Answer:
b
Explanation:
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Answer:
t = 1.27 x 10⁹ s
Explanation:
First, we will find the volume of the wire:
Volume = V = AL
where,
A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²
L = Length of wire = 150 km = 150000 m
Therefore,
V = 47.12 m³
Now, we will find the number of electrons in the wire:
No. of electrons = n = (Electrons per unit Volume)(V)
n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)
n = 3.97 x 10³⁰ electrons
Now, we will use the formula of current to find out the time taken by each electron to cross the wire:
[tex]I =\frac{q}{t}[/tex]
where,
t = time = ?
I = current = 500 A
q = total charge = (n)(chareg on one electron)
q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)
q = 6.36 x 10¹¹ C
[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]
Therefore,
t = 1.27 x 10⁹ s
Which circuit has the larger equivalent resistance: a circuit with two 10 ohm resistors connected in parallel or a circuit with two 10 ohm resistors connected in series?
Answer:
A circuit with two 10 ohm resistors connected in series.
Explanation:
The formula for the equivalent resistance for resistors in parallel is
[tex]\frac{1}{Rt} = \frac{1}{R1} + \frac{1}{R2}[/tex] So if R1=R2= 10 [tex]\frac{1}{Rt} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} <=> Rt =\frac{10}{2} =5 ohm[/tex]
The formula for the equivalent resistance for resistors in series is
Rt = R1 + R2 So Rt= 10 + 10 = 20
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25
Answer:
100Hz
Explanation:
In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz
Answer:
HZ 100 is the right answer hope you like it
derive expression for pressure exerted by gas
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity
The final speed of the ball when it reaches the floor is 7.10 m/s.
What is the conservation of energy?The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.
This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.
Here in the Question,
We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.
Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.
ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J
Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:
KEi + ΔPEi = KEf + ΔPEf
where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.
Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:
KEi = KEf + ΔKE
where ΔKE is the change in kinetic energy due to friction.
We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:
Wfric = ΔKE
where Wfric is the work done by friction.
The work done by friction can be expressed as:
Wfric = ffricd
where ffric is the force of friction and d is the distance traveled by the ball on the incline.
The force of friction can be expressed as:
ffric = μmg
where μ is the coefficient of kinetic friction, and m and g have their usual meanings.
Putting it all together, we get:
KEi = KEf + ffricd
KEi = KEf + μmgd
(1/2)mv^2 = (1/2)mu^2 + μmgd
v^2 = u^2 + 2gd
where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.
Plugging in the given values, we get:
v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)
v^2 = 50.405
v = 7.10 m/s
Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.
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In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first
Answer:
solid cylinder
Explanation:
the object that arrives first is the object that has more speed, let's use the concepts of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point
Em_f = K = ½ mv² + ½ I w²
since the body has rotational and translational movement
how energy is conserved
m g h = ½ mv² + ½ I w²
linear and angular velocity are related
v = w r
w = v / r
we substitute
m g h = ½ mv² + ½ I (v/r) ²
mg h = ½ v² (m + I /r²)
v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]
the tabulated moments of inertia for the bodies are
solid cylinder I = ½ m r²
hollow cylinder I = m r²
we look for the speed for each body
solid cylinder
v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]
v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]
let's call v₀ = [tex]\sqrt{2gh}[/tex]
v₁ = 0.816 v₀
hollow cylinder
v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]
v₂ = v₀ √½
v₂ = 0.707 v₀
Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.
Answer:
[tex]F_b= 0.720 N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=600N[/tex]
Average density [tex]\rho=1.20kg/m^3[/tex]
Mass
[tex]m=\frac{W}{g}[/tex]
[tex]m=\frac{600}{9.81}[/tex]
[tex]m=61.22kg[/tex]
Generally the equation for Volume is mathematically given by
[tex]V =\frac{ mass}{density}[/tex]
[tex]V= \frac{61.22}{1000}[/tex]
[tex]V=0.06122 m^3[/tex]
Therefore
Buoyant force [tex]F_b[/tex]
[tex]F_b=\rho*V*g[/tex]
[tex]F_b= rho_air*V*g[/tex]
[tex]F_b= 0.720 N[/tex]
What is the meant of by renewable energy and non-renewrable with example of each.
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0?
Check attached photo
Check attached photo
Answer:
Explanation:
1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.
2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.
The magnitude of A and B must be equal
An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown
Answer: [tex]283.2\times 10^{-9}\ nC[/tex]
Explanation:
Given
Cross-sectional area [tex]A=0.4\ cm^2[/tex]
Dielectric constant [tex]k=4[/tex]
Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]
Distance between capacitors [tex]d=5\ mm[/tex]
Maximum charge that can be stored before dielectric breakdown is given by
[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]
Answer:
The maximum charge is 7.08 x 10^-8 C.
Explanation:
Area, A = 0.4 cm^2
K = 4
Electric field, E = 2 x 10^8 V/m
separation, d = 5 mm = 0.005 m
Let the capacitance is C and the charge is q.
[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]
two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential
Answer:
B has greater potential
Explanation:
We know;
Potential Energy (PE) = mgh
where, m=mass of body
g=acceleration due to gravity
h=height of body
From the formula,
PE is directly proportional to the mass of the body
so the body with greater mass has greater potential.
A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?
Answer:
28 j
Explanation:
because when you add you get 28
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth
Your friend would have a better experience of this event, than you .
What is an eclipse?An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,
there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.
For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.
The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.
To learn more about the eclipse from here, refer to the link;
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A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Answer:
3.55 T
Explanation:
Applying,
F = BILsin∅.............. Equation 1
Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)
Substitute these values into equation 2
B = 2.45/(0.03×23×sin90)
B = 2.45/0.69
B = 3.55 T
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V
At what angle torque is half of the max
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
