Hi there!
We can use work and energy to solve this problem.
We know that:
Ei = Ef
Ei = Potential energy = mgh
Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²
The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:
I (hollow cylinder) = mr²
I (disk) = 1/2mr²
Calculate the moment of inertias of each.
Since the mass on the base is one-fourth of its side:
x = mass of side
x + x/4 = 15
4x + x = 60
5x = 60
x = 12 kg
end mass = 3 kg
Solve for each moment of inertia:
Side: (12)(0.4²) = 1.92 Kgm²
Bottom: 1/2(3)(0.4²) = 0.24 Kgm²
Side + bottom = 2.16 Kgm²
We can now solve:
mgh = 1/2mv² + 1/2(2.16)v²/r²
(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²
4851 = 14.25v²
v = 18.45 m/s
a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.
Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.
The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
The given parameters;
length of the cubical box, L = 0.368number of moles of the gas, n = 1.98temperature, T = 298 Kmass of the gas, m = 3.35 x 10⁻²⁶ kgThe average kinetic energy of the gas molecules is calculated as follows;
[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]
The average speed of the gas molecules is calculated as follows;
[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]
The time of collision of the gas molecules with the walls of the container is calculated as follows;
[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]
The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;
[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]
Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
Learn more here:https://brainly.com/question/17566463
A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
Visit: https://brainly.com/question/14232630
Which of the vectors in the graph below is the negative of the vector v
A. a
B. d
C. c
D. b
Answer:
Option c.
Explanation:
Swim swim swim swim swim swim swim swim swim swim swim swim.
A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.
The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.
Answer:
Explanation:
Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.
In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.
That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.
I NEED THE ANSWER ASAPP
Answer:
Explanation:
a) The spring force will equal the weight.
b) If up is positive
kx - mg = 0
mg = kx kx = 25 N
c) m = kx/g = 25/10 = 2.5 kg
Which properties make a metal a good material to use for electrial wires
Answer:
Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.
A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
A friend has suggested that you go swimming in a pool having water of temperature 350 K. What would this temperature be on the Fahrenheit scale?
109°F
123°F
170°F
202°F
This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.
The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.
the relation between Kelvin scale and Fahrenheit scale is given by:
(F - 32)/180 = (K - 273)/100
F - 32 = (350 - 273)(9/5)
F = 32 + (350 - 273)(9/5)
F = 170
Hence, this temperature would be 170° F on the Fahrenheit scale.
Learn more about temperature here:
https://brainly.com/question/11464844
#SPJ2
A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.
We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."
Answers:
Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex] Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]
From the question we are told
The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s
A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube
Therefore, the mass flow rate of CCI_4 at point A
= [tex]5.86*10^{-13} kg/s[/tex]
B) From Fick's law
[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]
Then,
[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]
For more information on this visit
https://brainly.com/question/23379286
Please help me as quick as possible!!!!!!! Please, please please!!!!!!
Answer:
23
Explanation:
3x-4
A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?
Hi there!
[tex]\large\boxed{E_{total} = 8.0 \text{ J}}[/tex]
For a mass undergoing SHM, the total energy of the system is given as:
[tex]ME = \frac{1}{2}kA^2[/tex]
Where:
k = Spring constant (N/m)
A = amplitude (m)
There is no quantity of mass in the equation, so the total mechanical energy of the system is NOT impacted by the object's mass.
Thus, the energy of the system will still be 8.0 J.
It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t)=Ccos(ωt)+Ssin(ωt), where C, S, and ω are constants.
A) Using the general equation for x(t) given in the problem introduction, express the initial position of the block xinit in terms of C, S, and ω (Greek letter omega).
b) Find the value of S using the given condition that the initial velocity of the block is zero: v(0)=0.
c)What is the equation x(t) for the block? Express your answer in terms of t, ω, and xinit.
d)Find the equation for the block's position xnew(t) in the new coordinate system.
Express your answer in terms of L, xinit, ω (Greek letter omega), and t.
The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.
x = C cos wt + S sin wt
Where x is the displacement C and S are constants. W the angular velocity and t the time.
A) The initial position of the body occurs when the time is zero, t = 0
We substitute.
x = C cos 0 + S sin 0
[tex]x_{init}[/tex] = C
B) The velocity of the particle is defined.
[tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]
The initial velocity occurred for time zero t = 0
v = - S w
It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.
S = 0
C) The equation for the block remains.
x (t) = [tex]x_{init} \ cos \ wt[/tex]
D) In some cases it is measured with respect to another reference system, the most common are:
For maximum compression it is the zero of the system. The maximum extension is the zero of the system.
