A thin rod of length 2.15 m and mass 13.1 kg is rotated at an angular speed of 3.95 rad/s around an axis perpendicular to the rod and through its center of mass. Find the magnitude of the rod's angular momentum.
Answer:
L = 239.2 J.s
Explanation:
The angular momentum of the rod is given by the following formula:
L = mvr
where,
L = Angular Momentum = ?
m = mass = 13.1 kg
r = radius of circle of rotation = length of rod = 2.15 m
v = linear speed of rod = r(angular speed of rod) = (2.15 m)(3.95 rad/s)
v = 8.5 m/s
Therefore,
L = (13.1 kg)(8.5 m/s)(2.15 m)
L = 239.2 J.s
The angular momentum of the thin rod will be L = 239.2 J.s
What is angular momentum?Whenever any object has rotational motion then its angular momentum will be the product of the linear velocity of the object and the radius of the circle of rotation of the object.
The angular momentum of the rod is given by the following formula:
[tex]L= MVr[/tex]
where,
L = Angular Momentum = ?
m = mass = 13.1 kg
r = radius of circle of rotation = length of rod = 2.15 m
w=3.95 rad/s
Now the angular momentum will be calculated as:
[tex]L= MVr[/tex]
The linear velocity will be
[tex]V=wr=3.95\times 2.15=8.5\ \frac{m}{s}[/tex]
Now put the value in the formula we get:
[tex]L=13.1\times 8.5\times 2.15=239.2\ J[/tex]
[tex]L=239.2\ J[/tex]
Hence the angular momentum of the thin rod will be L = 239.2 J.s
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Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .
Answer:
Quantum theory gives the concept of
The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v = [tex]\frac{dx}{dt}[/tex]
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
[tex]t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}[/tex]
t = [tex]\frac{18 \pm 42}{30}[/tex]
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.
Answer:
imma try and fail again and again
A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{\rm total}=\Delta K[/tex]
or
[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]
Compute the work performed by friction:
[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]
[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]
[tex]\implies \boxed{\mu\approx0.45}[/tex]
The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
What is coefficient of friction?Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{total}=\Delta K[/tex]
or
[tex]W_{friction}+W_{spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]
Compute the work performed by friction:
[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{friction}=-f(0.20)[/tex]
[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]
[tex]\mu =0.45[/tex]
Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
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b. The stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. . 3 At what distance x does the stream strike the floor?
Answer:
34.64 cm
Explanation:
Given that:
The depth of the hole h = 10 cm
height of the water holding in the tank H = 40 cm
For a stream of flowing water, the distance (x) at which the stream strikes the floor can be computed by using the formula;
[tex]x = 2 \sqrt{h(H-h)}[/tex]
[tex]x = 2 \sqrt{10(40-10)}[/tex]
[tex]x = 2 \sqrt{10(30)}[/tex]
[tex]x = 2 \sqrt{300}[/tex]
[tex]x = 2 \times 17.32[/tex]
x = 34.64 cm
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. Assume the spring is compressed 6 cm and released from rest. (c) Determine the speed of the object as it passes the point 1.9 cm from the equilibrium position
The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
W = ∆K = K
where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
W = 1/2 mv ²
where m is the mass of the object and v is the speed you want to find. Solving for v, you get
v = √(2W/m) ≈ 0.46 m/s
A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.
Water stands at a depth H in a tank whose side wails are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor?
Answer:
x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]
Explanation:
This is an exercise in fluid mechanics, let's find the speed of the water in the hole
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole
the pressure inside and outside the tank is the same
P₁ = P₂
we must measure the distance from the same reference point, let's locate it on the surface of the water
y₁ = 0
y₂ = h
Suppose the gap is small compared to the diameter of the tank
v₁ «v₂
v₂² = 2 g (0-h)
This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.
