Answer: kooi kooi
how r u and thanks for the free points :)
A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose that the man moves quickly to the center of the canoe and sits down there. How far does the canoe move in the water
Answer:
the canoe moved 1.2234 m in the water
Explanation:
Given that;
A man whose mass = 69 kg
A woman whose mass = 52 kg
at opposite ends of a canoe 5 m long, whose mass is 20 kg
now let;
x1 = position of the man
x2 = position of canoe
x3 = position of the woman
Now,
Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3
= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20
= (0 + 260 + 50 ) / ( 141 )
= 310 / 141
= 2.19858 m
Centre of mass is 2.19858 m
Now, New center of mass will be;
52 × 2.5 / ( 69 + 52 + 20 )
= 130 / 141
= 0.9219858 m { away from the man }
To get how far, the canoe moved;
⇒ 2.5 + 0.9219858 - 2.19858
= 1.2234 m
Therefore, the canoe moved 1.2234 m in the water
The canoe move in the water will be 1.2234 m. The canoe move depending on the center of mass of the bodies.
What is the center of mass?The center of mass of an item or set of objects is a place specified relative to it. It's the average location of all the system's components, weighted by their mass.
The centroid is the location of the center of mass for simple rigid objects with homogeneous density. The center of mass of a uniform disc shape, for example, would be at its center.
The given data in the problem is;
m₁ is the mass of man = 69 kg
m₂ is the mass of woman whose= 52 kg
m₃ is the mass of canoe = 20 kg
L is the length of canoe = 5 m
x₁ is the position of the man
x₂ is the position of the canoe
x₃ is the position of the woman
The center of mass will be;
[tex]\rm COM= \frac{[m_1x_1 + m_2x_2 + m_3x_3]}{ m1 + m2 + m3} \\\\ \rm COM= \frac{[69 \times 0 +52 \times 5 + 20 \times 2.5]}{ 69+ 52 + 20} \\\\ \rm COM= (0 + 260 + 50 ) / ( 141 )\\\\ \rm COM = 310 / 141 \\\\ \rm COM = 2.19858 m[/tex]
The new center of mass is;
[tex]\rm COM= \frac{52 \times 2.5 }{69+52+20} \\\\ \rm COM=\frac{130}{141} \\\\ \rm COM= 0.9219 m[/tex]
The distance to find how the canoe moved will be found by;
[tex]\rm x= 2.5+0.9219-2.1985 = 1.2234[/tex]
Hence the canoe move in the water will be 1.2234 m.
To learn more about the center of mass refer to the link;
https://brainly.com/question/8662931
The body mass of Asaiah is 70 Kg.
(a) What is his weight on Earth?
(b) If he goes to the Moon,
(i) What is his mass?
(ii) What is his weight?
Answer:
A I hope its not wrong I hope u do good
The certain forest moon travels in an approximately circular orbit of radius
14,441,566 m with a period of 6 days 10 hr, around its gas giant exoplanet host. Calculate the mass of the exoplanet from this
information. (Units: kilograms)
Answer:
Mass of Exoplanet = 0.58 kg
Explanation:
First, we will calculate the speed of the forest moon:
[tex]speed = v = \frac{Circumference}{time}\\[/tex]
circumference = 2πr = 2π(14441566 m) = 90739035.3 m
time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)
time = 554400 s
Therefore,
[tex]v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s[/tex]
We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:
[tex]Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}[/tex]
Mass of Exoplanet = 0.58 kg
Mention two ways in which the effects of friction can be minimised
Answer:
Polishing the rough surface.
Oiling or lubricating with graphite or grease the moving parts of a machine.
Providing all bearings or wheels between the moving parts of a machine or vehicles reduce friction and allow smooth movement as rolling friction is less than sliding friction.
Explanation:
Two identical positive charges exert a repulsive force of 6.4x10^-9 N when separated by a distance of 3.8x10^10 m. Calculate the charge of each.
Answer:F = kq2/d2 ⇒
q = √(Fd2/k)
q = d √(F/k)
d = 3.8 x 10-10 m
F = 6.4 x 10-9 N
Look up k in your physics book in appropriate units, and plug in the numbers. You should get q in coulombs.
Explanation:
The value of each charge will be 1.64 ×10⁻⁴. The concept of the columb force is used.
What is the charge?When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
Charges that are similar repel each other, whereas charges that are dissimilar attract each other. The term "neutral" refers to an item that has no net charge.
