Answer:
The magnitude of the angular acceleration is 8.39 1/(s^2)
Explanation:
The complete question is:
Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. What is the magnitude of the angular acceleration of the salad spinner as it slows down?
The frequency (f) at the beginning is calculated as follows:
f = 20 rotations/5 seconds = 4 1/s
The angular frequency at the beginning (ωi) is calculated as follows:
ωi = 2*π*f = 2*π*4 = 8π 1/s
At the end angular frequency is zero (the spinner stops moving). ωf = 0
The angular displacement in 6 times is:
θ = 6*2π = 12π
The angular acceleration (α) can be obtained from the following equation of rotational motion:
ωf^2 = ωi^2 + 2*α*θ
α = (ωf^2 - ωi^2)/(2*θ)
α = [0 - (8π)^2]/(2*12π)
α = 8.39 1/(s^2)
The angular acceleration of the salad spinner as it slows down can be found using the rotational kinematic equation ω = ωo + αt. A negative acceleration value indicates a slowdown. An example calculation was provided for better understanding.
Explanation:In physics, the concept of angular acceleration comes into play when dealing with rotational motion, such as that of a salad spinner slowing down. Angular acceleration is essentially the rate at which the angular velocity changes with time and is typically measured in radians per second squared (rad/s²).
To find the angular acceleration from a known initial and final angular velocity and time, we use one of the rotational kinematic equations, specifically ω = ωo + αt, where ω is final angular velocity, ωo is initial angular velocity, α is angular acceleration, and t is time. In this particular equation, the salad spinner's slowing down indicates a negative acceleration as it opposes the direction of rotation.
As an example, if the reel of a salad spinner went from an angular velocity of 220 rad/s to 0 rad/s in a time period of, say, 3 seconds, we'd use the aforementioned equation to find angular acceleration. Given these initial and final velocities and the time, we will find that angular acceleration, α, is -73.3 rad/s².
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You are directed to set up an experiment in which you drop, from shoulder height, objects with similar surface areas but different masses, timing how long it takes for each object to hit the floor. Of the following explanations, which best describes your findings?
a. the most dense object hits the ground first
b. the less dense object hits the ground first
c. they will hit the ground at the same time
Answer: c. they will hit the ground at the same time
Explanation:
The volume of both objects is almost the same, so the force of friction will be the same in each one, so we can discard it.
Now, when yo drop an object, the acceleration of the object is always g = 9.8m/s^2 downwards, independent of the mass of the object.
So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.
So the correct option is c
The man jumps from the window of a burning hotel and lands in a safety net that stops him fall in 1 mm. Determine the average force that the net exerts on the man if he enters the net at a speed of 28 m/sm/s. Assume that the man's mass is 64 kgkg.
Final answer:
The average force that the net exerts on the man is approximately -5.02 x 10^7 N. The negative sign indicates that the force is in the opposite direction to the motion.
Explanation:
To determine the average force that the net exerts on the man, you can use the equation:
Force = mass x acceleration
First, calculate the acceleration of the man using the formula:
acceleration = change in velocity / time taken
In this case, the final velocity is 0 m/s, the initial velocity is 28 m/s, and the time taken is the distance the net stops the man divided by his initial velocity. Since the net stops the man within a distance of 1 mm (0.001 m), the time taken is:
time taken = distance / initial velocity = 0.001 m / 28 m/s = 3.571 x 10-5 s
Now, you can calculate the acceleration:
acceleration = (0 - 28) m/s / 3.571 x 10-5 s = -7.838 x 105 m/s2
Next, calculate the force:
force = mass x acceleration = 64 kg x (-7.838 x 105 m/s2)
force = -5.02 x 107 N
The average force that the net exerts on the man is approximately -5.02 x 107 N. The negative sign indicates that the force is in the opposite direction to the motion.
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or out of the gas per cycle?
123 J transfers out of the gas
123 J transfers into the gas
246 J transfers into the gas
0 J (no heat transfers)
Answer:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that
[tex]\Delta U = 0[/tex]
so from 1st law of thermodynamics we have
[tex]dQ = W + \Delta U[/tex]
[tex]dQ = W[/tex]
so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2u = 2 cos ωt, u(0) = 0, u'(0) = 2 (a) Determine the steady state part of the solution of this problem.
