Answer:
a) v = 34.5 mph, b) v = 50 mph
Explanation:
The average speed is defined.
v = (x₂ -x₁) / (t₂ -t₁)
If the movement is with uniform speed we can also obtain the average speed, averaging the speeds with a time weight factor, if the time is equal for each speed
v = (v₁ + v₂) / 2
a) For the outward journey
v = (69.0 +0) / 2
v = 34.5 mph
b) For the return trip
v = (69.0 + 31.0) / 2
v = 50 mph
If you put two 60 W bulbs in series across a 120 V outlet, how much power would each consume if its resistance were constant?
Answer:15 watts
Explanation:
To get the current
Power P=iV
(Where P is power of bulb 60watts, i is current, and V is the voltage 120V
i=P/V
i = 60/120
i = 1/2A
But
V=iR where R is the resiance
Therefore R=V/i
=120÷1/2
=120×2
=240ohms
For series connection, each bulb draws the same current since,
Since the two bulbs are 60-watt bulbs, they will have the same resistance, therefore the voltage across each bulb is the same and equals halve of the applied voltage, 120/2 = 60 volts.
Taken the resistance 240 ohm is constant across the series, we can roughly estimate the current flow and calculate power dissipation in the series connection.
We have 120 V/(240 + 240 ) ohms = 1/4 A.
The power dissipated in each bulb is (1/4) × 60 = 15 watts.
When two 60 W bulbs are connected in series across a 120 V outlet, each bulb consumes 30 W of power because they share the same current, and the voltage is divided equally among them due to their series connection.
When two 60 W bulbs are connected in series across a 120 V outlet, they share the same current because they are in a series circuit. In a series circuit, the current is constant throughout, and the voltage is divided among the components. Since power is related to both voltage and current, we can use the formula P = VI to find out how much power each bulb consumes.
Total Resistance (R_total): In a series circuit, the total resistance is the sum of the individual resistances. Let's assume the resistance of each bulb is R.
Since P = VI, we can rearrange the formula to find current (I):
I = P / V
The power of each bulb is 60 W, and the voltage is 120 V, so the current passing through the circuit is:
I = 60 W / 120 V = 0.5 A
Voltage Across Each Bulb: In a series circuit, the voltage is divided among the components. Each bulb receives a portion of the total voltage. So, the voltage across each bulb is 60 V.
Power Consumed by Each Bulb: Now, we can use the formula P = VI to find the power consumed by each bulb. We know the voltage (60 V) and the current (0.5 A) passing through each bulb:
P = (0.5 A) * (60 V) = 30 W
Each bulb consumes 30 W of power when connected in series across the 120 V outlet.
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A solid conducting sphere with radius 0.75 m carries a net charge of 0.13 nC. What is the magnitude of the electric field at a point located 0.50 from the sphere's center 0.25 beneath the sphere's surface)?
Answer:
Explanation:
given that
Radius =0.75m
Cnet=0.13nC
a. Electric field inside the sphere located 0.5m from the center of the sphere.
The electric field located inside the sphere is zero.
b. The electric field located 0.25m beneath the sphere.
Since the radius is 0.75m
Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m
Then
E=kq/r2
K=9e9Nm2/C2
q=0.13e-9C
r=1m
Then,
E= 9e9×0.13e-9/1^2
E=1.17N/C. Q.E.D
Answer:
Magnitude of Electric field E at at point 0.50m which is within the sphere is Zero( i.e E = 0)
Explanation:
It is understand from Guass' law, the electric field in a region enclosed by a conducting sphere is Zero.
It is given that the radius of the sphere is 0.75m, therefore, a point located at 0.50m from the sphere centre 0.25m before the sphere surface still falls inside the sphere, therefore making the electric field at that point zero in magnitude.
If the fundamental frequency of a violin string is 440 HzHz, what is the frequency of the second harmonic?
Answer:
880Hz
Explanation:
A violin string is an example of an open pipe. An open pipe is open at both ends.
The fundamental frequency is the first harmonic.In an open pipe the second harmonic is twice the fundamental frequency.
