Answer:
D. an ellipse with the Sun at one focus
Explanation:
Based on Kepler's work, the orbit of a planet around the Sun is an ellipse with the Sun at one focus.
What is Kepler's law of planetary motion?The sun is at one of the foci of the planets' elliptical orbits around the sun, according to Kepler's first law. Perihelion, the name for the point at which a planet is closest to the sun, and aphelion, the name for the point at which a planet is farthest from the sunThe radius vector extended from the sun to the planet sweeps out equal areas in equal amounts of time, according to Kepler's second law.The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.Hence according to Kepler's first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one focus.
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How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
A.
0.0658 s
B.
15.2 s
C.
3.11 s
D.
23700 s
Answer: B. 15.2s
Explanation: 600/39.5 = about 15.2
The moon is a projectile that ...
A. Is big enough to keep its momentum around the earth
B. Is far enough to avoid the gravity of the earth
C. Has enough velocity to fall around the earth
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Which statement(s) is/are TRUE about photons?
SELECT ALL THAT APPLY
a Photons have greater mass than electrons.
b Photons are particles of light.
с
Photons carry a very specific quantum of
energy.
d Photons have zero mass.
Answer:
Not D, every substance in existence has mass and is made from it in any given form.
B and C
Explanation:
The statements which are true about photons are:
Choice B: Photons are particles of light.Choice C: Photons carry a very specific quantum of energy.Choice D: Photons have zero mass.Discussion:
The photon is a type of elementary particle. It is the quantum of the electromagnetic field including electromagnetic radiation such as light and radio waves, and the force carrier for the electromagnetic force.
Additionally, Photons are massless, so they always move at the speed of light in vacuum, 3 × 10⁸ m/s.
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Whenever I visit one of the big membership-driven bulk wholesalers in town, without fail I end up using a junk cart with bum wheels that not only drag, but pull to the left. If a cart requires 20 lb of force forward to overcome the wheel’s drag, but there are also around 3 lb of force pulling it to the left, with what force and in what direction do I need to push in order to make the cart go forward?
Required:
Measure your angle from the forward direction and specify whether it’s to the left or right.
Answer: Force of magnitude 20.22lbf to the right at angle 8.5°
Explanation: The image named "Untitled" below shows the forces acting on the cart.
The second image, "Untitled2", demonstrate the vector diagram of the forces acting on the kart, in which:
f is force forward
f(left) is pull to the left
f(R) is the resulting force necessary for the kart to move
The drag force is not draw on the vector diagram because its value is already part of the force forward.
As we can see, the diagram forms a right triangle with f(R) as hypotenuse.
Using Pytagorean Theorem:
[tex][f(R)]^{2}=f^{2}+[f(left)]^{2}[/tex]
[tex][f(R)]^{2}=20^{2}+3^{2}[/tex]
[tex][f(R)]^{2}=409[/tex]
[tex]f(R)=\sqrt{409}[/tex]
f(R) = 20.22
Force forward equals 20.22lbf to the right.
The angle formed form the forward direction and the resulting force is
[tex]cos\theta=\frac{20}{20.22}[/tex]
cosθ = 0.989
θ = cos⁻¹(0.989)
θ = 8.5°
The forward force is to the right at angle 8.5° and magnitude 20.22lbf.
Can someone please help ASAP :((
Answer:
Motion map #3
Explanation:
because with velocity negative the arrows have to be the same size in the reverse direction.
The mass of a fully loaded Boeing 747 is about 4,082,331.33 kg. If it is cruising eastward at a
velocity of 253 m/s, what is its momentum?
How do astronomers determine the surface temperature of stars? *
Answer:
To the extent that Stellar spectra look like blackbodies, the temperature of a star can also be measured amazingly accurately by recording the brightness in two different filters. To get a stellar temperature: Measure the brightness of a star through two filters and compare the ratio of red to blue light.
Explanation:
Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.
Answer:
137.2J
Explanation:
Ep= mgh
Given,
m=2kgh=7mand we know, g = 9.81 N/kg
Ep= 2 × 9.81 × 7
Ep= 137.2J
Do electromagnetic waves occur in a continuous spectrum based on their wavelength?
Answer: Yes
Explanation: Because Electromagnetic waves form a continuous spectrum and all types of electromagnetic wave travel at the same velocity through a vacuum (space) or air. The waves that form the electromagnetic spectrum are grouped in terms of their wavelength and their frequency.