Answer:
Therefore the required solution is
[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]
Explanation:
Given vibrating system is
[tex]u''+\frac{1}{4}u'+2u= 2cos \omega t[/tex]
Consider U(t) = A cosωt + B sinωt
Differentiating with respect to t
U'(t)= - A ω sinωt +B ω cos ωt
Again differentiating with respect to t
U''(t) = - A ω² cosωt -B ω² sin ωt
Putting this in given equation
[tex]-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t[/tex]
[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t[/tex]
Equating the coefficient of sinωt and cos ωt
[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2[/tex]
[tex]\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0[/tex].........(1)
and
[tex]\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0[/tex]
[tex]\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0[/tex]........(2)
Solving equation (1) and (2) by cross multiplication method
[tex]\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}[/tex]
[tex]\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]
[tex]\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex] and [tex]B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]
Therefore the required solution is
[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]
A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Answer:
(a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
Explanation:
Given that,
Spring stiffness = 205 N/m
Mass = 0.6 kg
Compression of spring = 13 cm
Initial speed = 3 m/s
(a). We need to calculate the maximum stretch during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2[/tex]
[tex]x_{m}=\sqrt{\dfrac{4.43\times2}{205}}[/tex]
[tex]x_{m}=20.7\ cm[/tex]
(b). Maximum speed comes when stretch is zero.
We need to calculate the maximum speed during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2[/tex]
[tex]v'=\sqrt{\dfrac{4.43\times2}{0.6}}[/tex]
[tex]v'=3.84\ m/s[/tex]
(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.6}{205}}[/tex]
[tex]T=0.33\ sec[/tex]
We need to calculate the energy
Using formula of energy
[tex]E=\dfrac{P}{t}[/tex]
Put the value into the formula
[tex]E=\dfrac{0.02}{0.33}[/tex]
[tex]E=0.060\ Watt[/tex]
Hence, (a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
In a certain material there is a current of 16 A flowing through a surface to the right, and there is an equal amount of positive and negative charge passing through the surface producing the current. How much negative charge passes through the surface?
Answer:
Explanation:
Give that
I=16A
The current is flowing to the right
Equal amount of positive and negative charges.
i.e, total charge is q= Q+Q=2Q. i.e magnitude
Then, the rate of charge that pass though is dq/dt
Give that,
q=it
2Q=it
Let differentiate with respect to t
2dQ/dt=i
Then, dQ/dt=i/2
Since i=16A
Then, dQ/dt=16/2
dQ/dt= 8 A/s. Toward the left
When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the weight lost by the object compare to the weight of the water displaced?
Answer:
Explanation:
- The volume of water displaced by immersing the object is equal the amount of water spilled and caught by Dr. Hewitt.
- The amount of water is proportional to the volume of object of fraction of object immersed in water will lead to the same fraction of water displaced and caught by Dr. Hewitt.
- When the object is immersed the force of Buoyancy acts against the weight and reducing the scale weight.
- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.
explain why a law is accepted as facr, but a theory is not
Answer:
Explanation:
A law and theory are distinct levels in the scientific method. They do not lead to one another.
A law is a description of an observed phenomenon in the natural world. Laws are always true and do not provide explanations as to why they hold true.
A theory is an explanation of an observed phenomenon. It is usually based on experimental evidence and bounded most time in the limits of available data.
A law cannot be deposed. It is a fact and holds true at all times. Theories can be discarded even with new technological advancements that provides a new way of study. This is why it is not a fact.A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) How many electrons are accelerated per pulse? (b) What is the average current for a machine operating at 500 pulses/s? If the electrons are accelerated to an energy of 50 MeV, what are the (c) average power and (d) peak power of the accelerator?
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
[tex]N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1[/tex]
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
[tex]P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}[/tex]
= 1250W
d) Final peak=
P= Ik/e
= [tex]= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W[/tex]
P = 2.5 × 10⁷W
Explanation:
Below is an attachment containing the solution.
When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length of the rod. If a silver rod has a resistance of 1.70 Ω at 21.0°C, what is its resistance when it is heated to 180.0°C? The temperature coefficient for silver is α = 6.1 ✕ 10−3 °C−1, and its coefficient of linear expansion is 18 ✕ 10−6 °C−1. Assume that the rod expands in all three dimensions.