Let's use the projectile launch ratios
vₓ = [tex]\sqrt{2g |h|}[/tex]
y = y₀ + v₀ t - ½ g t²
the height when reaching the floor is y = 0,
the initial height is measured from the floor therefore y₀ = H-h
0 = (H-h) + 0 - ½ g t²
t = [tex]\sqrt{\frac{2(H-h)}{g} }[/tex]
we look for the distance x traveled
x = vₓ t
x = [tex]\sqrt{2g |h| } \ \sqrt{ 2(H-h)/g}[/tex]
x = [tex]\sqrt{4 h (H-h)}[/tex]
x = 2h [tex]\sqrt{\frac{H}{h} -1 }[/tex]
A device is turned on and 3.20 A flows through it 0.130 ms later. What is the self-inductance of the device (in mH) if an induced 140 V emf opposes this
Answer:
the self-inductance of the device is 5.69 mH
Explanation:
Given;
change in current, ΔI = 3.2 A
change in time, Δt = 0.13 ms = 0.13 x 10⁻³ s
induced emf, E = 140
The self-inductance is calculated as;
[tex]E = L\frac{\Delta I}{\Delta t} \\\\where;\\\\L \ is \ the \ self-inductance\\\\\L = \frac{E\Delta t}{\Delta I} \\\\L =\frac{140 \times 0.13 \times 10^{-3}}{3.2} \\\\L = 0.00569 \ H\\\\L = 5.69 \ mH[/tex]
Therefore, the self-inductance of the device is 5.69 mH
explain why an equation may be homogenous with respect to its unit but still be incorrect
In the Homogenous equation with respect to its unit if the constants has been removed, then the equation is still homogenous but it is incorrect.
What is Homogenous equation?Homogenous equation is one of the type of the differential equations. These equation have zero on the right hand side of the equality sign. If the equation has the independent variable on the right side of equal sign then it is said to be non - homogenous equation. The homogenous and non - homogenous equations are the two types of linear equations.
An example of Homogenous equation is,
The formula for kinetic energy can be represented as,
K. E = 1/2 (mv²)
Here, 1/2 can be removed as it was a constant.
K. E = mv².
This equation can be homogenous but still it is incorrect.
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A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
Ayuda Porfavor es URGENTE
Consulta los dibujos adjuntos
define density in thermodynamics
Answer:
total mass (m) of that substance divided by the total volume (V) occupied by that substances
What is the difference between muscular endurance and muscular strength? A B or C
Answer:
Muscular endurance is how many times you can move a weight without getting tired.
Muscular strength is the amount force you can put out.
Which statement below is correct about this planet?
Everything on the planet will fall twice faster than on Earth.
Everything on the planet will fall twice slower than on Earth.
Everything on the planet will move twice faster than on Earth.
Everything on the planet will move twice slower than on Earth.
Answer:
Everything on the planet will move twice faster than on Earth.
Explanation:
Uranus has a mass of about 8.681 × 10^25 kg. That is its larger than earth and is also has a faster rotation peroid of 17 hours. Its has a gravitational potential less than earth.An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
energy of fossil fuel is also derived from solar energy why
Explanation:
All the energy in oil, gas, and coal originally came from the sun captured through photosynthesis. In the same way that we burn wood to release energy that trees capture from the sun, we burn fossil fuels to release the energy that ancient plants captured from the sun.
how do I learn French fast for an examination
Explanation:
It isn't possible to learn an entire language fast. For an examination, memorise some of the easy words. Practice mew minutes speaking French and keep speaking it till you know it.
Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 40 times as great as the angular speed of the tire itself. Each tire has a radius of 0.330 m. The coil consists of 163 turns, has an area of 3.23 x 10-3 m2, and rotates in a 0.0948-T magnetic field. The bicycle starts from rest and has an acceleration of 0.601 m/s2. What is the peak emf produced by the generator at the end of 6.72 s
Answer:
[tex]Emf=24.4V[/tex]
Explanation:
From the question we are told that:
Angular speed [tex]\omega=40*\omega'[/tex]
Radius [tex]r=0.33m[/tex]
No. Turns [tex]N=163[/tex]
Area [tex]A= 3.23 * 10^{-3} m^2[/tex]
Magnetic field. [tex]B=0.0948T[/tex]
Acceleration [tex]a=0.60m/s^2[/tex]
Time [tex]t=6.72s[/tex]
Generally the equation for momentum is mathematically given by
[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]
[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]
[tex]W=12.22rad/s^2[/tex]
Therefore
[tex]\omega=Aw[/tex]
[tex]\omega=12.22*40[/tex]
[tex]\omega=488.7rads/s[/tex]
Generally the equation for Peak emf is mathematically given by
[tex]Emf=NBA \omega[/tex]
[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]
[tex]Emf=24.4V[/tex]
3) A fahrenheit thermometer shaded from the sun on a hot day reads 101⁰ what is the temperature on the centigrade scale?
