Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
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 time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Answer:
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Explanation:
The radius of the electron's motion in a uniform magnetic field is given by
[tex]R = \frac{MV}{qB}[/tex]
where;
m is the mass of the electron
q is the charge of the electron
B is the magnitude of the magnetic field
V is speed of the electron
R is the radius of the electron's
Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.
The magnitude of the electron's acceleration inside the field is given by;
[tex]a_c = \frac{V^2}{R}[/tex]
where;
[tex]a_c[/tex] is centripetal acceleration of electron
Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.
The time the electron is in the magnetic field is given by;
[tex]T = \frac{2\pi M}{qB}[/tex]
The time of electron motion will not change
The magnitude of the net force acting on the electron inside the field will not change;
[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]
Therefore, the correct options are "C" and "D"
The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.
Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
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]
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 find its approximate location at time t = 1.02.
Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer:
Final energy = Uf = initial energy × d₂/d₁
Explanation:
Energy is the ability to do work.
capacitor is an electronic device that store charges
where
V is the potential difference
d is the distance of seperation between the two plates
ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.
A = cross sectional area
U =¹/₂CV²
C =ε₀A/d
C × d=ε₀A=constant
C₂d₂=C₁d₁
C₂=C₁d₁/d₂
charge will 'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced
Energy=U =(1/2)q²/C
U₂C₂ = U₁C₁
U₂ =U₁C₁ /C₂
U₂ =U₁d₂/d₁
Final energy = Uf = initial energy × d₂/d₁
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point where the stone strikes the water is given by which formula? R=v(2h/g)2 None of these are correct. R = 2mv sqrt(2h/g) R = v sqrt(2h/g) R=(1/2)gt2
Answer:
R = v √(2h / g)
Explanation:
This exercise can be solved using the concepts of science, projectile launching
let's calculate the time it takes to get to the water
y = y₀ +[tex]v_{oy}[/tex] t - ½ g t²
as the stone is skipped the vertical speed is zero
y = y₀ - ½ g t²
for y=0
t = √ (2y₀ / g)
the horizontal distance it covers in this time is
R = v₀ₓ t
R = v₀ₓ √(2 y₀ / g)
let's call the horizontal velocity as v and the height is h
R = v √(2h / g)
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) At the moment contact is made with the battery the voltage across the capacitor is
Answer:
(a) D. Zero.
(b) C. Equal to the battery's terminal voltage.
Explanation:
The question is incomplete, see the complete question for your reference and information.
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage
across its terminals. At the moment contact is made with the battery
(a) the voltage across the capacitor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero.
(b) the voltage across the resistor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero
A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.
basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R
How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.
Answer:
2.5×10¹⁹
Explanation:
4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second
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 department store expects to have 225 customers and 20 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. The average rate of heat generation from people doing light work is 115 W, and 70% of it is in sensible form.
Answer:
The contribution of people to the cooling load of the store is 19722.5 W
Explanation:
Total amount of customers = 225
Total amount of employees = 20
Total amount of people in the store at that instant n = 245 people
Average rate of heat generation Q = 115 W
percentage of these heat generated that is sensible heat = 70%
Sensible heat raises the surrounding temperature. Latent heat only causes a change of state.
The total heat generated by all the people in the store = n x Q
==> 245 x 115 = 28175 W
but only 70% of this heat is sensible heat that raises the temperature of the store, therefore, the contribution of people to the cooling load of the store = 70% of 28175 W
==> 0.7 x 28175 = 19722.5 W
An individual is moving out of his apartment and has several boxes to carry down the stairs. Some of the boxes are light and some of the boxes are heavy. 3. Using the concept of motor units, describe how he generates more or less power to move the boxes.
Answer:
Motor unit is made up of motor neurons. Group of Motor Units work together to coordinate contraction of single muscle. When an individual needs to carry several boxes down stairs then his brain gives signal to neurons who innervates a group of skeletal muscle. When neurons receive signal they stimulate all muscle fibers in that particular motor unit.
Explanation:
Motor unit is made up of motor neurons. Group of Motor Units work together to coordinate contraction of single muscle. When an individual needs to carry several boxes down stairs then his brain gives signal to neurons who innervates a group of skeletal muscle. When neurons receive signal they stimulate all muscle fibers in that particular motor unit.