Answer:
[tex]3.34\Omega[/tex]
Explanation:
The resistance of a metal rod is given by
[tex]R=\frac{\rho L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity
L is the length of the rod
A is the cross-sectional area
The resistivity changes with the temperature as:
[tex]\rho(T)=\rho_0 (1+\alpha (T-T_0))[/tex]
where in this case:
[tex]\rho_0[/tex] is the resistivity of silver at [tex]T_0=21.0^{\circ}C[/tex]
[tex]\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}[/tex] is the temperature coefficient for silver
[tex]T=180.0^{\circ}C[/tex] is the current temperature
Substituting,
[tex]\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0[/tex]
The length of the rod changes as
[tex]L(T)=L_0 (1+\alpha_L(T-T_0))[/tex]
where:
[tex]L_0[/tex] is the initial length at [tex]21.0^{\circ}C[/tex]
[tex]\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}[/tex] is the coefficient of linear expansion
Substituting,
[tex]L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0[/tex]
The cross-sectional area of the rod changes as
[tex]A(T)=A_0(1+2\alpha_L(T-T_0))[/tex]
So, substituting,
[tex]A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0[/tex]
Therefore, if the initial resistance at 21.0°C is
[tex]R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega[/tex]
Then the resistance at 180.0°C is:
[tex]R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega[/tex]
A pen contains a spring with a spring constant of 257 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.1 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.1 mm. How much work is done by the spring force to ready the pen for writing
Final answer:
The work done by the spring force to ready the pen for writing is 0.404 J.
Explanation:
To find the work done by the spring force, we can use the formula for work: work = (1/2)k(x²), where k is the spring constant and x is the displacement of the spring from its unstressed length.
In this case, the spring is initially compressed 5.1 mm and then compressed an additional 6.1 mm. So the total displacement is 5.1 mm + 6.1 mm = 11.2 mm.
Substituting the values into the formula, we have work = (1/2)(257 N/m)((11.2 mm / 1000)²) = 0.404 J. The calculated work of 0.404 J represents the energy transferred to or from the spring as it undergoes the specified compressions, providing insight into the mechanical behavior of the system under the influence of the spring force.
A small space telescope at the end of a tether line of length L moves at linear speed v about a central space station. What will be the linear speed of the telescope if the length of the line is changed to x*L ? x = 2.8; v = 2 m/s?
Answer:
v' = 0.714 m/s
Explanation:
Solution:
- Assuming no external torque is acting on the system then the angular momentum is conserved for the system.
- The initial momentum angular Mi and final angular momentum Mf are as follows:
Mi = Mf
m*L*v = m*x*L*v'
Where,
m : mass of the telescope
L : Length of teether line
v: Initial speed
v' : Changed speed.
- Then we have:
L*v = x*L*v'
v' = v / x
v' = 2 / 2.8
v' = 0.714 m/s
Answer:
The answer to the question is
The linear speed of the telescope will be 5.6 m/s if the length of the line is changed to x*L where x = 2.8; and initial velocity v = 2 m/s
Explanation:
Speed = v₁ = ωL = 2 m/s
When the line is changed to x*L where x = 2.8 the linear speed will be
v₂ = 2.8 × L× ω = 2.8× 2 = 5.6 m/s
The linear speed varies with the angular speed following the relation v/r =ω where
ω = angular speed
v = linear speed and
r = radius of the path of travel of the object at the vertex
A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0 ∘ below the positive x axis. A second vector, B⃗ , has a magnitude of 75.0 m and points in a direction 50.0 ∘ above the positive x axis.
a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C
Answer:
Explanation:
Check attachment for solution
The analytical method of the components allows to find the results for the questions about the sum vector are:
a) In the attached we have a scheme of the vectors
b) the modulus is C = 96.3 m
c) The angle is θ = 27.0º
Given parameters
Vector A with modulus A = 40.0 m and an angle of θ₁ = -20º Vector B has a modulus B = 75.0 m and an angle of θ₂ = 50ºTo find
a) Draw the vectors and their sum.
b) The module.
c) the adirection.
The sum of vectors has several methods of resolution:
Graphic. In this case the vectors are drawn and the second is placed on the tip of the first and the resulting vector goes from the origin of the first to the tip of the last. This method is complicated when there are several vectors. Analytical. This method is very precise and is the most used when there are many vectors.
The analytical method consists:
Decompose each vector into its components with respect to a given reference frame, using trigonometry. Find the sum of each component Construct the final vector using trigonometry and the Pythagorean theorem.
