Hi there!
With the work-energy theorem for oscillating springs:
ME = KE + PE
[tex]ME = \frac{1}{2}mv^2 + \frac{1}{2}kx^2[/tex]
Where:
m = mass (kg)
v = velocity (m/s)
k = Spring Constant (N/m)
x = displacement from equilibrium (m)
If the object is at the equilibrium position, there is NO potential energy since:
[tex]\frac{1}{2}k(0^2) = 0 J[/tex]
Thus:
[tex]ME = \frac{1}{2}mv^2[/tex]
Plug in the given values:
[tex]ME = \frac{1}{2}(0.50)(1.5^2) = \boxed{0.5625 \text{ J}}[/tex]
Pendulum makes 12 complete swings in 8 seconds, what are its frequency and period on earth
Hi there!
We can begin by finding the period of the pendulum.
[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]
The frequency is simply the reciprocal of the period, so:
[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]
g What is the CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk
Answer:
Explanation:
A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m
Lets call the outer and inner radii R and r respectively
Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables
I = ½m(R² + r²)
then use the parallel axis theorem to shift the position of the axis
I = ½m(R² + r²) + md²
where d is the distance of the shift. In this case d = R
I = ½m(R² + r²) + mR²
I = m(1.5R² + 0.5r²)
If we select a mass of say 20 grams
I = 0.020(1.5(0.060²) + 0.5(0.0075²))
I = 0.0001085625 kg•m²
Five ramps lead from the ground to the second floor of a workshop, as sketched below. All five ramps have the same height; ramps B, C, D and E have the same length; ramp A is longer than the other four. You need to push a heavy cart up to the second floor and you may choose any one of the five ramps.Assuming no frictional forces on the cart, which ramp would require you to do the least work?
The mechanical advantage of ramp A is greater than others and it will require the least force to move the load to greater distance.
Let the height of the ramp = hLet the length of ramp B, C, D and E = LLet the length of the ramp A = 2LThe mechanical advantage of the ramp is calculated as follows;
[tex]M.A = \frac{L}{h}[/tex]
The mechanical advantage of the ramp B, C, D and E is calculated as;
[tex]M.A = \frac{L}{h} \\\\[/tex]
The mechanical advantage of the ramp A is calculated as follows;
[tex]M.A = \frac{2L}{h} \\\\M.A = 2(\frac{L}{h} )[/tex]
Since the length of the ramp A is greater than other ramps, the mechanical advantage will be greater and it will require the least force to move the load to greater distance.
Learn more about mechanical advantage of ramps here: https://brainly.com/question/20367895
calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
Clothes stick together when you pull them out of the dryer because
clothing is a conductor.
clothing is an inductor.
they are not charged.
of static electricity.
What causes an astigmatism?
A. damaged lens
B. retina not focusing the image
C. cornea being wavy or not spherical
D. sclera not refracting light properly
Answer:
c) cornea being wavy or not spherical
Which region of electromagnetic spectrum will provide photons of the least energy
Answer:
Explanation:
Radio waves
Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.44 m/s2. Determine the orbital period of the satellite.
Explanation:
The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as
[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]
where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.
Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write
[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]
[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]
where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is
[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]
[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]
Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,
[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]
or
[tex]4\pi^2r = gT^2[/tex]
Solving for T, we get
[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]
We can further simplify the above expression into
[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]
Plugging in the values for r and g, we get
[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]
Please help me with this problem
Answer:
Summertime
Explanation:
the sun never sets south of the Antarctic circle in the summertime.
MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is
1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X
Answer:
ma*XsinB
option 2 is correct
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
PLEASE HELP I DONT GET THISS
Answer:
I feel like its the second one but I'm not completely sure..
Explanation:
A car was traveling at 25 m/s when it slammed on the brakes and came to a complete stop in 3 seconds. What is the cars INITIAL/FINAL VELOCITY?
Answer:
Explanation:
Initial velocity 25 m/s
final velocity 0 m/s
The ratio of the two is undefined as dividing by zero is wonky.
At the molecular level, as the kinetic energy increases, what happens to the temperature?
decreases
increases
stays the same
Answer: increases
Explanation:
Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.
2 W' is the symbol of a) antimony b) gold c) polonium d) tungsten
Answer:
D. Tungsten
Explanation:
W - Wolfram
What happens to the iron in the coilgun if the electricity in the coil was turned on
Elevations on the tongue are called
sulci
taste buds
papillae
gyri
Answer:
Papillae is correct
Explanation:
hope it helps you
Answer:
Papillae is the correct answer of this question
Objects 1 and 2 attract each other with a gravitational force of 178 units. If the mass of object 1 is one-fourth the original value AND the mass of object 2 is tripled AND the distance separating objects 1 and 2 is halved, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
now, several numbers change.
Fgravitynew = G*((1/4)*mass1*3*mass2)/(1/2 * D)² =
= G*((3/4)*mass1*mass2)/(D²/4) =
= (3/4)* (G*(mass1*mass2)/D²) *4 =
= 4*(3/4)* (G*(mass1*mass2)/D²) =
= 3* (G*(mass1*mass2)/D²) = 3* Fgravity
the new gravitational force will be 3×178 = 534 units.
