A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?
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]
The linear distance traveled by a wheel of radius 50cm after 99 complete revolutions is?
1)99m
2)210m
3)311
4)433
Answer:
3) 311 m
Explanation:
Circumference = 2πR = π m/rev
99 rev(π m/rev) = 99π m or about 311 meters
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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The oscillation of the 2.0-kg mass on a spring is described by x = 3.0 cos (4.0 t) where x is in centimeters and t is in seconds. What is the force constant k of the spring?
The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the block, m = 2 kg,
The oscillation of spring, x = 3 cos 4t,
Calculate the omega by comparing the standard equation given below,
[tex]x = A cos \omega t[/tex]
ω = 4
Calculate the spring constant by the formula given below,
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
4² = k / 2
k = 32 N / m
Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
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What is the relationship between balancing equations and the law of conservation of matter
What is sixth state of matter?
A train slows its speed from 52 kilometers per hour to 46 kilometers per hour in 0.04 hour. What is the acceleration o the train during this time?
Answer: here you go i have to put 20 letters in so just ignore this and look at the link.
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
A 6.5 N ball is thrown with an initial velocity of 20 m/s at a 35° angle from a height of 1.5 m, what is the velocity if it is caught at 1.5 m?
Answer:
20 m/s at -35°
Explanation:
Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.
Suppose a ball is thrown vertically upward (positive direction) from an initial height LaTeX: h_0 with initial velocity LaTeX: v_0. Find the position function LaTeX: s(t) of the ball after LaTeX: t seconds assuming the gravitational acceleration LaTeX: g is a positive constant pointing downward (negative direction).
After time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The given parameters;
initial velocity of the ball, = [tex]v_0[/tex]initial position of the ball, = [tex]h_0[/tex]acceleration due to gravity, = gThe position function of the ball after time t, is calculated as follows;
[tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.
Thus, after time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
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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]
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A 1300 watt hair blow dyer is designed to operate on 120 Volts. How much current does the dryer require
Answer:
10.83 Amperes
Explanation:
if A ⇒ current
W = VA
1300 = 120 x A
1300 / 120 = A
10.83 = A
A small, free-to-rotate magnet is placed in a strong magnetic field. In what orientation will it come to rest
Answer:
South-North
Explanation:
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
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
help :”)
a skydiver jumps out of a plane and falls for 45 seconds before deploying his parachute. how far did he fall?
Answer:
200 feet
Explanation:
identify the following prefixes:
1) Di-
2) Tetra-
3) Deca-
4) Hepta-
Explanation:
Di -. 2
Tetra. -3
deca. -. 10
Hepta. -- 7
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)
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
plz help me on this question thank you
Answer:
D
Explanation:
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.
science thanks sa points
Answer: Are these free point?
Explanation:
How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves away from a stationary sound source
Answer:
If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.
At which type of boundary is new oceanic crust created?
A. a convergent plate boundary
B. a divergent plate boundary
C. a subduction plate boundary
D. a transform plate boundary
Answer:
c.
Explanation:
If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.
When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….
Answer:
a divergent plate boundary
How are a series and parallel car different?
What is discordant characteristic ?
[tex] \: \: \: \: [/tex]
being at variance; disagreeingor incongruous: discordant opinions. disagreeable to the ear; dissonant; harsh.hope it helps[tex] \: \: \: \: \: [/tex]
dissimilar with respect to one or more particular characters
difference between speed and velocity
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
Explanation:
A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
Question 2 options:
0.5 m/s2
13.5 m/s2
6 m/s2
2 m/s2
The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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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
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, C represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]
The steepness of the slope is therefore;
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]
Where;
[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m
[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]
[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]
[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]
The maximum steepness of the slope where the truck can be parked is 54.55 %.
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