00 kg object that moves from an initial position
\text{A force $\mathbf{F}=(3.00 \mathbf{i}+7.00 \mathbf{j}) \mathrm{N}$ acts on a $2.00 \mathrm{~kg}$ object that moves from an initial position}
r1=(3.00i−2.00j)m to a final position r2=(5.00i+4.00j)m in 4.00 s.
\text{$\mathbf{r}_1=(3.00 \mathbf{i}-2.00 \mathbf{j}) \mathrm{m}$ to a final position $\mathbf{r}_2=(5.00 \mathbf{i}+4.00 \mathbf{j}) \mathrm{m}$ in $4.00 \mathrm{~s}$.}
What is the average power due to the force during that time interval?
\text{What is the average power due to the force during that time interval?}
Vavg=Δtr2−r1=4.002.00i+6.00j=0.50i+1.50j
\displaystyle \vec{V}_\text{avg}=\frac{\vec{r}_2-\vec{r}_1}{\Delta t}=\frac{2.00 {\bf{ i}}+6.00 \bf{ j}}{4.00}=0.50 {\bf i}+1.50{\bf j}
Pavg=F.Vavg=3.00×0.50+7.00×1.50=12.00 W
\text{$P_\text{avg}=\vec{F}.\vec{V}_\text{avg}$}=3.00\times0.50+7.00\times1.50=12.00 \text{ W}
Answer C)
\text{Answer C)}
C h a p t e r 7 \text{Chapter 7} C h a p t e r 7 \text{Chapter 7} C h a p t e r 7 \text{Chapter 7} C h a p t e r 7 \text{Chapter 7} D r . A d e l A b b o u t \textit{Dr. Adel Abbout}