Chapter II
\text{Chapter II}
Chapter II
\text{Chapter II}
Chapter II
\text{Chapter II}
Chapter II
\text{Chapter II}
Problem 1
\text{Problem 1}
%The position of a particle $x(t)$ as a function of time $(t)$ is described by the equation: $x(t)=2.0+3.0 t-t^3$, where $x$ is in $\mathrm{m}$ and $\mathrm{t}$ is in $\mathrm{s}$. What is the maximum positive position of the particle on the $x$ axis?}
The position of a particle x(t) as a function of time (t)is described bythe equation: x(t)=2.0+3.0t−t3, where x is in m and t is in s.
What is the maximum positive position of the particle on the x axis?
A) 5.0 m
B) 2.0 m
C) 3.0 m
D) 1.0 m
E) 4.0 m
Problem 2
\text{Problem 2}
Two cars are 150 km apart and traveling toward each other. One car is \text{ moving at $60 . \mathrm{km} / \mathrm{h}$ and the other is moving at 40 . $\mathrm{km} / \mathrm{h}$ }.\text{ In how many hours will they meet?}$$
\begin{align}
d\\
d
\end{align}
(1 nano =10−9)
1 shake=10−8 seconds.
Find out how many nano seconds (ns) are there in 1 shake.
Problem 2
\text{Problem 2}
Answer E
\text{Answer E}
How many molecules of water are there in a cup containing 250 cm3 of water?
Molecular mass of H2O=18 g/mole
Density of water =1.0 g/cm3
Avogadro s number =6.02×1023molecules/mole
A) 6.0×1023
B) 8.4×1024
C) 1.9×1026
D) 3.7×1028
E) 2.5×103
Problem 3
\text{Problem 3}
How many molecules of water are there in a cup containing 250 cm3 of water?
Molecular mass of H2O=18 g/mole
Density of water =1.0 g/cm3
Avogadro s number =6.02×1023molecules/mole
A) 6.0×1023
B) 8.4×1024
C) 1.9×1026
D) 3.7×1028
E) 2.5×103
Problem 3
\text{Problem 3}
Answer B
\text{Answer B }
Problem 4
\text{Problem 4}
Problem 4
\text{Problem 4}
Answer A
\text{Answer A}
Problem 5
\text{Problem 5}
Problem 5
\text{Problem 5}
Answer D
\text{Answer D}
[v]=[A][t]2 and [v]=[A][B][t]
\displaystyle [v]=[A][t]^2\text{ and }[v]=\frac{[B]}{[A]}[t]
LT−1=[A]T2 and [v]=[A][B]T
\displaystyle LT^{-1}=[A]T^2\text{ and }[v]=\frac{[B]}{[A]}T
In a sum or an equality, all the terms have the same dimension
\text{In a sum or an equality, all the terms have the same dimension}
LT−1=[A]T2
\displaystyle LT^{-1}=[A]T^2
LT−1=[A][B]T
\displaystyle LT^{-1}=\frac{[B]}{[A]}T
and
\text{ and }
⇒
\Rightarrow
LT−3=[A]
\displaystyle LT^{-3}=[A]
⇒
\Rightarrow
[A]LT−2=[B]
\displaystyle [A]LT^{-2}=[B]
L2T−5=[B]
\displaystyle L^2T^{-5}=[B]
⇒
\Rightarrow
The transverse displacement of an oscillating string is given by y=ymcos(ωt+ϕ)
\text{The transverse displacement of an oscillating string is given by } y=y_m \cos(\omega t+\phi)
where y, is in cm and t in ms.
\text{where }y,\text{ is in cm and }t \text{ in ms}.
What is the dimension of ω?
\text{What is the dimension of }\omega?
Problem 7
\text{Problem 7}
The transverse displacement of an oscillating string is given by y=ymcos(ωt+ϕ)
\text{The transverse displacement of an oscillating string is given by } y=y_m \cos(\omega t+\phi)
where y, is in cm and t in ms.
\text{where }y,\text{ is in cm and }t \text{ in ms}.
What is the dimension of ω?
\text{What is the dimension of }\omega?
The term inside a trigonometric function has no dimension. Therefore, we can write:
\text{The term inside a trigonometric function has no dimension. Therefore, we can write:}
[ω][t]=1⇒[ω]=T−1
[\omega][t]=1\Rightarrow [\omega]=T^{-1}
Problem 7
\text{Problem 7}
Problem 8
\text{Problem 8}
Problem 8
\text{Problem 8}
Answer C
\text{Answer C}
Problem 9
\text{Problem 9}
Problem 9
\text{Problem 9}
Answer E
\text{Answer E}
Problem 10
\text{Problem 10}
Answer B
\text{Answer B}
Problem 11
\text{Problem 11}
Problem 11
\text{Problem 11}
Answer D
\text{Answer D}
C h a p t e r I I \text{Chapter II} C h a p t e r I I \text{Chapter II} C h a p t e r I I \text{Chapter II} C h a p t e r I I \text{Chapter II}
Chapter 2-Phys101
By smstry
Chapter 2-Phys101
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