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
The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.20 × 10-4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.20 × 10-4 m2. Find (a) the force from the left tire, (b) the pressure from the left tire, (c) the force from the right tire, (d) the pressure from the right tire that each tire applies to the ground.
Answer:
Explanation:
Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N
Pressure on the left
PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa
Pressure on the right
PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa
A) The force from the left tire is; FL = 342 N
B) The pressure from the left tire is; PL = 551613 N/m²
C) The force from the right tire is; FR = 342 N
D) The pressure from the right tire is; PR = 371739 N/m²
We see that;
FL and FR are upward forces
W is the downward force.
We know that in equilibrium;
Sum of upward forces = sum of downward forces
Thus;
FL + FR = W
We are given W = 684 N
Since W is at the center, it means that FL = FR. Thus;
FL = FR = 684/2
FL = FR = 342 N
We are given;
Contact area of left tire; AL = 6.2 × 10⁻⁴ m²
Contact area of right tire; AR = 9.2 × 10⁻⁴ m²
Formula for pressure is;
Pressure = Force/Area
Pressure on the left tire;
PL = FL/AL
PL = 342/(6.2 × 10⁻⁴)
PL = 551613 N/m²
Pressure on the t right tire;
PR = FR/AR
PR = 342/(9.2 × 10⁻⁴)
PR = 371739 N/m²
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How much distance does a car travel with a speed of 2m/s in 15 min?
Explanation:
1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2
If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
How much energy is consumed by a 12 W night light left on for 10 hr?
Answer:
Energy consumed is 0.00033 Joules.
Explanation:
the formula of Energy is:
Energy = power/ time.
Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.
Answer:
d. The truck that left later was travelling faster
Explanation:
Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;
They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;
The only variable that can account for this difference is speed;
The one that left later, therefore, must have been going faster.
When 587.9 nm passes through a single slit 0.73 mm wide, it creates a diffraction pattern. (a) What distance away is the wall if the first minimum is 0.86 mm from the central maximum
From Young's single slit experiment, the distance away from the wall will be 1.068 m
Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.
From the question, the following parameters are given:
The wavelength of the light λ = 587.9 nm
The width of the slit a = 0.73 mm
Fringe width X = 0.86 mm
The distance away from the wall D = ?
The fringe width is related to the wavelength of the light source by the equation:
X = Dλ ÷ a
Substitute all the parameters into the formula
0.83 × [tex]10^{-3}[/tex] = 587.9 × [tex]10^{-9}[/tex] D ÷ 0.73 ×
Cross multiply
587.9 × [tex]10^{-9}[/tex] D = 6.278 × [tex]10^{-7}[/tex]
make D the subject of the formula
D = 6.278 × [tex]10^{-7}[/tex] ÷ 587.9 × [tex]10^{-9}[/tex]
D = 1.068 m
Therefore, the distance away from the wall is 1.068 m
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If the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
Given the data in the question;
Wavelength; [tex]\lambda = 587.9nm = 5.879*10^{-7}m[/tex]Width of slit; [tex]a = 0.73mm = 0.00073m[/tex] First minimum; [tex]y = 0.86mm = 0.00086m[/tex]Since its first, order number; [tex]m = 1[/tex]Distance; [tex]L = \ ?[/tex]From Thomas Young's single slit experiment:
[tex]\frac{a*y}{L} = m * \lambda[/tex]
Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.
We substitute our values into the equation
[tex]\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m[/tex]
Therefore, if the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
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Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
In which state of matter are molecules fastest?
A solid such as Sugar molecules or a liquid like in water molecules or gas molecules such as oxygen and nitrogen molecules in air. Since gas molecules have the weakest intermolecular forces than other molecules in the other two states then they will be the fastest.
Answer:
gas
Explanation:
since gas is in the air I think the answer is gas
A football is kicked with an initial velocity of 50.0 m/s, 60° above the horizontal line. Find the following: The time it takes to reach the maximum height; The maximum height reached by the projectile; The time of flight; and The range of projectile.
