Answer:
Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.
In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.
6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor? Explain your answer.
Answer:
Explanation:
Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it. if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it
For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.
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.
What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?
Answer:
The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]
Explanation:
From the question we are told that
The order of maximum diffraction is m = 2
The wavelength is [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]
The angle is [tex]\theta = 62.5^o[/tex]
Generally the condition for constructive interference for diffraction grating is mathematically represented as
[tex]dsin \theta = m * \lambda[/tex]
where d is the distance between the lines on a diffraction grating
So
[tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]
substituting values
[tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]
[tex]d = 1.747 *10^{-6} \ m[/tex]
Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?
Answer:
The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Explanation:
We are told that the particle moves back and forth along a straight line with velocity v(t).
Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.
Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe
Answer:
0.0127m
Explanation:
Using
Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m
A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2
Answer:
650W/m²Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 275 km above the Earth
Answer:
s_400 = 16.5 m , s_700 = 29.4 m
Explanation:
The limit of the human eye's solution is determined by the diffraction limit that is given by the expression
θ = 1.22 λ / D
where you lick the wavelength and D the mediator of the circular aperture.
In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.
by introducing these values into the formula
λ = 400 nm θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad
λ = 700 nm θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad
Now we can use the definition radians
θ= s / R
where s is the supported arc and R is the radius. Let's find the sarcos for each case
λ = 400 nm s_400 = θ R
S_400 = 6 10⁻⁵ 275 10³
s_400 = 16.5 m
λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³
s_700 = 29.4 m
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.
Answer:
work done in pumping the entire fuel is 1399761 J
Explanation:
weight per volume of the gasoline = 6600 N/m^3
diameter of the tank = 3 m
length of the tank = 6 m
The height of the tractor tank above the top of the tank = 5 m
The total volume of the fuel is gotten below
we know that the tank is cylindrical.
we assume that the fuel completely fills the tank.
therefore, the volume of a cylinder =
where r = radius = diameter ÷ 2 = 3/2 = 1.5 m
volume of the cylinder = 3.142 x x 6 = 42.417 m^3
we then proceed to find the total weight of the fuel in Newton
total weight = (weight per volume) x volume
total weight = 6600 x 42.417 = 279952.2 N
therefore,
the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)
work done in pumping = 279952.2 x 5 = 1399761 J
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:
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The spring scale [tex]F_2[/tex] reads [tex]F_2 = 2.4225 \ N[/tex]
Explanation:
From the question we are told that
The first force is [tex]F_1 = 10.5 \ N[/tex]
The acceleration by which the cart moves to the right is [tex]a = 2.50 \ m/s^2[/tex]
The mass of the cart is m = 3.231 kg
Generally the net force on the cart is
[tex]F_{net} = F_1 - F_2[/tex]
This net force is mathematically represented as
[tex]F_{net} = m * a[/tex]
So
[tex]m* a = 10 - F_2[/tex]
[tex]F_2 = 10.5 - 2.5 (3.231)[/tex]
[tex]F_2 = 2.4225 \ N[/tex]
Which has more mass electron or ion?
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?
Answer:
a
[tex]F =326.7 \ N[/tex]
b
[tex]M = 6[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 200 \ kg[/tex]
The length of the small object from the rock is [tex]d = 2 \ m[/tex]
The length of the small object from the branch [tex]l = 12 \ m[/tex]
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
[tex]W = m_r * g[/tex]
substituting values
[tex]W = 200 * 9.8[/tex]
[tex]W = 1960 \ N[/tex]
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
[tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]
Here [tex]\theta[/tex] is very small so [tex]cos\theta * l = l[/tex]
and [tex]cos\theta * d = d[/tex]
Hence
[tex]F * l - W * d = 0[/tex]
=> [tex]F = \frac{W * d}{l}[/tex]
substituting values
[tex]F = \frac{1960 * 2}{12}[/tex]
[tex]F =326.7 \ N[/tex]
The mechanical advantage is mathematically evaluated as
[tex]M = \frac{W}{F}[/tex]
substituting values
[tex]M = \frac{1960}{326.7}[/tex]
[tex]M = 6[/tex]
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air
Answer:
The critical angle is [tex]i = 41.84 ^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the sugar solution is [tex]n_s = 1.5[/tex]
The index of refraction of air is [tex]n_a = 1[/tex]
Generally from Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]
Note that the angle of incidence in this case is equal to the critical angle
Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]
So
[tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]
[tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]
[tex]i = 41.84 ^o[/tex]
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 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Answer:
The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
Explanation:
The magnetic field at the center of the solenoid is given by;
B = μ(N/L)I
Where;
μ is permeability of free space
N is the number of turn
L is the length of the solenoid
I is the current in the solenoid
The rate of change of the field is given by;
[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]
The induced emf in the shorter coil is calculated as;
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
where;
N is the number of turns in the shorter coil
A is the area of the shorter coil
Area of the shorter coil = πr²
The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m
Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²
[tex]E = NA\frac{\delta B}{\delta t}[/tex]
E = 14 x 0.000491 x 0.02514
E = 1.728 x 10⁻⁴ V
Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V
The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]
Induced EMF:The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:
[tex]E=-\frac{\delta\phi}{\delta t}[/tex]
where E is the induced emf,
[tex]\phi[/tex] is the magnetic flux through the coil.
