Answer:
x = -3 cm
Explanation:
The electrical potential is the sum of the potentials of each charge
V = k ∑ [tex]q_{i} / r_{i}[/tex]
let's apply this to our case where the potential is V = 0 for x = 0
0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))
in our case
q₁ = + 2.0 10⁻⁶ C
q₂ = - 6.0 10⁻⁶ C
q₃ = + 3.0 10⁻⁶ C
x₁ = -1.0 cm = 1.0 10⁻² m
x₂ = +2.0 cm = 2.0 10⁻² m
we substitute in the equation
0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)
3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴
3 10⁻⁶ / x = -1 10⁻⁴
x = - 3 10⁻² m
x = -3 cm
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even possible from a conservation of energy standpoint. Also, can you relate this behavior to the transient (natural) response of the circuit that you observed in the previous lab
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + ([tex]X_{C}[/tex] -[tex]X_{L}[/tex])² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL
A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward
Answer:
The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s
Explanation:
Given;
speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s
speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s
Determine the speed of the ball relative to the ground if the ball is thrown forward.
If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.
[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]
Therefore, the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.
1/1
2. Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the
volume flow rate of the fluid in m^3/s *
Answer:
9.42*10^-4 m^3/s
Explanation:
d=3/100 m
=0.03 m
A=3.14*0.03^2/6
=4.71*10^-4 m^2
Volume flow rate V = A * s
= 4.71*10^-4 * 2
= 9.42*10^-4
So, the volume flw rate of fluid is 9.42*10^-4 m^3/s
An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which of the following statements regarding the circuit are correct?
a) The capacitor charges to its maximum value in one time constant and the current is zero at that time.
b) The potential difference across the resistor and the potential difference across the capacitor are always equal.
c) The potential difference across the resistor is always greater than the potential difference across the capacitor.
d) The potential difference across the capacitor is always greater than the potential difference across the resistor
e) Once the capacitor is essentially fully charged, There is no appreciable current in the circuit.
Answer:
e)
Explanation:
In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.
At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.
An aluminum cup of mass 150 g contains 800 g of water in thermal equilibrium at 80.0°C. The combination of cup and water is cooled uniformly so that the temperature decreases by 1.50°C per minute. At what rate is energy being removed by heat? Express your answer in watts.
Answer:
Heat Flow Rate : ( About ) 87 W
Explanation:
The heat flowing out of the system each minute, will be represented by the following equation,
Q( cup ) + Q( water ) = m( cup ) [tex]*[/tex] c( al ) [tex]*[/tex] ΔT + m( w ) [tex]*[/tex] c( w ) [tex]*[/tex] ΔT
So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,
150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,
800 grams = .8 kilograms
Now remember that the specific heat of aluminum is 900 J / kg [tex]*[/tex] K, and the specific heat of water = 4186 J / kg [tex]*[/tex] K. Therefore let us solve for " the heat flowing out of the system per minute, "
Q( cup ) + Q( water ) = .15 [tex]*[/tex] ( 900 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5 + .8 [tex]*[/tex] ( 4186 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5,
Q( cup ) + Q( water ) = 5225.7 Joules
And the heat flow rate should be Joules per minute,
5225.7 Joules / 60 seconds = ( About ) 87 W
An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it
Answer:
V = 451.47 volts
Explanation:
Given that,
Distance, d = 0.21 m
Initial speed, u = 0
Time, t = 33.3 ns
Let v is the final velocity. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0
So,
[tex]d=\dfrac{1}{2}(v-u)t[/tex]
[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]
Now applying the conservation of energy i.e.
[tex]\dfrac{1}{2}mv^2=qV[/tex]
V is voltage
[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]
So, the voltage is 451.47 V.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
Which of the following is an element? A. Fire B. Carbon C. Salt D. Water
Answer:
OPTION B is correct
Carbon
Explanation:
element can be defined as a pure substance which cannot be broken down by into smaller units through a chemical method, an element has atoms with identical numbers of protons in their atomic nuclei
Each element is composed of its own type of atom. And this gives the reason why chemical elements are all very different from each other. And all substance on Earth has atoms of at least one of this elements.
There about 118 elements and all arranged in a row and colomn of the periodic table .This elements of the periodic table are arranged by their atomic number, which helps with the chemical properties. Example of elements are; Hydrogen, Oxygeñ, carbon.
Therefore, among the option only carbon is an element because it cannot be broken down into smaller unit unlike water which is made up of oxygen and hydrogen. Also salt is a compound containing more elements.
The substance which represents an element given the following option is carbon (option B)
What is an element?An element is a pure substance that consist of identical atoms.
An element can not be broken down into simple substances by ordinary methods.
The period table consist of a large number of elements. Some of which are:
HydrogenHeliumLithiumBerylliumBoronCarbonNitrogenOxygenFluorineNeonWe must also understand that when two or more elements are chemically combined together it is called a compound and when they are not chemically combined together, it is called a mixture.
Thus, we can conclude that the correct answer to the question is Carbon (option B)
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. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.
Let us start with explaining the waves of TV and radio.
The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.
Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.
Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.