In the attached we have a diagram of each vector and its sum. Let's use trigonometry to find the component of each vectors.
Vector A
cos θ₁ = [tex]\frac{A_x}{A}[/tex]
sin θ₁ = [tex]\frac{A_y}{A}[/tex]
Aₓ = A cos θ₁
[tex]A_y[/tex] = A sin θ₁
Aₓ = 40 cos (-20) = 37.59 m
[tex]A_y[/tex]= 40 sin (-20) = -13.68 m
Vector B
cos θ₂ = [tex]\frac{B_x}{B}[/tex]
sin θ₂ = [tex]\frac{B_y}{B}[/tex]
Bₓ = B cos θ₂
[tex]B_y[/tex] = B sin θ₂
Bₓ = 75 cos 50 = 48.21 m
[tex]B_y[/tex] = 75 sin 50 = 57.45 m
we add the components.
Cₓ = Aₓ + Bₓ
[tex]C_y = A_y + B_y[/tex]
Cₓ = 37.59 + 48.20 = 85.8 m
Cy = -13.68 + 57.45 = 43.8 m
b) We use the Pythagorean theorem to find the modulus of the resultant vector.
C² = Cₓ² + [tex]C_y^2[/tex]
C = [tex]\sqrt{85.8^2 + 43.8^2 }[/tex]
C = 96.3 m.
c) We use trigonometry to find the angle of the resultant vector.
tan θ =[tex]\frac{C_y}{C_x}[/tex]
θ = tan⁻¹ [tex]\frac{C_y}{C_x}[/tex]
θ = tan⁻¹ [tex]\frac{43.8}{85.8}[/tex]
θ = 27.0º
In conclusion, using the analytical method of the components we can find the results for the questions about the sum vector are:
a) In the attached we have a scheme of the vectors
b) the modulus is C = 96.3 m
c) The angle is θ = 27.0º
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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes from least kinetic energy to greatest kinetic energy. If two sets have the same amount of kinetic energy, place one on top of the other. 1. mass: m speed: v2. mass: 4 m speed: v3. total mass: 2 m speed: 1/4v4. mass: 4 m : speed: v5. total mass: 4 m speed: 1/2v
mass₃<mass₁=mass₅<mass₂=mass₄
Explanation:
Given data :-
1. mass: m speed: v
2. mass: 4 m speed: v
3. mass: 2 m speed: ¼ v
4. mass: 4 m speed: v
5. mass: 4 m speed: ½ v
We know that Kinetic energy (KE) = ½ mv²
Where m= mass of the body
v=velocity of the body
Substituting the values of respective mass and velocity from the above given data-
KE of Body 1(mass₁) = ½*m*v² = mv²/2
KE of Body 2(mass₂) = ½*4m*v² = 2mv²
KE of Body 3(mass₃) = ½*2m*(1/4v)² = mv²/16
KE of Body 4(mass₄) = ½*4m*v ² = 2mv ²
KE of Body 5(mass₅) = ½*4m*(1/2v)² = mv²/2
The spring of a spring gun has force constant k =400 N/m and negligible mass. The spring is compressed6.00 cm and a ball with mass 0.0300 kg isplaced in the horizontal barrel against the compressed spring. Thespring is then released, and the ball is propelled out the barrelof the gun. The barrel is 6.00 cm long, so the ballleaves the barrel at the same point that it loses contact with thespring. The gun is held so the barrel is horizontal.
Calculate the speed with which the ballleaves the barrel if you can ignore friction.
Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel.
For the situation in part (b), at whatposition along the barrel does the ball have the greatest speed?(In this case, the maximum speed does not occur at the end of thebarrel.)
What is that greatest speed?
a) 6.9 m/s
b) 4.9 m/s
c) After 4.50 cm
d) 5.2 m/s
Explanation:
a)
In this case, there is no resistive force. Therefore, according to the law of conservation of energy, all the initial elastic potential energy stored in the spring when it is compressed is converted into kinetic energy of the ball as it leaves the barrel; so we can write:
[tex]\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]
where:
k = 400 N/m is the spring constant
x = 6.00 cm = 0.06 m is the compression of the spring
m = 0.03 kg is the mass of the ball
v is the velocity of the ball as it leaves the barrel
Solving the equation, we can find the speed of the ball:
[tex]v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(400)(0.06)^2}{0.03}}=6.9 m/s[/tex]
b)
In this case, there is a constant resistive force of
[tex]F_r = 6.00 N[/tex]
acting on the ball as it moves along the barrel.