The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:
[tex]\phi = \frac{\mu_oNIA}{L}[/tex]
so;
[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]
where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.
[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]
The emf produced in the coil at the center of the solenoid which has 14 turns will be:
[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]
Learn more about induced emf:
https://brainly.com/question/16765199?referrer=searchResults
Two parallel plates have charges of equal magnitude but opposite sign. What change could be made to increase the strength of the electric field between the plates
Answer:
The electric field strength between the plates can be increased by decreasing the length of each side of the plates.
Explanation:
The electric field strength is given by;
[tex]E = \frac{V}{d}[/tex]
where;
V is the electric potential of the two opposite charges
d is the distance between the two parallel plates
[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]
Where;
ε₀ is permittivity of free space
L is the length of each side of the plates
From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.
Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates
An apple falls from a tree and hits your head with a force of 9J. The apple weighs 0.22kg. How far did the apple fall?
Answer:
The apple fell at a distance of 4.17 m.
Explanation:
Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.
When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.
In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.
The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.
Work=Force*distance* cosine(angle)
On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:
F=m*a
where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]
In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:
Work= 9 JF=m*a=0.22 kg*9.81 m/s²= 2.1582 Ndistance= ?angle=0 → cosine(angle)= 1Replacing:
9 J= 2.1582 N* distante* 1
Solving:
[tex]distance=\frac{9J}{2.1582 N*1}[/tex]
distance= 4.17 m
The apple fell at a distance of 4.17 m.
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.
A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.5 ss later.
How deep is the ocean at this point? (Note: Use the bulk modulus method to determine the speed of sound in this fluid, rather than using a tabluated value.)
_____ m
Answer:
1248m
The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s
The speed of sound in seawater is 1560 m/s
Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km
A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is
Answer:
0.0081T
Explanation:
The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e
B ∝ I [tex]\frac{N}{L}[/tex]
B = μ₀ I [tex]\frac{N}{L}[/tex] ----------------(i)
Where;
μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²
Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by
L = 2π r
L = 2π (0.042)
L = 0.084π
The number of turns N = 1000
The current in the toroid = 1.7A
Substitute these values into equation (i) to get the magnetic field as follows;
B = 4π x 10⁻⁷ x 1.7 x [tex]\frac{1000}{0.084\pi }[/tex] [cancel out the πs and solve]
B = 0.0081T
The magnetic field along the central radius is 0.0081T
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
Learn more about work here:
https://brainly.com/question/18094932
#SPJ2
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.
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v= 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answer:
The wait time is [tex]t_w = 3.4723 \ s[/tex]
Explanation:
From the question we are told that
The distance of the hot air balloon above the ground is [tex]z = 50 \ m[/tex]
The distance of the hot air balloon from the target is [tex]k = 100 \ m[/tex]
The speed of the wind is [tex]v = 15 \ m/s[/tex]
Generally the time it will take the balloon to hit the ground is
[tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]
where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]
[tex]t = 3.194 \ s[/tex]
Now at the velocity the distance it will travel before it hit the ground is mathematically represented as
[tex]d = v * t[/tex]
substituting values
[tex]d = 15 * 3.194[/tex]
[tex]d = 47.916 \ m[/tex]
Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as
[tex]b = k - d[/tex]
substituting values
[tex]b =100 -47.916[/tex]
[tex]b = 52.084 \ m[/tex]
Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as
[tex]t_w = \frac{b}{v}[/tex]
substituting values
[tex]t_w = \frac{52.084}{15}[/tex]
[tex]t_w = 3.4723 \ s[/tex]
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
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