[tex]\rm F =K \frac{q_1Q_2}{d^2} \\\\ F = K\frac{q^2}{d^2} \\\\ q = \sqrt{\frac{dF}{k} } \\\\ q = \sqrt{\frac{3.8 \times 10^{10}\times 6.4 \times 10^{-9}}{9 \times 10^9} } \\\\ q=1.64 \times 10-4[/tex]
Hence the value of each charge will be 1.64 ×10⁻⁴. The concept of the columb force is used.
To learn more about the charge refer to the link;
https://brainly.com/question/24391667
Which device or set of devices is contained in a mobile telephone?
Un teléfono móvil o teléfono celular es un dispositivo portátil que puede hacer o recibir llamadas a través de una portadora de radiofrecuencia, mientras el usuario se está moviendo dentro de un área de servicio telefónco. El enlace de radiofrecuencia establece una conexión con los sistemas de conmutación de un operador de telefonía móvil, que proporciona acceso a la red telefónica pública conmutada (PSTN). La mayoría de los servicios de telefonía móvil modernos utilizan una arquitectura de red celular, y por lo tanto los teléfonos móviles son, con frecuencia, llamados celulares, especialmente en Hispanoamérica. En España, se utiliza más el término móvil.
What causes coastal erosion
La erosión costera es la pérdida o desplazamiento de tierra, o la remoción a largo plazo de sedimentos y rocas a lo largo de la costa debido a la acción de olas, corrientes, mareas, agua impulsada por el viento, hielo transportado por el agua u otros impactos de marejadas ciclónicas.
Brainliest!!! Write: Forces are all around us. Imagine that your teacher has asked you to teach a lesson to your peers about forces. Explain, in detail, how you experience forces in your everyday life.
Are the orbits of the planets on the same plane?
Yes, more or less
No, they're all over the place
Answer: yes
Explanation:
A child makes a ramp to push his toy dump truck up to his sandbox. If he uses 5 newtons of force to push the 12-newton truck up the ramp, what is the mechanical advantage of his ramp?
Answer:
m = 2.4
Explanation:
Given that,
Input force, [tex]F_i=5\ N[/tex]
Output force, [tex]F_o=12\ N[/tex]
We need to find the mechanical advantage of the ramp. The ratio of output force to the input force is equal to mechanical advantage. So,
[tex]m=\dfrac{12}{5}\\\\m=2.4[/tex]
So, the mechanical advantage of his ramp is 2.4.
It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 100m above a raging river and attach a 35-m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 43N/m . Assume that your mass is 79kg . After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation? h=
Answer:
h = 47 m
Explanation:
First, we will calculate the force on the cord due to the weight:
[tex]Force = F = Weight\\F = mg\\F = (79\ kg)(9.81\ m/s^2)\\F = 775\ N[/tex]
Now, we will calculate the elongation by using Hooke's Law:
[tex]F = k \Delta x[/tex]
where,
k = spring constant = 43 N/m
Δx = elongation = ?
Therefore,
[tex]775\ N = (43\ N/m)\Delta x\\\\\Delta x = \frac{775\ N}{43\ N/m}\\\\\Delta x = 18\ m\\[/tex]
So, the final length of the cord will be:
[tex]Final\ Length = Initial\ Length + \Delta x\\Final\ Length = 35\ m + 18\ m\\Final\ Length = 53\ m\\[/tex]
Hence, the height from water (h) can be found using the following formula:
[tex]h = Height\ of\ Bridge - Final\ Length\ of\ cord\\h = 100\ m - 53\ m\\[/tex]
h = 47 m
Two people pull on the wagon each with a constant 20N force. Both people pull to the left. What is the Net Force on the wagon?