A 4.0 kg mass is attached to a spring whose spring constant is 950 N/m. It oscillates with an amplitude of 0.12 m. What is the maximum velocity of the mass
Answer:
velocity = 2.62m/s
Explanation:
950= (4 x A)/0.12
950 x 0.12 = 4 x A
114 = 4 x A
A = 114/4
A = 28.5m/s²
U²=2asU² = 2 x 28.5 x 0.12U² = 6.84U = √6.84U = 2.62m/sThe maximum velocity of the mass is equal to 1.85,/s when the amplitude of oscillation is 0.12m.
What is the spring force?The spring force will be acting on the spring when the spring is stretched or compressed, which opposes the load force. These springs are divided into many types based on how this load force is applied to them.
F = -kx
where k is the spring constant and x is the displacement of the spring attached with mass.
Given, the mass attached to the spring, m = 4.0 Kg
The value of spring constant, k = 950 N/m
The amplitude of oscillation, A = 0.12m
The maximum velocity can be calculated as:
[tex]\frac{mv_{max}^2}{2} =\frac{kA^2}{2}[/tex]
[tex]v_{max} =\sqrt{\frac{kA^2}{m} }[/tex]
Substitute the values of the m, k, and A in the above equation:
Vmax = [tex]=\frac{950N/m(0.12m)^2}{4Kg}[/tex]
Vmax = √3.42 m/s
Vmax = 1.85m/s
Therefore, the maximum velocity of the mass is equal to 1/85 m/s.
Learn more about spring force, here:
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A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will have a value that is
Answer:
The equivalent is 6R/11Explanation:
We know that the equivalent resistance of resistors connected in parallel is expressed as
[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]
the L.C.M is 6R
[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]
An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30o with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is:_________.
a. 24 m/s.
b. 11 m/s.
c. 15 m/s.
d. 5.3 m/s.
e. 17 m/s.
Answer:
The speed will be "16.67 m/s".
Explanation:
The given values are:
Distance
= 72 m
Angle
= 30°
Acceleration
= [tex]g(sin \theta-ucos \theta)[/tex]
= [tex](9.8\times sin30^{\circ}) - (0.53\times cos30^{\circ})[/tex]
= [tex]1.929 \ m/s^2[/tex]
Let the speed be "v".
⇒ [tex]v^2=u^2+2as[/tex]
⇒ [tex]v^2=0(2\times 1.929\times 72)[/tex]
⇒ [tex]v^2=277.226[/tex]
⇒ [tex]v=\sqrt{277.776}[/tex]
⇒ [tex]v=16.67 \ m/s[/tex]
Consider 4 charges placed at the corners of a square with side 1.25m as shown. What are the magnitude and direction of the electrostatic force on Q1 resulring from other three charges?
(Note: Please put your final answer in 4 decimal places.). pls answer:)
Answer:
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
F = 3.0226 10⁻² N , θ = 21.16º
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r₁₂²
We also use that the force is a vector magnitude, so we must calculate each component of the force , see the adjoint for the direction of the vectors
X axis
Fₓ = -F₁₄ + F₁₃ₓ
Y axis
[tex]F_{y}[/tex] = F₁₂ -F_{13y}
let's look for the expression for each force
where the side of the square is a = 1.25 m
F₁₂ = k Q₁Q₂ / a²
F₁₄ = k Q₁Q₄ / a²
the distance between 1 and 3 is
d = √(a² + a²) = a √2
F₁₃ = k Q₁Q₃ / d²
let's use trigonometry to find the components
cos 45 = F₁₃ₓ / F₁₃
F₁₃ₓ = F₁₃ cos 45
F₁₃ₓ = k Q₁Q₃ / 2a²
sin 45 = F_{13y} / F₁₃
F_{13y} = F₁₃ sin 45
F_{13y} = k Q₁Q₃ / 2a² sin 45
Taking all terms, we substitute in the force for each axis
X axis
Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45
Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2 cos 45)
Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25² (- 4.5 10⁻⁶ + 3.5/2 cos 45 10⁻⁶)
Fₓ = 8.64 10³ (3.2626 10⁻⁶)
Fₓ = 2.8189 10⁻² N
Y axis
F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a² sin 45
F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)
F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25² (2.5 10⁻⁶ - 3.5/2 sin 45 10⁻⁶)
F_{y} = 8.64 10³ (1.26256 10⁻⁶)
F_{y} = 1.0909 10⁻² N
The result can be given in two ways
1) F = Fₓ i ^ + F_{y} j ^
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry
F = √ (Fₓ² + F_{y}²)
F = 10⁻² √ (2,8189² + 1,0909²)
F = 3.0226 10⁻² N
tan θ = F_{y} / Fx
θ = tan⁻¹ (F_{y} / Fₓ)
θ = tan⁻¹ (1.0909 / 2.8189)
θ = 21.16º
If an inductor, a capacitor, and a resistor are connected in series with a sine wave generator, what quantity will be common to all three components
Answer:
CURRENTExplanation:
For series connected elements (an inductor, a capacitor, and a resistor) in a simple AC circuit, the same current will flows through the elements since are are no presence of nodes between the elements. The total current from the source is what will flow through all of them.