Answer:
Explanation:
Initial vertical velocity
vy₀ = 50.0sin60 = 43.3 m/s
This initial velocity is reduced to zero by gravity in a time of
t = v/a = 43.3/9.81 = 4.41 s
h(max) = ½gt² = ½(9.81)4.41² = 95.6 m
The ball will return to earth in the same amount of time
t(max) = 2(4.41) = 8.82 s
The horizontal velocity is
vx = 50.0cos60 = 25.0 m/s
d = vt = 25.0(8.82) = 221 m
That 's one heck of a kick! No air resistance of course.
A 25-kg box of books is dropped on the floor from a height of 1.1 m and comes to rest. What impulse did the floor exert on the box
The impulse the floor exert on the box is 116 kgm/s.
The given parameters;
mass of the books, m = 25 kgheight of the books, h = 1.1 mThe final velocity of the box when it dropped to the floor is calculated as follows;
[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 1.1} \\\\v = 4.64 \ m/s[/tex]
The impulse the floor exert on the box is calculated as follows;
the impulse the floor exert on the box is equal to change in momentum of the book
[tex]J = \Delta P\\\\J = \Delta Mv\\\\J = M(v_f - v_0)\\\\J = 25(4.64 - 0)\\\\J = 116 \ kgm/s[/tex]
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PLEASE HELP ON THIS QUESTION
[tex]r = 1.29×10^8\:\text{m}[/tex]
Explanation:
According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is
[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]
where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get
[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]
Taking the square root of the equation above, we get
[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]
Plugging in the values, we get
[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]
[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]
A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is
W = 25 J
Explanation:
Work done on an object is defined as
[tex]W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}[/tex]
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Can someone solve by showing the steps?
This question involves the concepts of work done and the frictional force.
a. Work done by the person is "692.82 N".
b. Work done by the frictional force is "490.5 N".
a.
Work done by the person can be given by the following formula:
[tex]W=FdCos\theta[/tex]
where,
W = work done by the person = ?
F = Force applied by the person = 80 N
d = distance traveled = 10 m
θ = angle between force and motion = 30°
Therefore,
[tex]W=(80\ N)(10\ m)Cos30^o[/tex]
W = 692.82 N
b.
Work done by the frictional force is given by the following formula:
[tex]W_f=fd\\W_f=\mu mgd[/tex]
where,
[tex]W_f[/tex] = work done by the frictional force = ?
μ = coefficient of friction = 0.5
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]
[tex]W_f=490.5\ N[/tex]
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Disk A, with a mass of 2.0 kg and a radius of 40 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 20 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
After the collision, what is magnitude of their common angular velocity (in rev/s)?
Hi there!
For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:
I₁w₁ + I₂w₂ = (I₁ + I₂)wf
We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:
Disk 1: 1/2(2)(0.40²) = .16 kgm²/s
Disk 2: 1/2(2)(0.20²) = .04 kgm²/s
Plug in the values. Let counterclockwise be positive.
.16(-50) + .04(50) = (.16 + .04)wf
Solve:
wf = -30 rev/s
Which statement best describes how light behaves with liquids, gases, and solids?
A. Light is unable to travel through liquids but travels easily through solids and some gases.
B. Light is unable to travel through gases but does travel through liquids and solids.
C. Light travels easily through liquids and gases, as well as through some solids like
glass.
D. Light travels easily through solids but is unable to travel through liquids and gases.
(20 points!)
Answer:
C number is write i think
PLEASE HELP ME WITH THISSSS
Answer:
she will move in the same direction at the same speed forever.
Explanation:
If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady
3. A circular section of copper cable has a resistance of 0.50. What will be the resistance of a
copper cable of the same length but of twice its diameter?
A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).
We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"
Answer:
Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]
From the question we are told
This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
A) Resistivity is given as,
[tex]p = \frac{RA}{l}[/tex]
where,
[tex]R = \frac{V}{I}[/tex]
Therefore,
[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]
B) Conductivity is given as,
[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]
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EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST
Explanation:
F = Icurrent×length×Bfieldstrength×sin(angle field to wire)
in our case
Icurrent = 10 A
length = 0.02km = 20 meters
B = 10^-6 T
angle = 30 degrees.