Hence, second harmonic = 2Fo
Fo=440Hz
Second harmonic = 2*440Hz
Second harmonic = 880Hz
Two workers push on a wooden crate. One worker push with a force of 543 N and the other with a force of 333 N. The mass of the wooden crate is 206 kg. What is the acceleration of the crate?
Answer:
[tex]a=4.2524\ m.s^{-2}[/tex]
Explanation:
Given:
two workers push a crate.
Force by first worker, [tex]F_1=543\ N[/tex]force by second worker, [tex]F_2=333\ N[/tex]mass of crate, [tex]m=206\ kg[/tex](Assuming that both the workers push in the same direction.)
We know that,
Acceleration is given as:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{F_1+F_2}{m}[/tex]
[tex]a=\frac{543+333}{206}[/tex]
[tex]a=4.2524\ m.s^{-2}[/tex]
Newton's second law A. describes how an object accelerates when a force is applied B. says that objects eventually stop unless a force is applied C. objects with mass attract each other D. forces come in action/reaction pairs E. an object will remain in uniform motion unless acted upon by a force F. like charges repel, opposite charges attract
Answer:
A. describes how an object accelerates when a force is applied
Explanation:
Newton's second law of motion concerns the behavior of objects that do not have a stability between all established forces. The second law states that an object's acceleration depends on two factors: the net force on the entity and the entity's mass.
A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg • m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
Answer:
0.26087 rad/s
Explanation:
mass of the child (m) = 40 kg
velocity (v) = 3 m/s
distance (r) = 1.5 m
moment of inertia (I) = 600 kg.m^{2}
rotational momentum of the child = Iω
where
moment of inertia of the child (I) = [tex]mr^{2}[/tex] = 40 x 1.5 x 1.5 = 90 kg/m^{2}angular velocity (ω) = velocity / distance = 3 / 1.5 = 2 rad/srotational momentum of the child = Iω = 90 x 2 = 180 kg[tex]m^{2}[/tex]/s
from the conservation of momentum the initial momentum of the child must be the same as the final momentum of the child
initial momentum of the child = final momentum of the child
180 = (90 + 600) ω
180 = 690 ω
ω = 180 / 690 = 0.26087 rad/s
The angular or rotational momentum is the conserved quantity that has both magnitude and direction. It is given by:
[tex]\rm L = mvr[/tex]
Where,
L = rotational/angular momentum, m = mass, v = velocity and r = radius
The angular speed will be 0.26087 rad/s.
The speed can be estimated as:
Given,
Mass of the child (m) = 40 kgVelocity (v) = [tex]3 \rm \;m/s[/tex]Distance (r) = 1.5 mMoment of inertia (I) = [tex]600 \;\rm kgm^{2}[/tex]Rotational momentum is calculated by Iω.
Where,
Moment of inertia of the child (I) = [tex]\rm mr^{2}[/tex][tex]\begin{aligned}&= 40 \times (1.5)^{2} \\&= 90 \rm kg/m^{2}\end{aligned}[/tex]
Angular velocity (ω) = [tex]\rm \dfrac{velocity}{distance}[/tex][tex]\begin{aligned}&= \dfrac{3 }{1.5} \\&= 2 \rm rad/s\end{aligned}[/tex]
Rotational momentum of the child = Iω
[tex]\begin{aligned} \rm I\omega &= 90 \times 2 \\\\&= 180 \;\rm kgm^{2}/s\end{aligned}[/tex]
The final and the initial momentum would be the same according to the conservation law.
Initial momentum = Final momentum
180 = (90 + 600) ω
180 = 690 ω
Solving further:
[tex]\begin{aligned}\omega &= \dfrac{180}{690}\\\\&= 0.26087 \;\rm rad/s\end{aligned}[/tex]
Therefore, 0.26087 rad/s is the angular speed.
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Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown solution?
Answer:
Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle
Explanation:
When you centrifuge precipitate it enables the nucleation to form.
Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.
Final answer:
Centrifuging out the precipitate formed in an unknown solution prevents it from interfering with the analysis of remaining ions. The separated supernatant can then be further tested to identify other ions while avoiding unwanted reactions and ensuring precise characterizations of the substances involved.