It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?
(A + B) / 30 s
(B – A) / 30 s
30 s / (A - B)
30 s / (B – A)
Answer:
Direction from A to B divided by time(30s)
Explanation:
What is a characteristic common to sound waves and light waves
Answer:
Both light and sound can be described in terms of wave forms with physical characteristics like amplitude, wavelength, and timbre. Wavelength and frequency are inversely related so that longer waves have lower frequencies, and shorter waves have higher frequencies
The ability of both to cause interference is the resemblance between sound waves and light waves.
What are sound waves and light waves?While sound waves are mechanical, light waves are electromagnetic. Unlike sound waves, which are longitudinal, light waves are transverse. Light waves can go across a vacuum. Since sound waves need a physical medium to go through, they cannot move in a vacuum.
Both sound and light waves adhere to the rule of reflection, which states that the angle of incidence and angle of reflection are equal. Therefore, choice 2 is the right one. While light waves are transverse, sound waves are longitudinal.
In that, they both have an amplitude, a wavelength, a period, and a frequency, light, and sound waves are comparable to one another.
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A horse slows from a velocity of 9.5 m/s to 5.5 m/s over a distance of 32 m. Find time.
Answer:
8
Explanation:
Finding time= t=change in velocity or distance/acceleration
so 32/9.5-5.5
32/4
8
Why are some substances dissolve both polar and nonpolar substances?
Answer:
Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.
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Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.
Amphiphilic molecules typically have a hydrophilic (water-attracting) or polar end and a hydrophobic (water-repelling) or nonpolar end. This dual nature allows them to interact with and dissolve in both polar and nonpolar solvents. Let's explore the reasons behind this ability:
Polar interactions: The hydrophilic end of the amphiphilic molecule can form hydrogen bonds and electrostatic interactions with polar solvents (like water) due to the presence of charged or partially charged atoms (such as oxygen or nitrogen). This makes them soluble in polar substances.
Nonpolar interactions: The hydrophobic end of the amphiphilic molecule is typically a long hydrocarbon chain or a nonpolar group. These nonpolar regions can interact with other nonpolar substances (like oils, fats, or hydrophobic portions of biomolecules) through van der Waals forces and London dispersion forces.
Emulsification: The presence of both polar and nonpolar parts in an amphiphilic molecule allows it to act as an emulsifier. It can stabilize the interface between polar and nonpolar substances, preventing them from separating. For example, in the formation of an oil-in-water or water-in-oil emulsion, amphiphilic molecules form a barrier between the two immiscible substances, keeping them dispersed and forming a stable mixture.
Examples of amphiphilic molecules include surfactants (detergents), phospholipids (essential components of cell membranes), and lipoproteins (transporters of lipids in the bloodstream). These substances play vital roles in various biological processes and industrial applications due to their ability to interact with both polar and nonpolar substances.
Hence, Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.
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A motorcycle traveling south reaches a speed of 38.0 m/s. It then begins uniform negative acceleration and comes to rest after 7.359s. How far did the motorcycle travel in this time?
Answer: 30.641
Explanation: If you get the ave which is 38.0 then the vi which is 7.359
38.0-7.359=30.641/0.0 which is 30.641
(Use 1.72×10-8 Ωm for the resistivity of copper.) Tries 0/20 For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 15.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current? Tries 0/20 What would be the voltage between the ends of the wire in the above problem? Tries 0/20 What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.) Tries 0/20 What is the drift velocity of the electrons when the wire is carrying the maximum allowable current? (The density of electrons in copper is 8.47×1028 m-3.)
Complete Question
A number 14 copper wire has a diameter of 1.628 mm. Calculate the resistance of a 37.0 m long piece of such wire.
(Use 1.72×10-8 ?m for the resistivity of copper.)
For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 15.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current?
What would be the voltage between the ends of the wire in the above problem?
What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.)
What is the drift velocity of the electrons when the wire is carrying the maximum allowable current?
(The density of electrons in copper is 8.47×1028 m-3.)
