The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.
So, E = E'
1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'
where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.
Substituting the values of the variables into the equation, we have
1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'
1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2
mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2
mR²ω²/4 + 1/2mv² = mgR/2
R²ω²/4 = gR/2 + 1/2v²
R²ω²/4 = (gR + v²)/2
ω² = 2(gR + v²)/R²
ω² = √[2(gR + v²)/R²]
ω = √[2(gR + v²)]/R
Since angular momentum L = Iω, the rolling disk's initial angular momentum is
L = 1/2mR² ×√[2(gR + v²)]/R
L = mR√[2(gR + v²)]/2
the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
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You spit a wad of paper horizontally
from a height of 1.8 m. The wad
leaves your mouth with a velocity of
6 m/s. How long does it take the
wad to hit the ground?
Goal statement write a positive statement that includes your desired outcome and your imagine preferred future after development
A goal statement may involve how to develop physical skills in a given outdoor activity (e.g., soccer) in order to reach my maximum potential. It involves a statement of purpose.
A goal statement is a statement of purpose that describes long-term professional and/or motivational objectives.
A goal statement involves writing a statement about how to reach personal objectives, i.e., how to accomplish a career, job title, sports skills, educational training, etc.
An excellent statement of purpose should ideally be clear, specific, acceptable, time-bound and realistic.
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The 6 strings on a guitar all have about the same length and are stretched with about the same tension. The highest string vibrates with a frequency that is 4 times that of the lowest string. 1)If the strings are made of the same material, how would you expect the diameters of the lowest and highest strings to compare
Answer:
not sure i need points
Explanation:
A solar collector for a hot water system absorbs solar radiation at the rate of 660 W/m^2
. Its collecting area is 3.8m^2 . Cold water at 15degrees enters the collector. What volume of
water per minute at an output temperature of 60 degrees can this collector deliver?
This question involves the concepts of the law of conservation of energy and specific heat capacity.
This collector can deliver "7.96 x 10⁻⁴ m³/min" of water at an output temperature of 60°C.
According to the law of conservation of energy:
Solar Energy = Energy Required to raise the temperature of the water
Solar Power = Energy Required to raise the temperature of the water
[tex]IA=\frac{mC\Delta T}{t}\\\\\frac{m}{t}=\frac{IA}{C\Delta}[/tex]
where,
[tex]\frac{m}{t}[/tex] = mass flow rate = ?
I = solar radiation = 660 W/m²
A = Area = 3.8 m²
ΔT = change in temperature = 60°C - 15°C = 45°C
C = specific heat capacity = 4200 J/kg.°C
Therefore,
[tex]\frac{m}{t}=\frac{(660\ W/m^2)(3.8\ m^2)}{(4200\ J/kg.^oC)(45^oC)}\\\\\frac{m}{t}=(0.0133\ kg/s)(\frac{60\ s}{1\ min})\\\\\frac{m}{t}=0.796\ kg/min[/tex]
Now, the volume flow rate will be:
[tex]\frac{V}{t}=\frac{(\frac{m}{t})}{(density\ of\ water)}=\frac{(0.796\ kg/min)}{(1000\ kg/m^3)}\\\\\frac{V}{t}=7.96\ x\ 10^{-4}\ m^3/min[/tex]
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The attached picture explains the law of conservation of energy.
Please help me as quick as possible!!!!!!! Please, please please!!!!!!
Answer:
23
Explanation:
3x-4
Peregrine falcons, which can dive at 200 mph (90 m/s), grab prey birds from the air. The impact usually kills the prey. Suppose a 480 g falcon diving at 75 m/s strikes a 240 g pigeon, grabbing it in her talons. We can assume that the slow-flying pigeon is stationary. The collision between the birds lasts 15 ms.
Answer:
What is the average force of collision?