Explanation:
It is necessary to centrifuge out any precipitate formed in the unknown solution during a chemical analysis for several reasons. The process of centrifugation uses inertia to separate particles in the fluid, so when the precipitate forms due to the reaction of different ions, it needs to be removed to isolate the remaining supernatant. The purpose is to ensure that subsequent testing only involves the dissolved substances, without interference from solids that have already reacted. Moreover, analysis of the residue is crucial after centrifugation and sometimes after supernatant evaporation to dryness, followed by reconstitution of the residue. This allows for a detailed study of the precipitate itself.
The remaining solution, after centrifugation, contains the supernate which may still contain ions or molecules of interest. If the precipitate is not removed, the solids could skew the results of further tests by hiding the presence of other ions or reacting further in unwanted ways. The separated liquid, or supernatant, can then be subjected to additional tests to identify other ions that may be present. For instance, if one is testing for the presence of barium sulfate in a mixture, tests like the precipitin ring test or radial immunodiffusion assay can be used, which demonstrates the presence of specific substances without the interference from the precipitate removed by centrifugation.
A stack of bricks weighs 170 KN (Kilo-newtons). The stack exerts 180 KPa (Kilo-pascals) of pressure on the ground. What is the area upon which this pressure is exerted (in square ft)?
Answer:
10.12square feet
Explanation:
Pressure exerted on the object is defined as the ratio of the force exerted on it to its unit area. Mathematically, Pressure = Force/Area
Given the force = 170kN
Pressure = 180KPa
Area = Force /Pressure
Area = 170kN/180KPa
Area = 0.94N/Pa
Note that 1Newton/Pascal = 10.764square feet
Therefore 0.94N/Pa = x
x = 0.94× 10.764
x = 10.12square feet
Therefore the area upon which this pressure is exerted is 10.12sqft.
The atmosphere of the sun consists mostly of hydrogen atoms (not molecules) at a temperature of 6000 K. What are (a) the average translational kinetic energy per atom and (b) the rms speed of the atoms
Explanation:
(a) Formula for average translational kinetic energy of a particle is as follows.
U = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
where, R = Reydberg's constant
T = absolute temperature
N = Avogadro's number
Therefore, we will calculate value of average translational kinetic energy as follows.
U = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
= [tex]\frac{3}{2}(\frac{8.314 J/mol K \times 6000 K}{6.023 \times 10^{23} mol^{-1}})[/tex]
= [tex]1.24 \times 10^{-19}[/tex] J
Therefore, value of average translational kinetic energy is [tex]1.24 \times 10^{-19}[/tex] J.
(b) Formula for average kinetic energy is as follows.
K.E = [tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex]
Here, M = molar mass = 1 kg/K mol
And, the average kinetic energy is equal to the average translational kinetic energy.
Hence, K.E = U
[tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex] = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
[tex]v^{2}_{rms} = \frac{3RT}{M}[/tex]
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
therefore, we will calculate r.m.s speed of the given atom as follows.
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
[tex]v = \sqrt{\frac{3 \times 8.314 J/mol K \times 6000 K}{1 kg/K mol}}[/tex]
= 386.84 m/s
Hence, value of r.m.s speed of the given atom is 386.84 m/s.
Following are the solution to the given points:
Given:
Temperature [tex]= 6000\ K[/tex]
To find:
kinetic energy =?
rms speed of the atoms=?
Solution:
For point a)
[tex]\to Avg \ KE =\frac{3}{2}\ KT \\\\[/tex]
[tex]\to KE_{avg} =1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\[/tex]
[tex]=1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=12.420 \times 10^{-20}\ J[/tex]
For point b)
[tex]\to[/tex] rms speed [tex]V_{rms}[/tex]:
[tex]\to v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
[tex]=\sqrt{\frac{3 \times 8314 \times 6000}{1}} \\\\=\sqrt{\frac{24942\times 6000}{1}} \\\\=\sqrt{149652000}\\\\=12233.233\\\\ = 12233\ \frac{m}{s}[/tex]
Therefore, the final answer is "[tex]12.420 \times 10^{-20} \ J\ \ and \ \ 12233 \ \frac{m}{s}[/tex]".
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Charge A is sitting in an electric field you know the following information:________
the magnitude of charge A and the magnitude of both the field and the potential at charge A position.