Answer:
I dont understand what you are trying to ask
Explanation:
You are comparing the beam waste for two different situations with the goal of using the smallest beam waste possible. A Nd-YAG laser system emits light at 532 nm and the beam is 8 mm in diameter. You also have a Ti-sapphire laser that emits at 855 nm and has a beam diameter of 6 mm. Compare the beam waist for both laser systems using a focusing lens with a focal length of 10 mm. Assume the light fills the lenses in each case
Answer:
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
Explanation:
Nd-YAG laser system : emits at 532 nm , beam diameter = 8 mm
Ti-sapphire laser system : emits at 855 nm , Beam diameter = 6mm
Comparing the beam waist for both lase systems using a focusing lens
Focal length = 10 mm
assumption : light fills lenses in each laser system
Beam waist radius ( W ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex]
β = wavelength , D = diameter illuminated , F = focal length
For
Nd-YAG laser system
β = 532 mm , D = 8 mm
hence ( Wn ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2*532 / π ) ( 10 / 8 ) = 2.117 μm
For
Ti-sapphire laser
β = 855 nm , D = 6 mm
hence ( Wt ) [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2* 855 ) / π ) ( 50 / 6 ) = 4.536 μm
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
The plates of a vacuum-gap parallel plate capacitor have a 100.0 mm2 area, a vacuum gap of 5.00 mm and are connected to a 1.5-volt battery. After the capacitor is charged, the battery is disconnected from the capacitor. After the battery is disconnected, the plates are pulled apart until the vacuum gap is 7.50 mm. a. What are the initial and final energies stored in the capacitor
Answer:
E₀ = 2.0*10⁻¹¹ J = 0.2 pJ
Ef = 3.0*10⁻¹¹ J = 0.3 pJ
Explanation:
The energy stored between the plates of a parallel plate capacitor can be expressed in terms of the capacitance C and the potential difference between plates V as follows:[tex]E = \frac{1}{2} * C * V^{2} (1)[/tex]
When the capacitor is fully charged, the potential difference between plates must be equal to the voltage of the battery, 1.5 V.In a parallel plate capacitor, the value of the capacitance is independent of the applied voltage, and depends only on geometric constants and the dielectric constant of the medium between plates, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} (2)[/tex]
We can find the initial value of C replacing in (2) by the givens below:A = 100.0 mm2d= 5.00 mmε₀ = 8.85*10⁻¹² F/m[tex]C_{o} = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*(10)^{-12} F/m)*(10^{-4} m2)}{5.0*(10)^{-3}m} = 1.77*10^{-13} F (3)[/tex]With this value of C₀, and the value of the initial potential difference between plates (1.5 V), we can find the initial charge on the capacitor, starting from the definition of capacitance:[tex]C =\frac{Q}{V} (4)[/tex]Solving for Q in (4):[tex]Q = C_{o}* V = 1.77*10^{-13} F * 1.5 V = 2.65*10^{-13} C (5)[/tex]Finally, we can find the initial energy stored in the capacitor, replacing (3) and V in (1):[tex]E_{o} = \frac{1}{2} * C_{o} * V_{o} ^{2} = \frac{1}{2} * 1.77*10^{-13}F*(1.5V)^{2} = 0.2 pJ (6)[/tex]
If we pull apart the plates until the vacuum gap is 7.50 mm, we will change the expression of C in (2), decreasing its value due to the expanded gap.Replacing in (2) the new value of the gap (7.50 mm), we can find the new value of C, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*10^{-12}F/m)*10^{-4} m2}{7.5*10^{-3}m} = 1.18*10^{-13} F (7)[/tex] In order to find the final energy stored in the capacitor, we need also the value of the final potential difference between plates.Once disconnected from the battery, the charge on any of the plates must remain the same, due to the principle of conservation of the charge.So, since we have the value of Q from (5) and the new value of C from (7), we can find the new potential difference between plates as follows:[tex]V_{f} = \frac{Q}{C_{f}} = \frac{2.7*10^{-13}C}{1.18*10^{-13}F} = 2.25 V (8)[/tex]With the values of Vf and Cf, we can find the value of the final energy stored in the capacitor, replacing these values in (1):[tex]E_{f} = \frac{1}{2} * C_{f} * V_{f} ^{2} = \frac{1}{2} * 1.18*10^{-13}F*(2.25V)^{2} = 0.3 pJ (9)[/tex]
During an experiment, Ellie records a measurement of 0.0034 m. How would
she write her measurement in scientific notation?
A. 3.4 x 10-3 m
B. 3.4 x 10-4 m
O C. 3.4 x 10-5 m
D. 3.4 x 10-2 m
Answer:
(A) She needs to move the decimal point by 3 places
When a skydiver is 1200 meters in the air they open the parachute. They are moving
cown for 400 seconds before landing. What is the skydiver's average velocity?
48 m/s down
48 m/s up
3 m/s up
3 m/s down
Answer:
3m/s up
Explanation:
Part one: Multiple choices
1) A person sitting in the compartment of moving train is:
a) in the state of rest with respect to surroundings of the compartment,
b) in the state of motion with respect to surroundings of the compartment.