In these cases, the change that must be made is
x = [tex]L - x_{min}[/tex] t
we substitute
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
L = [tex]x_{init}[/tex] (1 + cos wt)
In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
Learn more about simple harmonic motion here: brainly.com/question/17315536
A man applies a force of 540 N to the barrow in a direction 75 from the horizontal. He moves the barrow 30 m along the level ground. Calculate the work he does against friction?
The work done by the man against friction is 4,192.86 J.
The given parameters;
force applied, F = 540 Nangle of inclination, θ = 75⁰horizontal distance, x = 30 mThe work done by the man against friction is calculated as follows;
[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]
Thus, the work done by the man against friction is 4,192.86 J.
Learn more here:https://brainly.com/question/19498865
A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first
Answer:
The box arrives first.
Explanation:
Hope this helps!! :))
Learn more at :
https://brainly.com/question/20164917
According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
What is Friction?Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.
There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.
Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.
To learn more about Friction, refer to the link:
https://brainly.com/question/24338873
#SPJ6
For an object spinning around a central point, what will happen if its distance from the center is decreased?
A. Nothing will change.
B. Its acceleration will decrease.
C. Its acceleration will increase.
D. The centripetal force will decrease.
Answer:
Its acceleration will increase.
Explanation:
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity
This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves smaller radii and for lager radii acceleration will be less.
acceleration, a = v²/r
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
help me for a physics project please
Mister Brainly Please Help Me
Write 10 Information's About Sound
-cant travel through space since there's no molecules to travel through
-sound travels 4.3 times faster in water than air
-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules
-different types of sound like audible, inaudible, infrasonic, ultrasonic,
-sounds waves are either longitudinal, mechanical and pressure waves
-sound travels at 767 miles per hour
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
According to above question ~
Current (I) = 4 AmperesTime (t) = 3 seconds Charge (q) = ?Let's find the charge (q) by using formula ~
[tex]I = \dfrac{q}{t} [/tex][tex]4 = \dfrac{q}{3} [/tex][tex]q = 4 \times 3[/tex][tex]q = 12 \: \: coulombs[/tex]Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection? why
Answer:
Explanation:
The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity. The deflection is also inversely proportional to its mass. So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.
A truck moves 60 km West, and then 80 km North, and then
travels in a straight line back to its starting point. The distance
travelled by the truck is ____km and its displacement is _____km
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Distance travelled by the truck is ~
[tex] \boxed{240 \: \: km}[/tex]And it's displacement is ~
[tex] \boxed{0 \: \: km}[/tex][tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]
See the diagram in attachment for reference ~
Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.
The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~
According to Pythagoras theorem ~
hypotenuse² = sum of squares of other two legs
that is ~
[tex]h {}^{2} = 60 {}^{2} + 80 {}^{2} [/tex][tex] {h}^{2} = 3600 + 640 0[/tex][tex]h {}^{2} = 10000[/tex][tex]h = \sqrt{10000} [/tex][tex]h = \sqrt{100 \times 100}{}[/tex][tex]h = 100 \: \: km[/tex]So, the distance between the point A and O is 100 km
Now, The total distance is equal to the distance covered through actual path that is ~
60 km + 80 km + 100 km 240 kmAnd displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.
Lab report on velocity of sound
Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 13 m/s due north. Plane 2 taxies with a speed of 8.5 m/s in a direction 20 ∘ north of west.
Part A
What is the magnitude of the velocity of plane 1 relative to plane 2?
Part B
What is the direction of the velocity of plane 1 relative to plane 2?
Part C
What are the magnitude of the velocity of plane 2 relative to plane 1?
Answer:
Explanation:
Plane 2 is moving north at
8.5sin20 = 2.9 m/s
Plane 2 is moving west at
8.5cos20 = 8.0 m/s
Part A
v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s
Part B
θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E
Part C
13 m/s 52° S of W
relative velocity magnitude is independent of reference frame
A 1-kg mass at the Earth's surface weighs how much
Answer:
the answer is weight=10N
Answer:
[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]
Explanation:
Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.
It is calculated by multiplying the mass by the acceleration due to gravity.
[tex]F_g=mg[/tex]
The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.
m= 1 kg g= 9.8 m/s²Substitute the values into the formula.
[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]F_g= 9.8 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.
[tex]F_g= 9.8 \ N[/tex]
A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.
Convert 6 picoseconds into seconds.
Answer:
6e-12
Explanation:
divide the time value by 1e+12
jshshwjs sbwiwiw910mw s x djjskskekwkq
Answer:
jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb
Contrast the behavior of a water wave that travel by a stone barrier to a sound wave that travels through a door
calvin carter
Explanation:
here the file has everything
The second hand on a clock is 3.00 cm long. What is the speed of the outermost tip of that second hand
In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of
(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s