What would you do to get the potential energy of charge A?
Answer:
The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:
∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)
Explanation:
The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.
A hanging titanium wire with diameter 2.0 mm (2.0 × 10-3 m) is initially 2.5 m long. When a 9 kg mass is hung from it, the wire stretches an amount 0.605 mm. A mole of titanium has a mass of 48 grams, and its density is 4.51 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic bond.
To find the approximate value of the effective spring stiffness of the interatomic bond, calculate the stiffness constant of the titanium wire using Young's modulus.
Explanation:To find the approximate value of the effective spring stiffness of the interatomic bond, we need to calculate the stiffness constant of the titanium wire. The stiffness constant, or Young's modulus (Y), is given by the formula Y = F/A/L, where F is the force applied, A is the cross-sectional area of the wire, and L is the original length of the wire.
First, we need to find the force applied. The force can be calculated using the equation F = mg, where m is the mass and g is the acceleration due to gravity.
Next, we need to calculate the cross-sectional area of the wire. The cross-sectional area can be calculated using the formula A = π(r^2), where r is the radius of the wire.
Finally, we can substitute the values into the formula Y = F/A/L to find the Young's modulus of the titanium wire.
When water freezes, its volume increases by 9.05% (that is, ΔV V0 = 9.05 ✕ 10−2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.
Answer:
Δp=2.0×10⁴N/cm²
Explanation:
We can write the bulk modulus formula and we solve pressure difference Δp
So
B=Δp(V₀ / ΔV)
Δp=B( ΔV / V₀)
As ΔV / V₀ is given as 9.05×10⁻²
So
Δp=B(9.05×10⁻²)
Δp=(0.22×10¹⁰)(9.05×10⁻²)
Δp=2.0×10⁸ N/m²
Δp=2.0×10⁸ N/m²×(1m²/10⁴cm²)
Δp=2.0×10⁴N/cm²
Answer:
2.0×10⁸ N/m².
Explanation:
Bulk modulus, B is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume
B = -V₀ * (Δp/ΔV)
Given:
ΔV/V₀ = 9.05×10⁻²
Bw = 0.22×10¹⁰
Δp = -B * (ΔV/V₀)
= (0.22×10¹⁰) * (9.05×10⁻²)
= 2.0×10⁸ N/m².
The doppler effect is the change in observed frequency due to
Answer: relative motion between observer and the sound source.
Explanation: The Doppler effect states that when there is a relative motion between an observer and a sound source the frequency of sound perceived by the observer is different in frequency from the original from the source.
The mathematical back up for this claim is given below.
f' = (v+v') /(v-vs) × f
Where f' = observed frequency
v = speed of sound in air
v' = velocity of observer
vs = velocity of source
f = frequency of sound source.
From the formulae, it can be seen that a change in the value of the velocity of observer (v') and source (vs) produces different value of observed frequency (f').
Note, frequency of sound (f) is a constant.
The current through two identical light bulbs connected in a series is 0.25 amps. The voltage across both bulbs is 100 volts. Find the resistance of a single light bulb.
Answer: 200ohms
Explanation:
According to Ohm's law, the current (I) passing through a metallic conductor is directly proportional to the potential difference (v) between its ends at constant temperature and pressure. Mathematically, E=IRt
Where E is the voltage across both bulbs, I is the current, Rt is the total equivalent resistance
E = 100V, I = 0.25amps Rt = R+R(since they are 2 identical bulbs in series, we will add them together)
Rt = 2R
Substituting this values in the formula to get R, we have;
100 = 0.25(2R)
100 = 0.5R
R = 100/0.5
R = 200ohms
The resistance of a single light bulb will be 200ohms.
Note that the unit of resistance is ohms
As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 39.0 m/s.
How fast is the watermelon going when it passes Superman?
Answer:
The velocity of watermelon when it passes Superman is 78 m/s.