The work done by this force on the ball is:
[tex]W=-F_rd[/tex]
where
d = 6.00 cm = 0.06 m is the length of the barrel
And where the negative sign is due to the fact that the force is opposite to the motion of the ball
Substituting,
[tex]W=-(6.00)(0.06)=-0.36 J[/tex]
Therefore, part of the initial elastic potential energy stored in the spring has been converted into thermal energy due to the resistive force. So, the final kinetic energy of the ball will be less than before:
[tex]\frac{1}{2}kx^2+W=\frac{1}{2}mv^2[/tex]
And solving for v, we find the new speed:
[tex]v=\sqrt{\frac{kx^2}{m}+\frac{2W}{m}}=4.9 m/s[/tex]
c)
The (forward) force exerted by the spring on the ball is
[tex]F=k(x_0-x)[/tex]
where
k = 400 N/m
x is the distance covered by the ball
[tex]x_0=0.06 m[/tex] is the maximum displacement of the spring
While the (backward) resistive force is instead
[tex]F_r=6.0 N[/tex]
So the net force on the ball is
[tex]F=k(x-x_0)-F_r[/tex]
The term [tex]k(x-x_0)[/tex] prevails at the beginning, so the ball continues accelerating forward, until this term becomes as small as [tex]F_r[/tex]: after that point, the negative resistive force will prevail, so the ball will start delecerating. Therefore, the greatest speed is reached when the net force is zero; so:
[tex]k(x_0-x)-F_r=0\\x=x_0-\frac{F_r}{k}=0.06-\frac{6.00}{400}=0.045 m[/tex]
d)
After the ball has covered a distance of
x = 0.045 m
The work done by the resistive force so far is:
[tex]W=-F_r x =-(6.00)(0.045)=-0.27 J[/tex]
The total elastic potential energy of the spring at the beginning was
[tex]E_e=\frac{1}{2}kx_0^2 = \frac{1}{2}(400)(0.06)^2=0.72 J[/tex]
While the elastic potential energy left now is
[tex]E_e'=\frac{1}{2}k(x_0-x)^2=\frac{1}{2}(400)(0.015)^2=0.045 J[/tex]
So, the kinetic energy now is:
[tex]K=E_e-E_e'+W=0.72-0.045-0.27=0.405 J[/tex]
And by using the equation
[tex]K=\frac{1}{2}mv^2[/tex]
We can find the greatest speed:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.405)}{0.03}}=5.2 m/s[/tex]
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 80.0 m above the glacier at a speed of 120 m/s.How far short of the target should it drop the package?
Explanation:
Below is an attachment containing the solution
A pipe has a length of 1.29 m. Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz.
Answer:
265.9Hz
Explanation:
In an open pipe, both ends of the pipes are opened. The fundamental frequency in an open pipe is expressed as fo = V/2L where;
f is the frequency of the wave
V is the velocity of the wave = 343m/s
L is the length of the pipe = 1.29m
Substituting the value to get the fundamental frequency in the open pipe we have;
Fo = 343/2(1.29)
Fo = 343/2.58
Fo = 132.95Hz
Harmonics are integral multiples of the fundamental frequency e.g 2fo, 3fo, 4fo, 5fo...
The first harmonic in the open pipe will be f1 = 2fo
Since f1 =2(132.95)
f1 = 265.9Hz
The frequency of the first harmonic if the pipe is open at each end is 265.9Hz
Answer:
132.95 Hz.
Explanation:
Given:
v = 343 m/s
L = 1.29 m.
Since the pipe is open at both ends,
L = λ/2
λ = v/f = 2L
= 2 × 1.29
= 2.58 m
f = 343/2.58
= 132.95 Hz.
A ? is a conductor installed on the supply side of a service or separately derived system to ensure the required electrical conductivity between metal parts required to be electrically connected.
Answer:
Supply-side bonding jumper.
Explanation:
A supply-side bonding jumper is a conductor installed on the supply side of a service or separately derived system to ensure the required electrical conductivity between metal parts required to be electrically connected.
The supply side bonding jumper was referred to as the equipment bonding jumper prior to 2011 NEC when its name changed.
