All Solutions
Page 5: Practice Problems
– $R=50 Omega$ – the resistance of the lamp
– $V=9 text{V}$ – the potential difference across the conductor
**Required.**
In this problem, we are asked to determine the current passing through the lamp.
The relationship between the resistance of a conductor, the current passing through it, and the potential difference applied across the conductor is given by Ohm’s law. In math terms, Ohm’s law can be written as.
$$V = I cdot Rtag{1}.$$
From equation (1), for the current passing through the conductor we get.
$$I=dfrac{V}{R} tag2$$
Equation (2) states that the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit.
Explanation of terms for equations (1) and (2).
– $V$ – the potential difference across the conductor $[V]$,
– $I$ – the current passing through a conductor $[A]$,
– $R$ – the resistance of the conductor $[Omega]$.
#
**Calculation method.**
$$begin{align*}
I &= dfrac{V}{R}\\
&= dfrac{9 text{V}}{50 Omega}\\
&= boxed{0.18 text{A}}\
end{align*}$$
**Conclusion.**
Thus the current passing through the lamp is $0.18$ ampere.
From kinematics of motion the general equation of motion that relates the initial velocity $v_i$, final velocity $v_f$, acceleration $a$ and duration of motion $t$ is given by the following equation
$v_f = v_i + at tag{1}$
And since it is stated that the motion starts initially from rest, therefor the initial velocity is zero $v_i = 0$ which reduced the equation (1) to the following
$$
v_f = at tag{2}
$$
Which is the same equation stated in the problem, and since it is given that the object accelerates to a final speed of 7 m/s in 4 seconds, hence knowing both final velocity and the duration of motion we can find the acceleration of the body from equation (2).
Substituting the knowns in equation (2) to find the constant of acceleration $a$, we get
$$
begin{align*}
a &= v_f / t\
&= 7/4\
&= fbox{$1.75 ~ rm{m}/rm{s}^2$}
end{align*}
$$
Thus the constant of acceleration for the accelerating object is 1.75 m/s$^2$.
1.75 ~ rm{m}/rm{s}^2
$$
a = dfrac{7m/s}{4s}
$$
From kinematics of motion the general equation of motion that relates the initial velocity $v_i$, final velocity $v_f$, acceleration $a$ and duration of motion $t$ is given by the following equation
$v_f = v_i + at tag{1}$
And since it is stated that the motion starts initially from rest, therefor the initial velocity is zero $v_i = 0$ which reduced the equation (1) to the following
$$
v_f = at tag{2}
$$
Since it is given that the body is accelerated from rest with a constant of acceleration of 0.4 m/s$^2$ to a final speed of 4 m/s, thus using equation (2) we can find the duration of time by which the body should be accelerated to reach the stated final speed.
Substituting the knowns in equation (2) to find the duration of acceleration $t$, we get
$$
begin{align*}
t &= v_f / a\
&= 4/0.4\
&= fbox{$10 ~ rm{s}$}
end{align*}
$$
Therefor the body is being accelerated from rest for 10 seconds with an acceleration constant of 0.4 m/s$^2$ to reach a final speed of 4 m/s.
10 ~ rm{s}
$$
Pressure by definition is the amount of perpendicular applied force per unit area, which is given by the following equation
$$
P = dfrac{F}{A}
$$
Where,
$P$ The pressure of unit Pascal (N/m$^2$).
$F$ The applied force of unit N.
$A$ The area upon which the force is applied of unit m$^2$.
Which means that if we have an object of weight W with a base Area $A$, thus upon putting such object on a table for example, then that object will exerts a pressure which can be calculated using equation (1).
A woman wearing heels would exerts a much higher pressure on the ground compared to when not wearing a heels ” for example bare foot”, where the heels have much smaller base and thus from equation (1) as the woman have the same weight it would thus have a much higher pressure exerted on the ground.
Given that the weight (force) of the woman is 530 N and the pressure exerted on the floor is 32,500 N/m$^2$, thus from equation (1) we can find the area that would produce such pressure due to the given weight, which is the area of her shoes soles.
Calculating the area of the woman is shoes solesusing the givens in the problem in equation (1), we get
$$
begin{align*}
A &= F/P \
&= 530/ 32500 \
&= fbox{$0.0163 ~ rm{m}^2$}\
end{align*}
$$
$textbf{underline{textit{note:}}}$ The calculated area is the total area of the two shoes soles, if we wanted to find the area of only one shoe then we would divide the calculated area by two.
A = dfrac{520N}{32,500 N/m^2}
$$
750 kHz times dfrac{1MHz}{1,000kHz}
$$
dfrac{kHz cdot MHz}{kHz}
$$
dfrac{750 kHz cdot MHz}{1000kHz}
$$
Knowing that a 1 km is equal 1000 m and 1 m is 100 cm thus
$$
1 ~ rm{km} = 100,000 ~ rm{cm}
$$
Thus, the conversion factor to convert a 1 cm to 1 km is
$$
1 ~ rm{cm} = 10^{-5} ~ rm{km}tag{1}
$$
Which simply means that to convert some length of a unit cm we simply multiply the length by $10^{-5}$ to convert from the unit of cm to km.
Using the conversion factor (1), we convert the given length as follows
$$
begin{align*}
5021 ~ rm{cm} times 10^{-5} ~ dfrac{rm{km}}{cm} &= 5021 times 10^{-5} ~ rm{km}\
&= fbox{$0.05021 ~ rm{km}$}
end{align*}
$$
1 year times dfrac{366 days}{1year} times dfrac{24 hours}{1 day} times dfrac{3600 seconds}{1 hour}
$$
dfrac{yearcdot days cdot hours cdot seconds}{yearscdot days cdot hours}
$$
dfrac{1yearcdot 366 days cdot 24 hours cdot 3600 seconds}{1 yearcdot 1 day cdot 1hour}
$$
5.3 m/s times dfrac{1km}{1,000m} times dfrac{3,600seconds}{1hour}
$$
dfrac{scdot m cdot km}{hrcdot mcdot s}
$$
dfrac{3600scdot 5.3m cdot km}{hrcdot 1000mcdot s}
$$
b) 1.6km
$$
10.8g – 8.264g = 2.536g
$$
$$
2.536g rightarrow 2.5g
$$
However, we must apply the rules of significant digits to this problem. 10.8g is the least-precise value given to you; it only measures up to the tenths place (one space behind the decimal point).
Therefore, our answer rounds from 2.536g to 2.5g to match this.
4.75m – 0.4168 m = 4.3332m
\
4.3332m rightarrow 4.33 m
$$
a) 320 cm^2
$$
b)13.6km^2
$$
a) 1.22 g/mL
$$
b)4.1g/cm^3
$$