Explanation:
The velocity of the combined mass after impact is found by conservation of momentum
0.480(75) + 0.240(0) = (0.480 + 0.240)v
v = 50 m/s
An impulse results in a change of momentum
FΔt = mΔv
F = mΔv/Δt
for the pigeon
F = 0.240(50 - 0)/0.015
F = 800 N
for the falcon
F = 0.480(50 - 75)/0.015
F = -800 N
The final speed(v) of the Peregrine falcons and pigeon is 50 m/s
The objective of this question is to determine the final speed(v) of the Peregrine falcons and pigeon
From the information given:
the mass of the falcon m_f = 480 gthe speed of the falcon v_f = 75 m/sthe mass of the pigeon m_p = 240 gthe collision time = 15 ms = 0.015 sAccording to the conservation of momentum, we can say that the totality of momentum before and after the collision is the same. As such;
[tex]\mathbf{m_fv_f = (m_f+m_p) v}[/tex]
[tex]\mathbf{480 g \times 75 m/s = (480 + 240) \ g \times v}[/tex]
[tex]\mathbf{v = \dfrac{480 g \times 75 m/s}{(480 + 240) \ g}}[/tex]
[tex]\mathbf{v = \dfrac{36000 \ m/s}{ 720}}[/tex]
v = 50 m/s
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NO LINKS
According to "Patterns of Change," select the ways that rocks are changed naturally over time. Choose three answers. O A. water O B. location O C. animals D. wind O E. people O F temperature
Answer: A , D & F
Explanation:
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s
The energy in the system is given by the initial potential energy at the point 1.
The linear velocity at point 3, is approximately 33.59 m/s.
Reasons:
The parameters are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
At point 2, we have;
Change in potential energy = Kinetic energy
Which gives;
(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²
Which gives;
v ≈ 22.37 m/s
At point 3, the rotational kinetic energy remains constant while the
translational kinetic energy increases as follows;
K.E. at point 3 = Initial kinetic energy + Change in potential energy
Which gives;
K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8
[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]
v₃ ≈ √(1128.15) ≈ 33.59
The linear velocity at point 3, v₃ ≈ 33.59 m/s
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The probable question parameters as obtained from a similar question online are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
Determine the unbalanced force necessary to accelerate a 2.60 kg object at a rate of 14.0 m/s².
Answer:
Explanation:
F = ma
F = 2.60(14.0)
F = 36.4 N
An object will begin moving from rest when acted upon by which forces?
A. Forces that are slightly less than the force of friction
B. Forces that result in a net force of zero
C. Forces that are equal and act in opposite directions
D. Forces that are greater in one direction than in any other direction
Answer:
D
Explanation:
Process of elimnination + it's the only one that makes sense
An object will begin moving from rest when acted upon by forces that are greater in one direction than in any other direction. Hence, Option (D) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
When forces that are greater in one direction than in any other direction, resultant will be unbalanced forces. Unbalanced forces are those acting on a body when the net force acting on the body is greater than zero. The body alters its state of motion when unbalanced forces act on it.
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The element which does not show variable valency a) AI b)Fe c) Cu d) Hg
Answer:
None of these elements.
Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is changed to four times the original value (i.e., quadrupled), then the new gravitational force will be ______ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant throughout the universe.
D is the distance between the 2 objects.
the distance is now quadrupled.
Fgravitynew = G*(mass1*mass2)/(4D)² =
= G*(mass1*mass2)/(16D²) =
= (G*(mass1*mass2)/D²) / 16 = Fgravity/16
the new gravitational force will be 179/16 = 11.1875 units
Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that the red car has ____ mass as the blue car.
Answer:
equal
Explanation:
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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All circuits include
a battery, wires, and a switch.
an energy source, a resistor, and a battery.
a battery, a light bulb, and a switch.
an energy source, a load, and wires.
Answer:
a battery, wires, and a switch.
Explanation:
All circuits include?
Four small 0.600-kg spheres, each of which you can regard as a point mass, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane at point O.
Answer:
.192 kg x m^2
Explanation:
I= mass of a times radius of a squared + mass of b times radius of b squared +...
I= .6 kg x .4m^2 + .6 kg x .4m^2
= .192 kg x m^2
Hope this helps :)
Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.
Which two quantities will the students multiply together before and after the collision?
A. mass and velocity
B. distance and time
C. mass and acceleration
D. velocity and time
This question involves the concepts of the law of conservation of momentum, velocity, and mass.