Explanation:
Height of the building, d = 320 m
Speed of the superman, v = 39 m/s
We need to find the speed of watermelon when it passes Superman. Let t is the time taken by the watermelon. So,
[tex]d=ut+\dfrac{1}{2}gt^2[/tex]
Here u = 0
[tex]d=\dfrac{1}{2}gt^2[/tex]
[tex]39t=\dfrac{1}{2}\times 9.8\times t^2[/tex]
t = 7.95 seconds
Let v is the speed of the watermelon. It is given by :
[tex]v=gt[/tex]
[tex]v=9.8\times 7.95[/tex]
v = 77.91 m/s
or
v = 78 m/s
So, the velocity of watermelon when it passes Superman is 78 m/s.
An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
Answer:
12.24 m/s
Explanation:
Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.
Using the formula,
a = v/t................ Equation 1
Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.
making v the subject of the equation,
v = at ................. Equation 2
Given: a = 5.1 m/s², t = 2.4 s.
Substitute into equation 2
v = 5.1(2.4)
v = 12.24 m/s.
Hence, the speed of the sprinter = 12.24 m/s
Four point charges each having charge Q are located at the corners of a square having sides of length a.
(a) Find a symbolic expression for the total electric potential at the center of the square due to the four charges (Use any variable or symbol stated above along with the following as necessary: ke.)
Vtotal = ______
(b) Find a symbolic expression for the work required to bring a fifth charge p from infinity to the center of the square (Use any variable or symbol stated above along with the following as necessary: ke.)
W = _____
Answer:
Explanation:
Check attachment
The total electric potential in the center of the square due to 4 chargers at its corners is given by V_total= (4 * ke * Q) / sqrt( a^2/2 ), and the work needed to bring a fifth charge p from infinity to that point is W = p * (4 * ke * Q) / sqrt( a^2/2 ).
Explanation:The electric potential at a particular point (center in this case) because of a point charge follows the formula V = k*Q / r where:
V is the electric potentialk is Coulomb's constantQ is the charger is the distance from the point to the chargeFor four charges Q at corners of a square of side length a, the distance from the charge to the center of square would be sqrt(a^2/2), so:
V_total= (4 * ke * Q) / sqrt( a^2/2 )
The work done to bring a charge from infinity to a point is given by W = q * V_total where q is the charge being moved. So, under given circumstances:
W = p * (4 * ke * Q) / sqrt( a^2/2 )
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Two wires, each of length 1.3 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 640 Hz. However, the same frequency of 640 Hz is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire.
Answer:
Explanation:
Given
Length of each wire [tex]L=1.3\ m[/tex]
On wire A second harmonic frequency is given by
[tex]f_2_{a}=2\times (\frac{v}{2L})[/tex]
where f=frequency
v=velocity of wave
L=length of wire
[tex]v_a=f_2\times L[/tex]
[tex]v_a=640\times 1.3=832\ m/s[/tex]
For wire B third harmonic is given by
[tex]f_3_{b}=3\times (\frac{v}{2L})[/tex]
[tex]v_b=\frac{2L}{3}\cdot f_3_{b}[/tex]
[tex]v_b=\frac{2\times 1.3}{3}\times 640=554.66\ m/s[/tex]
A ball is thrown straight up from a bridge at a speed of 11.0 m/s. What will be its velocity (speed and direction) after 2.0 seconds?
Explanation:
Let upper direction be positive
We have equation of motion v = u + at
Initial velocity, u = 11 m/s
Final velocity, v = ?
Time, t = 2 s
Acceleration,a = -9.81 m/s²
Substituting
v = u + at
v = 11 + -9.81 x 2
v = -8.62 m/s
Speed of ball after 2 seconds is 8.62 m/s downward.
All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certain steam iron states that the iron carries a current of 6.00 A when connected to a source of 1.20 ✕ 102 V. What is the resistance of the steam iron?
The value of steam iron resistance is [tex]20ohm[/tex].
The relation between voltage, current and resistant shown below,
[tex]V=IR\\\\R=\frac{V}{I}[/tex]
Where V is voltage and I is current .
Given that, [tex]I=6A,V=1.20*10^{2}[/tex]
Substitute values in above relation.
[tex]R=\frac{1.2*10^{2} }{6}=20ohm[/tex]
Hence, the value of steam iron resistance is [tex]20ohm[/tex].