For example, let assume the total current flowing in the circuit is 3A, the amount of current that will flow through the inductor, capacitor and resistor will be the same 3A because of the nature of their connection (series). It is the voltage across each of them that differs.
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 850 m . After this point, its acceleration is that of gravity, downward.
Answer:
v = 73.75 m/s
Explanation:
It is given that,
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.
Let us assume we need to find the velocity of the rocket when it runs out of fuel.
Let v is the final speed. Using the third equation of kinematics as :
[tex]v^2-u^2=2as[/tex]
u = 0
[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]
So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s
A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire carrying a 3.0 A current intersects the x-axis at x=+2.3cm.
Required:
a. At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?
b. At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?
Answer:
a) v r = 0.7318 cm , b) r = 7.23 cm
Explanation:
The magnetic field generated by a wire carrying a current can be found with Ampere's law
∫ B. ds = μ₀ I
the length of a surface circulates around the wire is
s = 2π r
where r is the point of interest of the calculation of the magnetic field
B = μ₀ I / 2π r
In this exercise we have two wires, write the equation of the magnetic field of each one
wire 1 I = 5.8 A
B₁ = μ₀ 5,8 / 2π r₁
wire 2 I = 3.0 A
B₂ = μ₀ 3/2π r₂
the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field
Let's apply these expressions to our case
a) the two streams go in the same direction
using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero
B₁ - B₂ = 0
B₁ = B₂
μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂
5.8 / r₁ = 3 / r₂
5.8 r₂ = 3r₁
the value of r is measured from each wire, therefore
r₁ = 2.3 + r
r₂ = 2.3 -r
we substitute
5.8 (2.3 - r) = 3 (2.3 + r)
r (3 + 5.8) = 2.3 (5.8 - 3)
r = 2.3 2.8 / 8.8
r = 0.7318 cm
b) the two currents have directional opposite
with the right hand rule in the field you have opposite directions outside the wires
suppose it is zero on the right side where the wire with the lowest current is
B₁ = B₂
5.8 / r₁ = 3 / r₂
5.8 r₂ = 3 r₁
r₁ = 2.3 + r
r₂ = r - 2.3
5.8 (r - 2.3) = 3 (2.3 + r)
r (5.8 -3) = 2.3 (3 + 5.8)
r = 2.3 8.8 / 2.8
r = 7.23 cm
Dr. Stein's hypothesis is that excess sugar causes hyperactivity. He is interested in doing research.
Which research method would be the best to use?
Answer:
The correct answer would be - dependent independent variable experiment.
Explanation:
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.
Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, the correct answer would be - dependent independent variable experiment.
The best research method to use for the research of hyperactivity, would be dependent-independent variable experiment.
The given problem is based on the effect of sugar on hyperactivity. Hyper activity refers to the increased movement, impulse actions and a shorter attention span.
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, we can conclude that the best research method to use, would be - dependent-independent variable experiment.
Learn more about the hyperactivity here:
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Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you encounter a long, narrow depression on the ocean floor. Your passenger asks whether you think it is a submarine canyon, a rift valley, or a deep-ocean trench. How would you respond? Explain your response.
Answer:
I would say its a deep ocean trench
Explanation:
This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm2. What is the current in the wire
Answer:
5.95 A
Explanation:
From the question
R = ρL/A..................... Equation 1
Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.
Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m
Substitute these values into equation 1
R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶
R = 0.084 Ω.
Finally, using Ohm law,
V = IR
Where V = Voltage, I = current
Make I the subject of the equation
I = V/R............... Equation 2
I = 0.5/0.084
I = 5.95 A
We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will now dissipate the most heat?
Both wires will dissipate the same amount of heat.
A. The Nichrome wire (resistance 2.7)
B. The copper wire (resistance 0.1)
Answer: B. The copper wire (resistance 0.1)
Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law
i = [tex]\frac{V}{R}[/tex]
So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.
Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:
P = [tex]i^{2}*R[/tex]
As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.
In conclusion, the copper wire will dissipate more heat when connected in parallel.
A system is a group of objects that’s analyzed as one unit. Consider a car moving along a road that has a flat section and a hill. The energy of the car at any given time is equal to the energy that its engine provides minus the energy that the car. When the car moves along the flat section, all of its energy is , which is calculated from its velocity and . When the car moves uphill, some of its energy is transformed to , which is calculated from its gravity, height, and .
Answer:
a) Em= K +U, b) Em= K
Explanation:
The system in this case is formed by the mobilizes and the hill.
Let's write the expressions correctly and completely.
a) When the car moves in the path, the mechanical energy is the siua of the kinetic energy of the car and the potential energy of the car when going up the hill.
Em = K + U
be) when the car moves in the flat part all the mechanical energy is formed by its kinetic energy that is calculated with the mass and speed of the car
Em = K
c) When the car goes up the hill the energy the mechanical energy is conserved, but part of the kinetic energy is transformed into potential energy.
Answer:
leaves
kinetic energy
mass
potential energy
mass
Explanation:
Given that the mass of the Earth is 5.972 * 10^24 kg and the radius of the Earth is
6.371 * 10^6 m and the gravitational acceleration at the surface of the Earth is 9.81
m/s^2 what is the gravitational acceleration at the surface of an alien planet with
2.4 times the mass of the Earth and 1.9 times the radius of the Earth?
Although you do not necessarily need it the universal gravitational constant is G =
6.674 * 10^(-11) N*m^2/kg^2
9
Answer:
gₓ = 6.52 m/s²
Explanation:
The value of acceleration due to gravity on the surface of earth is given as:
g = GM/R² -------------------- equation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
M = Mass of Earth
R = Radius of Earth
Now, for the alien planet:
gₓ = GMₓ/Rₓ²
where,
gₓ = acceleration due to gravity at the surface of alien planet
Mₓ = Mass of Alien Planet = 2.4 M
Rₓ = Radius of Alien Planet = 1.9 R
Therefore,
gₓ = G(2.4 M)/(1.9 R)²
gₓ = 0.66 GM/R²
using equation 1
gₓ = 0.66 g
gₓ = (0.66)(9.81 m/s²)
gₓ = 6.52 m/s²
The horizontal surface on which the objects slide is frictionless. If F = 6.0 N and M = 1.0 kg, what is the magnitude of the force exerted on the large block by the small block?
The image is missing, so i have attached it.
Answer:
The force exerted on the large block by the small block = 8.4 N
Explanation:
From the image attached, the mass of the small block = 2M while the mass of the large block = 3M
Also,Force on small block = F and force on large block = 2F
Equilibrium of forces on the left gives;
2F - N = 3Ma
Thus,
Ma = (2F - N)/3 - - - - eq1
Also, on right hand side, Equilibrium of forces gives;
N - F = 2Ma
Ma = (N - F)/2 - - - - eq2
Equating eq(1) and eq(2) gives us;
(2F - N)/3 = (N - F)/2
Where N is the force exerted on the large block by the small block.
Making N the subject gives;
4F - 2N = 3N - 3F
5N = 7F
N = 7F/5
We are given F = 6N
Thus;
N = 7(6)/5
N = 8.4 N
A parallel combination of a 1.01 μF capacitor and a 2.93 μF capacitor is connected in series to a 4.75 μF capacitor. This three‑capacitor combination is connected to a 16.3 V battery. Determine the charge on each capacitor.
Answer:
A.16.5x10^-6C
B. 47.5x10^-6C
C.77x10^-6C
Explanation:
Pls see attached file
Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c