C) in the state of rest with respect to surroundings outside of the compartment
d)all of them
2) The motion of tuning fork prongs on vibration is:
a) Linear motion
b) periodic motion
c) circular motion
d) projectile motion
3) All the following are periodic motion except
a) moving car in straight line
b) Earth's rotation
c) pendulum
d) Swing
4) The rate of change of displacement is:
a) Acceleration
b) force
c) distance
d) velocity
5) When an object moves at negative acceleration in a straight line its:
a) displacement equals zero
b) velocity decrease
C) velocity increase
d) none of them
6) When the object speeds up its acceleration:
a) decreases
b) increases c) it has no acceleration d) All of them
7) The rate of change of velocity is:
a) force
b) variable velocity
c) instantaneous velocity
d) acceleration
8) If a train is moving in a straight line to cover a distance of 600 m in a minute its
velocity is:
a) 600 m/s
b) 60 m/s c) 100 m/s d) 480 m/s
9) The division between total displacement and total time is the:
a) variable velocity b) average velocity c) speed d) Instantaneous velocity
10) The rate of change of displacement at a given instant is called the
a) average velocity
b) instantaneous velocity
C) average velocity
d) instantaneous acceleration
11) A body completes one circular revolution in a roundabout whose diameter
140 m.
Find its displacement,
a) 439.6 m
b) 440 m
c) zero
d) 879.2 m
Answer:
1)
a) in the state of rest with respect to surroundings of the compartment,
2)
b) periodic motion
3)
a) moving car in a straight line
4)
d) velocity
5)
b) velocity decrease
6)
b) increases
7)
d) acceleration
8)
10 m/s
9)
b) average velocity
10)
b) instantaneous velocity
11)
a) 439.6 m
Explanation:
1)
With respect to the inside surrounding the person will be at rest. Because the person is not moving inside the compartment.
2)
The vibration motion follows periodic motion.
3)
The car moving in a straight line is an example of rectilinear motion and its wheels are in rotational motion. They are not in periodic motion.
4)
Definition of velocity.
5)
Acceleration is the rate of change of velocity. So negative acceleration means a decrease in velocity.
6)
Acceleration is the rate of change of velocity. So an increase in velocity means an increase in acceleration.
7)
Definition of acceleration.
8)
[tex]velocity = \frac{Distance}{Time}\\\\velocty = \frac{600\ m}{1\ min}\frac{1\ min}{60\ s}\\\\velocity = 10\ m/s[/tex]
Hence, none of the options is correct. The correct answer is 10 m/s.
9)
Definition of average velocity.
10)
Definition of instantaneous velocity.
11)
[tex]Displacement = Circumference = \pi d\\Displacement = \pi(140\ m)\\Displacement = 439.6\ m[/tex]
In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into castles. You are asked to study the motion of such a projectile for a group of local enthusiast planning a medieval war reenactment. Unfortunately an actual trebuchet had not been built yet, so you decide to first look at the motion of a thrown ball as a model of rocks thrown by a trebuchet. Specifically, you are interested in how the horizontal and the vertical components of the velocity for a thrown object change with time. 1. Make a large rough sketch of the trajectory of the ball after it has been thrown. Draw the ball in at least five different positions; two when the ball is going up, two when it is going down, and one at its maximum height. Label the horizontal and vertical axes of your coordinate system.
2. On the sketch, draw and label the expected acceleration vectors of the ball (relative sizes and directions) for the five different positions. Decompose each acceleration vector into its vertical and horizontal components.
3. On the sketch, draw and label the velocity vectors of the object at the same positions you chose to draw your acceleration vectors. Decomposes each velocity vector into its vertical and horizontal components. Check to see that the changes in the velocity vector are consistent with the acceleration vectors.
4. Looking at the sketch, how does someone expect the ball's horizontal acceleration to change with time? Could you give a possible equation giving the ball's horizontal acceleration as a function of time? Graph this equation. If there are constants in your equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
5. Looking at the sketch, how does someone expect the ball's horizontal velocity to change with time? Is it consistent with the statements about the ball's acceleration from the previous question? Could you give a possible equation for the ball's horizontal velocity as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
6. Could you give a possible equation for the ball's horizontal position as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph? Are any of these constants related to the equations for horizontal velocity or acceleration?
7. Repeat questions 4-6 for the vertical component of the acceleration, velocity, and position. How are the constants for the acceleration, velocity and position equations related?
Answer:
2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t
5) changes the horizontal speed, should change range
7) changes the vertical speed change the maximum height
Explanation:
1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.