The wire runs from the source of the separately derived system to the first disconnecting means.
What is the difference between porosity and permeability
Explanation:
Porosity is the percentage of spaces,(hollow) within a rock or a material that can contain air or fluid. This can be used in geology( study of rocks), soil mechanics, engineering and pharmaceutics. Industrial CT scanning can be used to test for porosity of a substance. While permeability is the ability of water or other fluids to flow through a rock or material that has spaces ( that is, that are porous). This is also applicable in the fields of chemical engineering and geology.
Porosity refers to the amount of void space in a material that can hold water, while permeability measures how well those spaces are connected, affecting water movement. Materials with high permeability have larger, well-connected pores, allowing water to flow easily; materials with low permeability do not. The hydraulic conductivity is a measure of permeability that accounts for both material and fluid properties.
Difference Between Porosity and Permeability
The difference between porosity and permeability concerns the storage and movement of water in subsurface materials like rock or sediment. Porosity refers to the percentage of open space within a material that can potentially hold water, expressed as a ratio of the volume of voids to the total volume of the material. Permeability, on the other hand, is about how well those pores are interconnected, determining the ease with which water can move through the material. Higher permeability indicates the presence of larger, well-connected pores, enabling water to flow with less friction. Conversely, materials with low permeability have fewer, smaller, and poorly connected pores, which restricts water movement.
Understanding both porosity and permeability is crucial when discussing groundwater storage and extraction, as they define an aquifer's ability to store and transmit water. Soil texture plays a significant role in determining these properties. Coarse-grained soils with larger pores tend to have both higher porosity and permeability, whereas fine-grained soils like clay can have high porosity but low permeability due to poorly connected pores.
Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the acceleration of the center of mass of spool B?
Answer:
The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.
Explanation:
From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.
And for the acceleration of both spools, we mention the net force.
The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.
For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.
With the net force and mass, for each spool equal, from
ΣF = ma, we get that a = ΣF/m
Meaning that the acceleration of the identical spools is equal also.
Hope this Helps!
A spring is hung vertically with a 425g mass attached to it. The mass is at rest. If the mass causes the spring to stretch 0.67 m, what is the spring constant?
Answer:
6.22 N/m
Explanation:
From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then
[tex]k=\frac {F}{x}[/tex]
We also know that the force F is equal to mg where m is the mass of an object and g is acceleration due to gravity hence substituting F with mg we get that
[tex]k=\frac {mg}{x}[/tex]
Substituting m with 425 g which is equivalent to 0.425 kg and g with 9.81 then 0.67 for x we get that
[tex]k=\frac {mg}{x}=\frac {0.425\times 9.81}{0.67}=6.222761194 N/m\approx 6.22\ N/m[/tex]
Therefore, the spring constant is approximately 6.22 N/m
A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = +20 mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.6 μT k^k^. What is the charge Q? (μ 0 = 4π × 10-7 T · m/A)
Answer:
The charge on the particle = -0.00075 C = -0.75 mC = -750 μC
Explanation:
The solution to this question is presented in the attached image to this answer.
The Biot Savart's formula for calculating magnetic field due to moving point charge is used in this calculation.
Hope this Helps!!!
A room that has an average ambient sound pressure level of 62 dBA and a maximum sound pressure level lasting more than a minute at 68 dBA must have a public mode signal that is at least ? .
Answer:
Explanation:
A fire alarm notification appliance is an active fire protection component of a fire alarm system. The primary function of the notification appliance is to alert persons at risk.
If want the audible public mode signal to be hear clearly then, we need to have a sound level that is at least 15dB above the average ambient sound level or 5dB above the maximum sound level of at least 1minute
In this case the,
The average ambient sound level is 62dB,
And the maximum sound level is 68dB
Then, the public mode signal should be at least
1. 62dB+ 15dB=77dB
Or
2. 68dB +5dB =73dB.
Then the public mode signal hearing must be at least 77dB.
Uranium is an element that is often used in nuclear power plants. Uranium atoms are very large, and the substance can be dangerous if it is not carefully contained. What is true about all uranium atoms?
Answer:
The answer for this is that they each have the same number of protons.
Explanation:
Plutonium is an element that is also used in nuclear power plants, because of the same amount of protons in it, plutonium is used in nuclear power plants.