Answer:
38.33°C
Explanation:
Applying,
180/100 = (F-32)/C............. Equation 1
Where F = Temperature of the hot day in fehrenheit, C = Temperature of the hot day in centigrade.
make C the subject of the equation
C = 100(F-32)/180.............. Equation 2
From the question,
Given: F = 101°F
Substitute into equation 2
C = 100(101-32)/180
C = 38.33°C
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz
Answer:
The correct answer is "121.36 meters and 2.40 MHz".
Explanation:
Given:
Red shift,
[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]
Wavelength,
[tex]\lambda = 125 \ meters[/tex]
The observed frequency will be:
⇒ [tex]f = \frac{c}{\lambda}[/tex]
[tex]=\frac{3\times 10^8}{125}[/tex]
[tex]=24\times 10^5[/tex]
[tex]= 2.40\times 10^6[/tex]
[tex]=2.40 \ MHz[/tex]
hence,
The Emitted wavelength will be:
⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]
[tex]\frac{125}{\lambda_0}-1 =0.03[/tex]
[tex]125=1.03 \ \lambda_0[/tex]
[tex]\lambda_0=\frac{125}{1.03}[/tex]
[tex]=121.36 \ m[/tex]
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Describe how a positively charged object can be used to give another object a negative charge. What is the name of this process
Answer:
induction using ground connection
Explanation:
This process is referred to as induction and it is done using ground process.
The process can be carried out as follows;
-An object that has a positive charge would be brought close to another object that is neutrally charged. then the electrons of the neutral object would be attracted and +ve charges would be repelled. This would cause an alignment at the two opposite corners.
- After this is done, the object would be grounded from the side that has the positive charges. This would bring about electron flowing from the ground to the earth. bringing about a neutralization of the positive charges.
-then the grounding would be taken away resulting in electrons leaving in majority.
-At the end, after the positively charged object has been removed, the charges would reassemble themselves. this would cause the object to be negatively charged.
therefore negative charge can be induced on a different object that is neutral using positive charge, through the induction process and grounding.
3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)
Answer:
Explanation:
Changing the resistance of a resistor means the resistance is either increased or decreased.
a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.
Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;
V = IR
6 = I x 30
I = 0.2 A
If the resistance is changed to 50 Ohm's, then:
I = 0.12 A
(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.
In the previous example, if the resistance is changed to 5 Ohm's, then:
V = IR
6 = I x 5
I = 1.2 A
(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.
In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.
Answer:
In the given circuit, R
2
,R
6
and R
4
are in series. So,
R
1
′
=7+5+12=24Ω
Now R
1
′
and R
5
′
are in parallel. So,
R
2
′
1
=
8
1
+
24
1
=
24
3+1
=
24
4
=
6
1
R
2
′
=6ohm.
Now R
2
′
,R
1
and R
3
are in series. So,
R=R
2
′
+R
1
+R
3
=6+3+2=11ohm.
We know i=
R+r
E
=
11+1
6
=
12
6
=
2
1
i=0.5amp.
Canopus, which is in the constellation of Carina and Argo Navis, is 310 ly away. You plan a sight-seeing vacation for Canopus and book a flight on the fastest spaceship in the universe, which travels at the speed of light. How many meters will you traverse to reach your vacation destination
Answer:
[tex]D'=2.933*10^{18}m[/tex]
Explanation:
From the question we are told that:
Distance [tex]D=310ly[/tex]
Where
[tex]1ly=9.46*10^{15}[/tex]
Generally the equation for meters will you traverse to reach your vacation destination is mathematically given by
[tex]D'=D*1ly[/tex]
[tex]D'=310*9.46*10^{15}[/tex]
[tex]D'=2.933*10^{18}m[/tex]