2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration
3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression
v_y = v_{oy} - gt
at the point of maximum height, vy = 0 is equal to the maximum height
4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed
In the graph it would be directed to the left, therefore the velocity would be
vₓ = v₀ₓ - ax t
5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.
the equations of motion are
x = v₀ₓ t
y = v_{oy} t - ½ g t²
7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,
the equations of motion are the same.
the radius of earth is about 6.38 x10^3 km. A 7.20 x10^3 N spacecraft travels away from earth. What is the weight of the spacecraft at the following distances from Earth's surface? a) 6.38 x 10^3 km
Answer:
[tex]1796.65\ \text{N}[/tex]
Explanation:
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
w = Weight of spracecraft at the surface = [tex]7.2\times10^3\ \text{N}[/tex]
m = Mass of spracecraft
R = Radius of Earth = [tex]6.38\times10^3\ \text{km}[/tex]
h = Elevation = [tex]6.38\times10^3\ \text{km}[/tex]
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{7.2\times 10^3}{9.81}\\\Rightarrow m=733.94\ \text{kg}[/tex]
From the gravitational law we have
[tex]w'=\dfrac{GMm}{(r+h)^2}\\\Rightarrow w'=\dfrac{6.674\times10^{-11}\times 5.972\times 10^{24}\times 733.94}{(6.38\times10^6+6.38\times10^6)^2}\\\Rightarrow w'=1796.65\ \text{N}[/tex]
The weight of the spacecraft at the given height is [tex]1796.65\ \text{N}[/tex]
Which soil is best for growing most plants?
Answer:
sandy loam
Explanation:
The best soil for most plants to ensure optimum growth is a rich, sandy loam. This soil is an even mixture of all three main types of soil. In most cases, you'll need to amend the soil with compost. Depending on how compact the soil is, you may need to add peat moss and sand.
Please help. It’s probably easy
Help me please I have other ones like this too on my page please help!
A car traveling at 10 m/s slams into a trash can on the side of the road. The trash can
sticks to the front of the car and they move as one object. If the momentum is
conserved, what would happen to the velocity of the car after the collision?
The velocity would decrease
It would remain the same
The velocity would increase
The car comes to rest
Answer:
The velocity would decrease
Explanation:
Let
[tex]m_1[/tex] = Mass of car
[tex]m_2[/tex] = Mass of trash can
[tex]u_1[/tex] = Initial velocity of car = 10 m/s
[tex]u_2[/tex] = Initial velocity of trash can = 0
v = Velocity of combined mass
From the question the momentum balance of the system will be
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{10m_1}{m_1+m_2}[/tex]
The velocity would decrease as now the mass of the car and trash will both be moving as one. The mass would increase resulting in a lower velocity of the combined mass.
Suppose a rocket in space is accelerating at 1.5 m/s2. If, at a later time, the rocket quadruples its thrust (i.e., net propelling force), what is the new acceleration?
Why is the city of Hoboken, NJ (20 minutes from Newark) and other coastal cities in the United suing ExxonMobile?
Answer:
is that youuuuuuu? fine.
An important problem that all electric circuits have is the fact that since current is flowing through the traces connected the components, the circuit is generating a complex magnetic field during operation. You can shield electric fields pretty easily using a Faraday cage, but how do you shield your device from the magnetic field it is generating itself
Answer:
grid or metal sheets or metal foams
Explanation:
Electric and magnetic fields are generated in electrical circuits that often cause interference in the measurements made.
To avoid these interferences, shields are generally used for the electric fields, Farday boxes are used, which are metal grids connected to earth.
For the magnetic field there is also a grid or metal sheets or metal foams,
It should be noted that the gaps in the screening must be smaller than the wavelength of the radiation to be screened.
The work-energy theorem states that the work done on an object is equal to a change in which quantity?
kinetic energy
displacement
potential energy
mass
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
Answer:
a kinetic energy
Explanation:
Two identical R = 6.77 ohm light bulbs are is series circuit
with a 17.8 volt battery. What is the power (watt) of both
glowing bulbs?
Answer:
the power of the two light bulbs is 23.4 W
Explanation:
Given;
the resistance of the two identical light bulbs in series, R₁ and R₂ = 6.77 ohms and 6.77 ohms respectively
battery voltage , V = 17.8 V
The equivalent resistance of the two light bulbs, R = R₁ + R₂
R = 6.77 + 6.77
R = 13.54 ohms
The power of the two light bulbs is calculated as follows;
[tex]P = IV = (\frac{V}{R} )V = \frac{V^2}{R} = \frac{17.8^2}{13.54} = 23.4 \ W[/tex]
Therefore, the power of the two light bulbs is 23.4 W
identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.
Answer:
Explanatation
1 is just walking
2 spinging
3 roatating
4 rolling
5 stomping there feet
Those should be right but if im wrong then just someone eles the brainly
The total kinetic energy of an object depends on two (2) factors. Select those factors from the list below.
Mass
Density
Volume
Velocity