The fact about all uranium atoms is that they have the same number of protons that make them very large and can be harmful if not carefully treated.
Answer:
same number of protons
Explanation:
Tech A says motor action occurs through the interaction of the magnetic fields of the field coils and the armature, which causes a rotational force to act on the armature, creating the turning motion. Tech B says two magnetic fields are required for motor action: one in the casing and the other in the rotating armature. Who is correct
Answer:
Both A and B
Explanation:
The interaction of magnetic fields and armature results into a rotational force of the armature hence turning motion. It's important to note that you will always need two magnetic fields in order to experience the force since one magnetic field is at the rotating armature and another at the casing. Considering the arguments of these two technicians, both of them are correct in their arguments.
______ Made geocentric model of the solar system using epicycles
Answer:
Ptolemy made geocentric model of the solar system using epicycles
Explanation:
Ptolemy made geocentric model of the solar system using epicycles.
This model accounted for the apparent motions of the planets in a very direct way, by assuming that each planet moved on a small circle, called an epicycle, which moved on a larger circle, called a deferent.
Therefore, Ptolemy is the answer.
Your roommate is working on his bicycle and has the bike upside down. He spins the 54.0 cm diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the period of the pebble stone?
Answer:
If the diameter is 54cm, then the radius is half that, the radius is:
r = 54cm/2 = 27cm
Then the perimeter of the wheel is p = 2*pi*27cm
this would mean that the pebble travels the distance:
d = 3*2*pi*27cm in one second.
Then the velocity of the pebble is:
speed = 3*2*3.1416*27cm/s = 509cm/s
Now, we know that the pebble does 3 cycles in a second, so does each cycle in 1/3 seconds, so the period would be the time that it needs to do one cycle, that we already find that is equal to 1/3 seconds.
Final answer:
The pebble stuck in the bicycle wheel tread has a speed of 5.091 m/s, and the period of its motion is 0.333 seconds.
Explanation:
To solve for the pebble's speed and the period of its motion, we will perform calculations based on the given wheel diameter and the frequency of the pebble's motion. The wheel's diameter is 54.0 cm, which gives us a radius (r) of 27.0 cm or 0.27 meters.
The speed (v) of the pebble on the tread of the wheel can be found using the formula for the circumference (C) of the wheel and multiplying by the frequency (f) of the pebble's motion:
C = 2πr
v = C × f
Plugging in the values we get:
C = 2 × π × 0.27 m = 1.697 m
f = 3 rev/s (since it goes by three times every second)
v = 1.697 m × 3 rev/s = 5.091 m/s
The period (T) is the inverse of frequency:
T = 1 / f
T = 1 / 3 rev/s = 0.333 s
Therefore, the pebble's speed is 5.091 meters per second, and the period of the pebble stone is 0.333 seconds.
What is the amount of electric field passing through a surface called? A. Electric flux.B. Gauss’s law.C. Electricity.D. Charge surface density.E. None of the above.
Answer:
A. Electric flux
Explanation:
Electric flux is the rate of flow of the electric field through a given area (see ). Electric flux is proportional to the number of electric field lines going through a virtual surface.
Electric flux has SI units of volt metres (V m), or, equivalently, newton metres squared per coulomb (N m2 C−1). Thus, the SI base units of electric flux are kg·m3·s−3·A−1.
If an athlete expends 3480. kJ/h, how long does she have to play to work off 1.00 lb of body fat? Note that the nutritional calorie (Calorie) is equivalent to 1 kcal, and one pound of body fat is equivalent to about 4.10 × 103 Calories.
Answer:
The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
Explanation:
1 lb of body fat = 4.1 k cal
1 k cal = 4.184 Kilo joule
1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule
Athlete expends 3480 Kilo joule in one hour
⇒ Time required to expand 3480 Kilo joule = 60 minute
⇒ Time required to expand 1 Kilo joule = [tex]\frac{60}{3480}[/tex] [tex]\frac{min}{KJ}[/tex]
⇒ Time required to expand 17.1544 Kilo joule = [tex]\frac{60}{3480}[/tex] × 17.1544 = 0.296 min
Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
Studying climate on other planets is important to understanding climate on earth because
Answer:
yes it is important.
Explanation:
It is important to study the climate of other planet to understand the climate of the earth because it is useful to know which things are harmful for our earth and which are important for our earth climate comparing with the other planets.