F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =
= 200 × 10^-6 = 2 × 10^‐4 N
I need the answer fast for c pleaseeee
Answer:
F = - K x for spring (note that that F here is given in grams, F = m g is correct)
K here is 100 g / cm for the spring constant
x = -420 g / 100 g/cm = -4.2 cm
The spring would compress 4.2 cm for a total length of 20 - 4.2 = 15.8
d) to compress the spring 6.8 cm one can see that the load would be 680 g
What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and Mars
The sun emits electromagnetic waves with a power of
4.0 ∗ 10 (26) W.
A cognitive bias is a kind of logical fallacy.
True or false?
Answer:
True
Explanation:
It definetly is because, if we are biased it sets us up to fail on some things.
c. Boat travels north then west
A boat travels 76.0 km due north in 8.0 hours then 56.0 km due west in 5.0 hours.
Determine the direction (as a bearing) of the average velocity (to 1 decimal places) of the boat in the 8 + 5 hour period.
PLEASE HELP!!
Answer:
Explanation:
θ = arctan(56.0/76.0) = 36.4° West of North
average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s
. A load of 250 kg is hung by a crane’s cable. The load is pulled by a horizontal force such that the cable makes a 300 angle to the vertical plane. If the load is in the equilibrium, calculate the magnitude of the tension in the cable.
If the load is in the equilibrium, the magnitude of the tension in the cable is equal to 1,414.5 Newton.
Given the following data:
Mass of load = 250 kgAngle of inclination = 30°Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To calculate the magnitude of the tension in the cable, if the load is in the equilibrium:
First of all, we would determine the tension caused by the horizontal component of the force:
[tex]\sum F_y ; Tcos\theta - mg=0\\\\Tcos\theta = mg\\\\T=\frac{mg}{cos\theta} \\\\T=\frac{250 \times 9.8}{cos30} \\\\T= \frac{2450}{0.8660}[/tex]
T = 2,829.1 Newton
The magnitude of the tension in the cable is given by:
[tex]\sum F_x ; F - Tsin\theta = 0\\\\F = Tsin\theta\\\\F = 2829.1sin30\\\\F = 2829.1 \times 0.5[/tex]
F = 1,414.5 Newton.
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How does a balanced chemical equation demonstrate the Law of Conservation of Mass? it shows that only physical changes follow the Law of Conservation of Mass it shows that only physical changes follow the Law of Conservation of Mass it shows that the properties of the elements stay the same after the reaction it shows that the properties of the elements stay the same after the reaction it shows that all compounds remain bonded after the reaction it shows that all compounds remain bonded after the reaction it shows that no atoms have been gained or lost during the reaction
Answer:
it shows that the properties of the elements stay the same after the reaction
it shows that the properties of the elements stay the same after the reaction
it shows that all compounds remain bonded after the reaction
it shows that all compounds remain bonded after the reaction
it shows that only physical changes follow the Law of Conservation of Mass
it shows that only physical changes follow the Law of Conservation of Mass
it shows that no atoms have been gained or lost during the reaction
it shows that no atoms have been gained or lost during the reaction
what is kinetic friction ? what causes it ? what does it generate ?
Answer:
What is kinetic friction?
Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement.
What causes it?
When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.
What does it generate?
When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire.
Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when it is subjected to tension of 50.0 N.
Hi there!
We can use the following equation to find the frequency of each harmonic:
[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]
n = nth harmonic
L = length of string (m)
T = Tension of string (N)
λ = linear density (kg/m)
Begin by converting the linear mass density to kg:
2.00g /m · 1 kg / 1000g = 0.002 kg/m
Now, we can use the equation to find the first three harmonics.
First harmonic:
[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]
Second harmonic:
[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]
Third harmonic:
[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]
The current in a resistor is 3.0 A, and its power is 60 W. What is the voltage?
Answer:
20 volts
Explanation:
Use the equation [tex]P=VI[/tex]
[tex]60=V(3)[/tex]
[tex]V=20[/tex]
Which statement describes electromagnetic waves with wavelengths grater than 700 nanometers
Answer:
They take the form of heat, I think thats it but I cant see if you put down any answers
Explanation:
Answer:a
Explanation:
they form hear
One of the great challenges of cosmology today is to ---
A
determine the amount of matter in the Universe
B
find intelligent signals emanating from outer space
С
look backward in time to before the Big Bang
D
locate wormholes to help define the structure of the Universe
Answer:
Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.
Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).