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An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron
Answer:
acting force is the answer
The direction of the magnetic force on the moving electron is upward.
The direction of the magnetic force on the electron can be determined by applying right hand rule.
This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.
Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.
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A motorcycle travels up one side of a hill over the top and down the other side. The crest of the hill can be considered to be a circular arc with radius of 45.0 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.
Answer:
The maximum speed of the motorcycle should be 21 m/s
Explanation:
Since the hill is considered to be a circular arc, the motorcycle will experience centripetal force that tends to flip it away from the center of the hill.
Since the motorcycle does not lose contact with the ground, it means that the weight of the motorcycle downwards just balances the centripetal force on the motorcycle.
we know that the centripetal force on the motorcycle is equal to
centripetal force = [tex]\frac{mv^{2} }{r}[/tex]
where m is the mass of the motorcycle,
v is the velocity of the motorcycle,
and r is the radius of the hill = 45.0 m
Also we now that the weight of the motorcycle is equal to
weight = mg
where m is still the mass of the motorcycle,
and g is the acceleration due to gravity = 9.81 m/s
Equating the both forces since they are equal, we'll have
[tex]\frac{mv^{2} }{r}[/tex] = mg
the mass of the motorcycle will cancel out, and we'll be left with
[tex]v^{2} = gr[/tex]
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{9.81*45}[/tex]
[tex]v = \sqrt{441.45}[/tex]
[tex]v[/tex] = 21 m/s
You want the output current from the secondary coil of a transformer to be 10 times the input current to the primary coil. The ratio of the number of turns N2/N1 must be:_____________.
A. 100
B. 10
C. 1
D. 0.1
Answer:
D. 0.1
Explanation:
Using transformer equation,
N2/N1 = I1/I2................... Equation 1
Where N2 = secondary coil, N1 = primary coil, I1 = input current, I2 = output current.
make I2 the subject of the equation
I2 = I1/(N2/N1)............ Equation 2
From equation 2 above, For the output current of the secondary coil to be 10 times the input current, N2/N1 = 0.1
Hence the right option is D. 0.1
If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g
Answer: Translational Kinetic Energy
Rotational Kinetic Energy
Explanation:
An object has translational kinetic energy when it is undergoing through a linear displacement.
Rotational energy is kinetic energy due to the rotation of an object .
Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.
Hence, the tires have been two kinds of energy : translational and rotational kinetic energy
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
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A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
Answer:
4.9x10^-6T
Explanation:
See attached file
A long solenoid (1500 turns/m) carries a current of 20 mA and has an inside diameter of 4.0 cm. A long wire carries a current of 2.0 A along the axis of the solenoid. What is the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire
Answer:
The magnitude of the magnetic field is 55μT
Explanation:
Given;
number of turns of the solenoid per length, n = N/L = 1500 turns/m
current in the solenoid, I = 20 mA = 20 x 10⁻³ A
diameter of the solenoid, d = 4 cm = 0.04 m
The magnetic field at a point that is inside the solenoid;
B₁ = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Given;
current in the wire, I = 2 A
distance of magnetic field from the wire, r = 1 cm = 0.01 m
The magnetic field at 1.0 cm from the wire;
[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]
The magnitude of the magnetic field;
[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]
Therefore, the magnitude of the magnetic field is 55μT
The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Given the following parameters from the question
Number of turns of the solenoid per length, n = N/L = 1500 turns/m current in the solenoid, I = 20 mA = 20 x 10⁻³ A Diameter of the solenoid, d = 4 cm = 0.04 mThe magnetic field at a point that is inside the solenoid is expressed according to the formula;
B₁ = μ₀nIWhere;
μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Next is to get the magnetic field strength in the second wire.
Current in the wire, I = 2 A Distance of magnetic field from the wire, r = 1 cm = 0.01 mThe magnetic field at 1.0 cm from the wireSubstitute into the formula:
[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]
Get the resultant magnetic field:
[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]
Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
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The length of a certain wire is doubled and at the same time its radius is also doubled. What is the new resistance of this wire
Answer:
R' = R/2
Therefore, the new resistance of the wire is twice the value of the initial resistance.
Explanation:
Consider a wire with:
Resistance = R
Length = L
Area = A = πr²
where, r = radius
ρ = resistivity
Then:
R = ρL/A
R = ρL/πr² --------------- equation 1
Now, the new wire has:
Resistance = R'
Resistivity = ρ
Length = L' = 2 L
Radius = r' = 2r
Area = πr'² = π(2r)² = 4πr²
Therefore,
R' = ρL'/πr'²
R' = ρ(2 L)/4πr²
R' = (1/2)(ρL/πr²)
using equation 1:
R' = R/2
Therefore, the new resistance of the wire is twice the value of the initial resistance.
In a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please help, and include the formula for the answer and a step by step explanation
Answer:
Mass of shot (m) = 4 kg
Explanation:
Given:
Velocity (v) = 15 m/s
Mechanical kinetic energy (K.E) = 450 J
Find:
Mass of shot (m) = ?