A rollercoaster car passes the hill which is 5.5m above the ground at speed 9.3m/s, and rolls over the second hill which is 2.5m above the ground, and heads toward the third hill which is 4.0 m higher than the first one. If the track is frictionless,
a. What maximum height will the car climb on the third hill? [h max = 9.9m, so car will climb the entire 9.5m hill]
b. Will the speed of the car on top of the hill 3 be lower or higher than its speed on the top of the hill one? [lower]
c. Calculate the speed of the car when it is 1m lower than the top of the third hill. [5.3m/s]
Would somebody kindly go over the questions :D
Answer:
Explanation:
Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.
Let m be the mass of the car and ground level is the origin.
on the 5.5 m hill, total energy is
E = PE + KE
E = mgh + ½mv²
E = m(9.8)(5.5) + ½m(9.3)² = 97m J
a) The maximum height will occur when the total energy is all potential energy.
E = mgh
h = E/mg
h = 97m/m(9.8) = 9.9 m
As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.
b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.
c) KE = E - PE
KE = 97m - m(9.8)(9.5 - 1.0)
KE = 97m = 83.3m
KE = 13.7m = ½mv²
v² = √(2(13.7)
v = 5.2345...
v = 5.2 m/s
A spherical ball of lead (density 11.3 g/cm 3) is placed in a tub of mercury (density 13.6 g/cm 3). Which answer best describes the result
The lead ball will float with about 17% of its volume above the surface of the mercury.
We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.
Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.
Learn more: https://brainly.com/question/12108425
Missing parts;
A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?
A.The lead ball will float with about 83% of its volume above the surface of the mercury.
B.The lead ball will float with about 17% of its volume above the surface of the mercury.
C.The lead ball will float with its top exactly even with the surface of the mercury.
D.The lead will sink to the bottom of the mercury.
E.none of the above
Seven friends equally split a restaurant bill that
comes to $93.17. How much does each person pay?
Answer:
$13.31
Explanation:
We know that the bill comes to $93.17 and that 7 people will split the bill equally
We can just use the equation
bill = $93.17/7
bill = $13.31
Find the time it takes for an object dropped from a building and reaches a final velocity of 20 m/s downward?
I need the formula
Answer:
Explanation:
v = at
t = v/a
t = 20 m/s / 9.8 m/s²
t = 2.0408163...
t = 2.0 s
A golfer hits golf ball, imparting to it an initial speed of 53 m/s directed 27° above the horizontal. Assuming that the
mass of the ball is 41g and the club and the ball are in contact for 1.0 ms. Find
b) the direction of the average force exerted on the ball by the club:
Answer:
Explanation:
An impulse results in a change of momentum
FΔt = mΔv
F = mΔv/Δt
F = 0.041(53 - 0) / 0.001
F = 2,173 N 27° above horizontal
give with an example a cause where the velocity of an object is zero but its acceleration is not zero .
Answer:
At the highest point when you toss a ball into the air.
Explanation:
At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.
Hi there!
An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.
When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).
Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.
Give an example of intense aerobics activity. Prompt must be accurate.
Answer:
Explanation:
An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.
An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration?
I also need the complete Formula (Nothing left out)
Answer:
Explanation:
s = s₀ + v₀t + ½at²
ASSUMING the acceleration is in the direction of initial motion.
s = 0 + 3.3(10) = ½(3.7)(10²)
s = 218 m
this is electricity in physics please help
Explanation:
a. (i) When the variable resistor is set at zero, the only resistance in the circuit is due to the lamp. So the current flowing through the circuit is
[tex]I = \dfrac{V}{R} = \dfrac{220\:\text{V}}{440\:Ω} = 0.5\:\text{A}[/tex]
(ii) The power output P of the lamp is given by
[tex]P = I^2R = (0.5\:\text{A})^2(440\:Ω) = 110\:\text{W}[/tex]
b. (i) The variable resistor is in a series connection to the lamp so when its value is set to its maximum value of 660 Ω, the total resistance of the circuit is simply the sum of the two resistances:
[tex]R_T = R_{vr} + R_L = 660\:Ω + 440\:Ω = 1100\:Ω[/tex]
Therefore, the current through the circuit is
[tex]I = \dfrac{V}{R_T} = \dfrac{220\:\text{V}}{1100\:Ω} = 0.20\:\text{A}[/tex]
(ii) Using the result in Part (ii), we can solve for the potential difference across the lamp as follows:
[tex]V_L = IR_L = (0.20\:\text{A})(440\:Ω) = 88\:\text{V}[/tex]
(iii) The power output of the lamp is
[tex]P = I^2R_L = (0.20\:\text{A})^2(440\:Ω) = 17.6\:\text{W}[/tex]
(iv) The rate at which electrical energy is supplied, i.e., the power output of the circuit is equal to the square of the current multiplied by the total resistance of the circuit:
[tex]P = I^2R_T = (0.20\:\text{A})^2(1100\:Ω) = 44\:\text{W}[/tex]
Describe the concept of energy quanta of EM radiation which was explained by Planck.
Answer:
Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.
Explanation:
Answer:
Energy does not occur in continuous amounts but in discrete amounts described by:
E = N h ∨ where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)
Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)
Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
Learn more about NEO here:
https://brainly.com/question/24157038
CAN SOMEONE PLEASE HELP ME
Answer:
she will eventually slow down and come to a stop