The two quantities, the students should multiply before and after the collision are "A. mass and velocity".
According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.
To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.
This situation is illustrated in the attached picture.
So, according to the law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.
======================================Proving Conservation of Momentum Quick Check - 5/5NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!
=======================================1. Mass & velocity
2. The total momentum after the collision is the same as the total momentum before the collision.
3. 0.54 kg⋅m/s
4. The system has external forces, such as friction and air resistance, acting on it.
5. 3.0 m/s
After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 88.50 kg
per meter of length, and the tension in the cable was =11650 N
. The crane was rated for a maximum load of 500 kg
. If =5.580 m
, =0.522 m
, =1.350 m
, and ℎ=2.070 m
, what was the magnitude of L
(the load on the crane) before the collapse? The acceleration due to gravity is =9.810 m/s2
The magnitude of the load L on the crane before the collapse is 3211.81 N
To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.
It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.
From the given information;
The tension in the cable = 11650 NThe length (d) = 5.580 mThe mass of the horizontal steel beam (M) = 88.50 kg/m (d)= 88.50 kg/m × 5.580 m= 493.83 kgDistance (s) = 0.522 mx = 1.350 m and h = 2.070 mAcceleration due to gravity = 9.81 m/s²From the question;
the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.
[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]
[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]
[tex]\mathbf{\theta =22.26^0}[/tex]
Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.
∴
[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]
By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:
[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]
[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]
[tex]\mathbf{W_L = 3211.81 \ N }[/tex]
Therefore, we can conclude that the magnitude of the load is 3211.81 N
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is a fuel cell a primary or secondary cell
Answer:
Enrol in our 50 studyscore masterclass. Click here! Fuel cells are a type of primary cell in that they are not recharged, However, they are unique because they never run out, if the reactants are constantly supplied.
A 20-N falling apple encounters a 4-N of air resistance. The magnitude of the net force on the apple is?
A. 0n
B.4n
C.16n
D. 20n
E. None of the above
Answer:
C
Explanation:
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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.
A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.
Answer:
A) -31.2 rad/s
B) 28.1 rad/s^2
Explanation:
What happens to the iron in the coilgun if the electricity in the coil was turned on
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation:
you are standing ata known distance from the statue of liberty describe how you could determine its height using only a meter stick
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty with angle of elevation, then, add the two heights.
If you are standing at a known distance from the statue of liberty, a meter stick can be used to measure your known distance away from the statue of liberty.
To determine its height using only a meter stick, the angle at which you look at the peak of statue of liberty must be measured or known. The height of the statue of liberty can be calculated if you know the angle of elevation at which you look at the peak of the statue, and the availability of the meter stick.
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty. That is,
Tan Ф = opposite / adjacent
Tan Ф = H/d
H = d x TanФ
Where
H = calculated height from the eyes level and above
Ф = angle of elevation
d = known distance away from the statue
Let h = Your measured height of your body to the eyes level
Then,
The height of the statue of liberty = H + h
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Jane plans to fly from Binghamton, New York, to Springfield, Massachusetts, about 280 km due east of Binghamton. She heads due east at 280 km/h for one hour but finds herself at Keene, which is 294 km from Binghamton in a direction 17.8 degrees north of due east. What was the wind velocity
Answer:
90m/h N
Explanation:
we consider janes velocity to the positive X axis so Vj=280km/s
and the final position a dot in on the graph r=294km, we find the x and y components:
294*cos(17.8)=280 X+ confirming our theory
294*sin(17.8)=90 Y+ or N and that's the winds' velocity.
An object moving at a constant velocity of 5.4 m/s travels for 12 s. How far will it move during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
we know that
s=vt
given
v=5.4 m/s
t=12 s
s=5.4 m/s*12 s=64.8m
Explanation:
Hope this helps:)
A 0,9 -kg object attached to the end of a string swings in a vertical circle (radius = 75 cm). At the top of the circle the speed of the object is 6,5 m/s. What is the magnitude of the tension in the string at this position?
Please help me with this problem
Answer:
Summertime
Explanation:
the sun never sets south of the Antarctic circle in the summertime.
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]