Computation:
Mechanical kinetic energy (K.E) = 1/2mv²
Mechanical kinetic energy (K.E) = [1/2](m)(15)²
450 = [1/2](m)(15)²
900 = 225 m
Mass of shot (m) = 4 kg
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assume the orbits are circular, and estimate the mass of the object, in units of the mass of the sun (MSun = 2x1030 kg). If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.
Answer:
The mass of the object is 745000 units of the sun
Explanation:
We know that the centripetal force with which the stars orbit the object is represented as
[tex]F_{c}[/tex] = [tex]\frac{mv^{2} }{r}[/tex]
and this centripetal force is also proportional to
[tex]F_{c}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]
where
m is the mass of the stars
M is the mass of the object
v is the velocity of the stars = 10^6 m/s
r is the distance between the stars and the object = 10^14 m
k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
We can equate the two centripetal force equations to give
[tex]\frac{mv^{2} }{r}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]
which reduces to
[tex]v^{2}[/tex] = [tex]\frac{kM}{r}[/tex]
and then finally
M = [tex]\frac{rv^{2} }{k}[/tex]
substituting values, we have
M = [tex]\frac{10^{14}*(10^{6})^{2} }{6.67*10^{-11} }[/tex] = 1.49 x 10^36 kg
If the mass of the sun is 2 x 10^30 kg
then, the mass of the the object in units of the mass of the sun is
==> (1.49 x 10^36)/(2 x 10^30) = 745000 units of sun
soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.
Answer:
Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection
Explanation:
Two identical wooden barrels are fitted with long pipes extending out their tops. The pipe on the first barrel is 1 foot in diameter, and the pipe on the second barrel is only 1/2 inch in diameter. When the larger pipe is filled with water to a height of 20 feet, the barrel bursts. To burst the second barrel, will water have to be added to a height less than, equal to, or greater than 20 feet? Explain.
Answer:
The 1/2 inch barrel will burst at the same height of 20 ft
Explanation:
The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.
From the equation of fluid pressure,
P = ρgh
where
P is the pressure at the bottom of the fluid due to its height
ρ is the density of the fluid in question
h is the height to which the water stand.
You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.
We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.
That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.
NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
The Hermes spacecraft is traveling at 0.1c(1/10 the speed of light past Mars and shines a laser in front of the ship. You would see the light traveling at c (the speed of light )away from your ship. According to Einstein's special relativity how fast with a person on Mars observe the light to be traveling?
Answer:
So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.
Explanation:
The special theory of relativity has two main postulates:
1- VALIDITY OF PHYSICAL LAWS
The laws of physics such as Newton's Laws and Maxwell's Equations are valid in all inertial frame of references.
2- CONSTANCY OF SPEED OF LIGHT
The speed of light in vacuum is the same for all observers in uniform translational relative motion, and it is independent of the motion of the source or the observer. Thus, speed of light is a universal constant and its value is c = 3 x 10⁸ m/s.
So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.
Which examples are simple machines?
Select all correct answers.
a hammer
an automobile
O a pulley
an inclined plane
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
Two ships of equal mass are 109 m apart. What is the acceleration of either ship due to the gravitational attraction of the other? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)
Answer:
The acceleration is [tex]a = 2.190 *10^{-7} \ m/s^2[/tex]
Explanation:
From the question we are told that
The distance of separation of the ship is [tex]r= 109 \ m[/tex]
The mass of each ship is [tex]M = 39,000 \ metric\ tons =39,000 * 1000 = 3.9 *10^{7}\ kg[/tex]
The gravitational force of attraction exerted on each other is mathematically represented as
[tex]F_g = \frac{ GMM}{r^2}[/tex]
Where G is the gravitational constant with value
substituting values
[tex]F_g = \frac{ 6.674 30 * 10^{-11} (3.9 *10^{7})^2}{109^2}[/tex]
[tex]F_g = 8.54 \ N[/tex]
This force can also be mathematically represented as
[tex]F_g = M * a[/tex]
=> [tex]a = \frac{F_g}{M}[/tex]
substituting values
[tex]a = \frac{8.544}{3.9 *10^{7}}[/tex]
[tex]a = 2.190 *10^{-7} \ m/s^2[/tex]
a large crane has a mass of 8500kg calculate the weight of the crane
Answer:
Weight is 83 385 N
Explanation:
Weight is calculated by multiplying the mass by the gravitational acceleration constant
Weight = mass* gravity
Assuming that the gravitational constant is 9.81 m/s^2
Weight = mass* gravity
Weight of crane = (8500 kg)*(9.81 m/s^2)
Weight = 83 385 kg*m/s^2 or 83 385 N
If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ????) of the system?
The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.
Given the following data:
Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]Work done = [tex]4.50 \times 10^2 \;kcal[/tex]Conversion:
1 kcal = 4.184 kJ
[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]
To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.
Mathematically, the first law of thermodynamics is given by the formula:
[tex]\Delta E = Q - W[/tex]
Where;
[tex]\Delta E[/tex] is the change in internal energy.Q is the quantity of heat released.W is the work done.Substituting the given parameters into the formula, we have;
[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]
Change in internal energy, E = -18823 Kilojoules
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