Answer:
a
[tex]R =0.3057 \ \Omega [/tex]
b
[tex]P = 68.78 \ W [/tex]
c
[tex]V = 4.5855 \ V[/tex]
d
[tex]J = 7.205 *10^{6}\ A/m^2 [/tex]
e
[tex]v_d = 0.0005316 \ m/s[/tex]
Explanation:
From the question we are told that
The resistivity of copper is [tex]\rho = 1.72* 10^{-8} \ \Omega \cdot m[/tex]
The current limit is [tex]I = 15 \ A[/tex]
The diameter is [tex]d= 1.628 mm= 0.001628 \ m[/tex]
The length of the wire is [tex]l = 37.0 m[/tex]
The density of electron is [tex]n = 8.47*10^{28} m^{-3}[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.001628}{2}[/tex]
=> [tex]r =0.000814 \ m [/tex]
Generally the resistance is mathematically represented as
[tex]R = \frac{\rho * l }{A}[/tex]
Here A is the area which is mathematically represented as
[tex]A = 3.142 * (0.00081)^2[/tex]
=> [tex]A = 2.082 *10^{-6}\ m^2 [/tex]
=> [tex]R = \frac{1.72* 10^{-8} * 37.0 }{2.082 *10^{-6}}[/tex]
=> [tex]R =0.3057 \ \Omega [/tex]
Generally the power dissipated is mathematically represented as
[tex]P = I^2 R[/tex]
=> [tex]P = (15)^2 * 0.3057 [/tex]
=> [tex]P = 68.78 \ W [/tex]
Generally the voltage is mathematically represented as
[tex]V = I * R[/tex]
=> [tex]V = 15 * 0.3057[/tex]
=> [tex]V = 4.5855 \ V[/tex]
Generally the charge density is mathematically represented as
[tex]J = \frac{I}{A}[/tex]
=> [tex]J = \frac{15}{2.082 *10^{-6}}[/tex]
=> [tex]J = 7.205 *10^{6}\ A/m^2 [/tex]
Generally the drift velocity is mathematically represented as
[tex]v_d = \frac{I}{ n * A * e}[/tex]
Here e is the charge on an electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]v_d = \frac{15}{ 8.47*10^{28} * 2.082 *10^{-6} * 1.60 *10^{-19} }[/tex]
=> [tex]v_d = \frac{15}{ 8.47*10^{28} * 2.082 *10^{-6} * 1.60 *10^{-19} }[/tex]
=> [tex]v_d = 0.0005316 \ m/s[/tex]
A book is sitting on a desk. What best describes the normal force acting on the book?
the upward force the desk pushes on the book
the upward force the book pulls on the Earth
the downward force the book pushes on the desk
the downward force the Earth pulls on the book
A book is sitting on a desk is due to the Gravitational force of Earth Gravity so D. the downward force the Earth pulls on the book.
What is the gravitational force of an object?The force of attraction between all masses that is objects in the universe, especially the attraction of the Earth's mass for bodies near its surface.
How gravity is formed?Einstein said that the shape of spacetime is responsible for the rise of force that we experience as gravity. A concentration of mass, such as the Earth or sun, bends space around it like a rock bends the flow of a river.
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To increase the current in a circuit, which can be increased?
voltage
resistance
interference
ohms
Answer:
voltage
Explanation:
the current in a circuit, the voltage can be increased.
What is ohm's law?According to Ohm's law, when all other physical parameters, including temperature, are held constant, the voltage across a conductor is directly proportional to the current flowing through it.
From ohm's law, we can state that voltage is directly proportional to current.
Looking at the following formula:
I = V/R
Where V is voltage,
I is current, and
R is resistance
In the above equation, Current and Voltage are in a direct relationship such that if I is increased, V is also increased, and vice versa.
Therefore, To increase the current, the voltage should be increased.
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if a 100-n net force acts upon a 50g car, what will the acceleration of the car be
Answer:
2m/s
Explanation:
The formula for acceleration can be found by [ a = f/m ]
We are given the force [ 100N ] and the mass [ 50g ].
We can use these values to solve for the acceleration.
a = 100/50
a = 2m/s
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Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10g≈10 m/s2 for the downward acceleration due to gravity
Let [tex]u[/tex] be the initial velocity of the soccer ball at an angle of inclination of [tex]\theta_0[/tex] with the positive x-axis.
Given that:
[tex]\theta_0=45^{\circ}[/tex]
The horizontal distance covered by the projectile=20 m
Time of flight, [tex]t_f=2[/tex] seconds
Acceleration due to gravity, [tex]g= 10 m/s^2[/tex] downward.
As "north" and "up" as the positive x ‑ and y ‑directions, respectively.
So, [tex]g= -10 m/s^2[/tex]
As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.
The x-component of the initial velocity, [tex]u_x=u\cos\theta_0[/tex].