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The resistance of the steam iron can be calculated using Ohm's Law, expressed as the formula R = V/I. With a current of 6.00 A and a voltage of 1.20 ✕ 102 V, the steam iron has a resistance of 20.0 Ohms.
Explanation:The electrical characteristics of a device, including resistance, can usually be determined using Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points. It can be expressed in the formula R = V/I, where R is resistance, V is voltage, and I is current.
In this case, our steam iron has a current (I) of 6.00 A and a voltage (V) of 1.20 ✕ 102 V. Substituting these values into the Ohm's Law formula gives us: R = (1.20 x 102 V) / 6.00 A.
This calculates to a resistance (R) of 20.0 Ohms for the steam iron.
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A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force the spring exerts on the car is given by the equation F=−Kx2, where K=5000 Nm2. The potential energy stored in the spring when the car is pushed against it is most nearly:________________
The potential energy stored in the spring when the car is pushed against it is most nearly is 0.026 J.
The calculation is as follows:Non linear spring is
[tex]F = -kx^2 \rightarrow force[/tex]
The potential energy
[tex]\frac{dU}{dX} = -F\\\\U = \int\limits {-Fdx}\\\\U = + \int\limits^x_0 {kx^2dx}\\\\U = \frac{kx^3}{3} \\\\U = 5000 \div 3 (\frac{2.5}{100}^)2\\\\U = \frac{5000 \times 2.5\times 2.5\times 2.5}{3\times 100\times 100\times 100}[/tex]
= 0.026 J
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Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickly, while telephone poles in the distance are passing by much more slowly. This is an example of
Answer: Relative motion
Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.
The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p
Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.
Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.
Motion of this nature is known as relative motion.
Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid
For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.
The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.
A student does 60. joules of work pushing a 3.0-kilogram box up the full length of a ramp that is 5.0 meters long. What is the magnitude of the force applied to the box to do this work?
Answer:
F= 12 N
Explanation:
Given that
Work done by student ,W= 60 J
The mass of the box ,m = 3 kg
Length ,x = 5 m
We know that ,The work done by a force a force F is given as
W= F .x
x=Displacement
F=Force
W=Work
Now by putting the values
60 = F x 5
[tex]F=\dfrac{60}{5}\ N[/tex]
F= 12 N
That is why the magnitude of the force will be 12 N.
The magnitude of the force applied to the box to do the work is 12 N
Definition of workWorkdone is defined as the product of force and distance moved in the direction of the force. Mathematically, it can be expressed as:
Workdone (Wd) = force (F) × distance (d)
Wd = Fd
With the above formula, we can obtain the force used to do the work.
Determination of the force•Workdone (Wd) = 60 J
•Distance (d) = 5 m
•Force (F) =?
Wd = Fd
60 = F × 5
Divide both side by 5
F = 60 / 5
F = 12 N
Thus, 12 N is required to do the work.
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Ask Your Teacher It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 340 m the field has magnitude 60.0 N/C. At an altitude of 220 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 120 m on edge, with horizontal faces at altitudes of 220 and 340 m. Neglect the curvature of Earth.
Answer:q=3.536+10^-6C
Explanation:
According to the Revere and Black (2003) article, processes that result in an error probability of 0.000070 should be recognized as achieving the Six Sigma standard. Group of answer choices True False
Answer:
False
Explanation:
The ‘sigma’ refers to the Greek letter used to denote standard deviation, so ‘six sigma’ means that the error rate lies beyond six standard deviations from the mean.
. Assume you have a dot grid with 36 dots per sq.in. How many acres are represented by each dot using the following map scales: (a) 330 ft. per in., (b) 25 chains per in., (c) 1 mile per in.
Answer:
a) 0.069 acres
b) 0.114acres
c) 17.78acres
Explanation:
1 dot=1/36sqin
a) 1 sqin= 330×330=108900ft^2
1 dot=108900/36 =3025ft^2
Converting to acre,divide by43560
3025/43560 =0.069acre
b) 1 chain =22 yards
25×22=550yards
Converting to acre divide by 4840
550/4840 =0.114acre
c)1 sqmile =640acres
(1/36) ÷ 640
640/36
17.78acres
To calculate the area represented by each dot on the dot grid, you need to consider the scale of the map. With a scale of 330 ft. per inch, 1 dot represents (330^2 / 43,560) acres. With a scale of 25 chains per inch, 1 dot represents ((25 chains)^2 / 10) acres. And with a scale of 1 mile per inch, 1 dot represents ((1 mile)^2 / 640) acres.