The horizontal distance covered by the projectile [tex]= u_x\times t_f[/tex]
[tex]\Rightarrow u_x\times t_f=20[/tex]
[tex]\Rightarrow u_x\times 2=20[/tex]
[tex]\Rightarrow u_x=10[/tex] m/s
So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).
Now, [tex]u\cos(45^{\circ})=10[/tex] [as [tex]u_x=u\cos\theta_0[/tex]]
[tex]\Rightarrow u=10\sqrt{2}[/tex] m/s.
The vertical component of the initial velocity,
[tex]u_y= u\sin\theta_0[/tex]
[tex]\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})[/tex]
[tex]\Rightarrow u_y=10[/tex] m/s
Let v be the vertical component of the velocity at any time instant t.
From the equation of motion,
[tex]v=u+at[/tex]
where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.
In this case, we have [tex]u=u_y, a= -10 m/s^2[/tex].
So at any time instant, t.
[tex]v=u_y+(-10)t[/tex]
[tex]\Rightarrow v=10-10t[/tex]
The vertical component of the velocity, v, is the function of time and related as [tex]v=10-10t[/tex].
This is a linear equation.
At 2 second, the vertical component of the velocity
v=10-10x2=-10 m/s.
The graph has been shown in figure (ii).
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
Answer:
the equipotential lines or surface are perpendicular to the electric field
Explanation:
In this exercise you are asked to explain the relationship between the direction of the equipotential lines and the electric field.
Electric field lines go from a positive charge to a negative charge, the specific shape of these lines depends on the charge distribution, let's look at two specific cases:
* in the case of point charge the lines are radial
* In the case of electric plates, they are lines perpendicular to the plates.
Equipotential surfaces are related to the elective field by
Ex = - dV / dx
Ey = -dV / dy
Ez = -dV / dz
E = Exi ^ + Ey j ^ + Ezk ^
remember electric potential is a scalar and the electric field is a vector quantity, therefore the equipotential lines or surface are perpendicular to the electric field. Specifically for
* point charge the equipotential surfaces are concentric with the charge
* For electrical plates equipotential lines are parallel to the plates.
A car had a constant acceleration of 6m/s? north for 2 seconds. What was thecar's change in velocity if it traveled in a straight line?
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A volleyball player jumps to hit a ball horizontally at 7.0 m/s straight on. If the
height at which the ball was hit is 3.0 m tall, how far did the ball go horizontally
before it hit the ground?
5.5 m
3.6 m
O 4.3 m
4.2 m
Answer:
5.5 is the correct answer
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A tiger leaps horizontally froma 7.5 m high rock with a speed of 3.0 m/s. How far from the base of the rock will she land?
Answer:
The tiger will land at 3.71 meters from the base of the rock
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
The tiger leaps from a height of h=7.5 m with a speed of v=3 m/s. Substituting the values into the formula:
[tex]\displaystyle d=3\cdot\sqrt{\frac {2\cdot 7.5}{9.8}}=3.71\ m[/tex]
The tiger will land at 3.71 meters from the base of the rock
If you have ever baked a cake or bread, you know that the ingredients that combine to make it taste different from the baked food. Why do you think that is?
Answer:
Well we all know that bread and cake taste different. We do this on purpose.
Explanation:
When we make cake we add plenty of sugar to make it the delectable dessert we all know and love while when making bread we focus more on the corn starch and yeast's.
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Answer:
0.75 meters per second; north
Explanation:
speed (velocity)= distance/time
15 meters= distance
20 seconds= time
15/20= 0.75 meters per second; north
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Two parallel plates having charges of equal magnitude but opposite sign are separated by 29.0 cm. Each plate has a surface charge density of 32.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density. kN/C (b) Determine the potential difference between the plates. V (c) Determine the kinetic energy of the proton when it reaches the negative plate. J (d) Determine the speed of the proton just before it strikes the negative plate. km/s (e) Determine the acceleration of the proton. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2 towards the negative plate (f) Determine the force on the proton. N towards the negative plate (g) From the force, find the magnitude of the electric field. kN/C (h) How does your value of the electric field compare with that found in part (a)
Answer:
(a) E = 3.6 x 10³ N/C = 3.6 KN/C
(b) ΔV = 1044 Volts
(c) K.E = 1.67 x 10⁻¹⁶ J
(d) Vf = 4.47 x 10⁵ m/s
(e) a = 3.45 x 10¹¹ m/s²
(f) F = 5.76 x 10⁻¹⁶ N
(g) E = 3.6 x 10³ N/C = 3.6 KN/C
(h) Both values are same in part (h) and (a)
Explanation:
(a)
Electric field between oppositely charged plates is given as follows:
E = σ/ε₀
where,
E = Electric Field Intensity = ?
σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²
ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²
Therefore,
E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)
E = 3.6 x 10³ N/C = 3.6 KN/C
(b)
E = ΔV/r
ΔV = Er
where,
r = distance between plates = 29 cm = 0.29 m
ΔV = Potential Difference = ?
ΔV = (3.6 x 10³)(0.29)
V = 1044 Volts
(c)
Kinetic Energy of Proton = Work done on Proton
K.E = F r
but, F = E q
K.E = E q r
where,
q = charge on proton = 1.6 x 10⁻¹⁹ C
Therefore,
K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)
K.E = 1.67 x 10⁻¹⁶ J
(d)
K.E = (1/2)m(Vf² - Vi²)
where,
m = mass of proton = 1.67 x 10⁻²⁷ kg
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]
Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)
Vf = √(20 x 10¹⁰ m²/s²)
Vf = 4.47 x 10⁵ m/s
(e)
2as = Vf² - Vi²
2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²
a = (20 x 10¹⁰ m²/s²)/0.58 m
a = 3.45 x 10¹¹ m/s²
(f)
F = ma
F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)
F = 5.76 x 10⁻¹⁶ N
(g)
E = F/q
E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)
E = 3.6 x 10³ N/C = 3.6 KN/C
(h)
Both values are same in part (h) and (a)
How much work is done when 25 N of force is used to push a crate 3.0 m
Answer:
The answer is 75 JExplanation:
The work done by an object can be found by using the formula
work done = force × distanceFrom the question
force = 25 N
distance = 3 m
We have
work done = 25 × 3
We have the final answer as
75 JHope this helps you
In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. Calculate the magnitude and direction of the net horizontal force on the rope.
Answer:
74 N to the right
Explanation:
the forces are going in opposite horizontal directions, meaning that they are directly opposing each other. this means that you can subtract the force applied in the direction that is greater from the direction that is less to get the net force for the greater direction
this means 182 N - 108 N = 74 N to the right
In a game of tug of war the magnitude and direction of the net horizontal force on the rope is 74 N.
What is force?A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. The magnitude and direction of the net horizontal force on the rope is,
F = 182 - 108
F = 74 N
The magnitude and direction of the net horizontal force on the rope is 74 N.
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3. An 18.0 N force pulls a cart against a 15.0 N frictional force. The speed of the cart increases 1.0 m/s every 5.0 s. What is the cart's mass?
Answer:
The total force on the cart is:
F = F(applied) - F(friction) = (18 N) - (15 N) = 3.0 N
The acceleration of the cart is:
a = (change in velocity)/(time) = (1.0 m/s)/(5.0 s) = 0.20 m/s^2
Using F = ma, the mass of the cart is:
m = F/a = (3.0 N)/(0.20 m/s^2) = 15 kg
Order the circuits according to their power usage, from highest to lowest. All batteries are the same voltage, and all light bulbs have the same resistance. A battery connected to three light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the third bulb. The third bulb connects to the negative battery terminal. Circuit A A battery connected to two light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit B A battery connected to two light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the negative battery terminal. Circuit C A battery connected to three light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit D Highest power usage Lowest power usage
Answer:
P_a = 3P₀ , P_b = 2 P₀ , P_c = 3 P₀
P_c> P_b> P_a
Explanation:
To determine which circuit consumes more energy than is given by the expression
P = V I
V = I R
I = V / R
P = V² / R
As all circuits have the same battery, the value of the resistance to which the battery is connected determines the consumption
Circuit A
In this circuit the three bulbs are in series so the total resistance is
R_total = 3 R
the power dissipated is
P_a = V² / 3R
if we call
P₀ = V² / R
we substitute
P_a = P₀/3
Circuit B
Two bulbs are connected in series
R_total = 2 R
power is
P_b = V2 / 2R
P_b = P₀/2
Circuit C
The 3 bulbs are connected, but in parallel, the resistance is
1 / R_totak = 1 / R + 1 / R + 1 / R
R_total = R / 3
P _c = V2 3 / R
P_c = 3 Po
By reviewing these results, we can sort the circuits
P_c> P_b> P_a