Explanation:
The conversion from dots to acres varies depending on the scale:
In the circuit described by the diagram, which pair of resistors is connected in parallel?
*answer choices attached*
Option 4 ( R2 and R3 ) is the correct answer.
Explanation:
In the below given diagram, we can see a circuit diagram that has four resistors such as R1, R2, R3, and R4.The opening of the circuit is noted as "a" and the ending is noted as "b".By observing the above diagram, we can clearly see that R2 and R3 are the pair of resistors that are connected in a parallel manner.Where all the other resistors such as R1 and R4 are neither connected in parallel nor in series.Hence we can conclude that Resistor R2 and R3 are the ones that are connected in parallel.
Match the following: (Statistical methods) Frequency counts Frequency distribution Measurements of central tendency Measurements of spread A. Applied to categorical values B. Applied to quantitative values C. Mean, median, and mode D. Max, min, percentiles, and standard deviation
Final answer:
Match the correct statistical methods to their definitions as follows: Frequency counts are A (applied to categorical values), Frequency distribution is B (applied to quantitative values), Measurements of central tendency are C (mean, median, and mode), and Measurements of spread are D (max, min, percentiles, and standard deviation).
Explanation:
When examining statistical methods, it is important to correctly match each concept with its definition to understand data analysis better. Here are the correct matches:
Frequency counts - Applied to categorical values (A)Frequency distribution - Applied to quantitative values (B)Measurements of central tendency - Mean, median, and mode (C)Measurements of spread - Max, min, percentiles, and standard deviation (D)These matches are crucial in the proper analysis and interpretation of data whether you are dealing with categorical or quantitative data. It's important to apply the correct statistical method for the level of measurement being dealt with, such as nominal, ordinal, interval, or ratio scales.
Converging circuits with excitation and inhibition are associated most closely with which step of the perceptual process?
Answer:
Neural processing
Explanation:
Neural processing- it is referred to as the unit that is responsible for the implementation of arithmetic logic which is crucial for the execution of machine learning algorithms.
it is used to operate the artificial neural network in designing and can be helpful to enhance the efficiency of machine learning applications like artificial language.
Suppose a spring with spring constant 9 N/m is horizontal and has one end attached to a wall and the other end attached to a mass. You want to use the spring to weigh items. You put the spring into motion and find the frequency to be 0.9 Hz (cycles per second). What is the mass? Assume there is no friction.
To find the mass attached to a spring when the frequency of oscillation is known, use the formula for the frequency of SHM (f = (1/2π) ∙ √(k/m)) and rearrange it to solve for mass.
Explanation:To determine the mass attached to a horizontal spring where the frequency of oscillation is known, we can use the formula for the frequency of a mass-spring system undergoing simple harmonic motion (SHM), which is f = (1/2π) ∙ √(k/m), where f is the frequency, k is the spring constant, and m is the mass. Given a spring constant 9 N/m and a frequency 0.9 Hz, we can rearrange the formula to solve for the mass (m = k / (2πf)^2). Plugging in the values, we get the mass m = 9 N/m / (2π ∙ 0.9 Hz)^2, which we can calculate to find the mass of the object attached to the spring.
Using the frequency equation for a mass-spring system, the mass can be determined to be approximately 0.28 kg.
To find the mass attached to the spring, we can use the equation for the frequency of a mass-spring system:
f = (1 / (2π)) * √(k / m)Where:
f is the frequency (0.9 Hz)k is the spring constant (9 N/m)m is the mass in kgThis equation can be rearranged to solve for mass:
m = k / (4π²f²)Plugging in the provided values:
m = 9 N/m / [4π² * (0.9 Hz)²]First, simplify the expression inside the brackets:
(0.9 Hz)² = 0.814π² * 0.81 ≈ 32.17Then:
m = 9 / 32.17 m ≈ 0.28 kgThus, the mass attached to the spring is approximately 0.28 kg.