The work done to move a positive test charge from a point at a radial distance of 5R from the center of a charged sphere to a point at a radial distance of 3R is given by the difference in the electric potentials at these points times the charge of the test charge.
Explanation:The work done, W, in moving a small positive test charge, q0, in an electric field produced by another charged object is given by the expression W = q0 x (Vf - Vi), where Vf and Vi are> the final and initial electric potentials, respectively. The electric potential, V, at a point located a distance r from the center of a conducting sphere carrying a charge q is V = 1/4πε0 x q/r. So, the work done to move the test charge from r = 5R to r = 3R is W = q0 x {[1/4πε0 x q/(3R)] - [1/4πε0 x q/(5R)]}, which simplifies to W = q0q/4πε0R x (5/15 - 3/25).
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The work required to move a small positive test charge from a radius of 5R to 3R on a conducting sphere of radius R carrying positive charge q is calculated to be W = q*q0/(30πϵ0R) using Coulomb's law and the work-energy theorem.
Explanation:The work needed to move a charge in an electric field is given by the integral of force times distance. The force on a charge in an electric field is given by Coulomb's law: F = k*q*q0/r^2, where k is Coulomb's constant = 1/4π*ϵ0. The work done in moving a charge from r1 to r2 is given by the integral from r1 to r2 of dr, which results in k*q*q0*(1/r1 - 1/r2).
In this case, with r1=5R and r2=3R, the work is W = k*q*q0*(1/5R - 1/3R), which simplifies to W = 2k*q*q0/(15R), or W = q*q0/(30πϵ0R) when substituting k = 1/4π*ϵ0.
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You a building a circuit that requires a 150 pF capacitor, but you only have a drawer full of 100 pF capacitors.
(How) can you make a 150 pF equivalent capacitor?
To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.
When capacitors are connected in series, they consecutively add their capacitance.
However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is [tex]\frac {1} {C}[/tex] where, C is the capacitance.
For the given case, it is best to connect two of these capacitors in series and one in parallel, so
We have three 100 pF capacitors, then
[tex]C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\[/tex]
Here we can see how two capacitors 1 and 2 are in series and the third in parallel.
[tex]\frac{1}{C_{serie}} = \frac{1}{100pF}+\frac{1}{100pF}[/tex]
[tex]\frac{1}{C_{serie}} = \frac{1}{50pF}[/tex]
Investing equality
[tex]C_{serie} = 50pF[/tex]
Adding it in parallel, then
[tex]C_{total} = C_{serie} +C_3[/tex]
[tex]C_{total} = 50pF+100pF[/tex]
[tex]C_{total} = 150pF[/tex]
A compound die is used to blank and punch a large washer out of 6061ST aluminum alloy sheet stock 3.2 mm thick. The outside diameter of the washer is 25.0 mm, and the inside diameter is 12.0 mm. Determine the punch and die sizes for (a) the blanking operation, and (b) the punching operation.
Answer:
24.616 mm
12.384 mm
Explanation:
[tex]D_b[/tex] = Blanking die diameter = 25 mm
[tex]D_h[/tex] = Punch diameter = 12 mm
a = Allowance = 0.06 for aluminium
t = Thickness = 3.2 mm
Clearance
[tex]c=at\\\Rightarrow c=0.06\times 3.2\\\Rightarrow c=0.192[/tex]
Diameter of punch is
[tex]d_p=D_b-2c\\\Rightarrow d_p=25-2\times 0.192\\\Rightarrow d_p=24.616\ mm[/tex]
Diameter of blanking punch is 24.616 m
For punching operation
[tex]d_d=D_h+2c\\\Rightarrow d_d=12+2\times 0.192\\\Rightarrow d_d=12.384\ mm[/tex]
The diameter of punching die is 12.384 mm
For the blanking operation, the punch diameter should be 12.0 mm and the die diameter should be 25.0 mm. For the punching operation, the punch diameter should be 12.0 mm and the die diameter should be slightly larger than 12.0 mm, typically by a clearance of 0.05 mm to 0.15 mm per side.
(a) Blanking Operation:
The blanking operation involves cutting the outside profile of the washer from the aluminum sheet stock. The punch and die sizes for this operation are determined by the outside and inside dimensions of the washer.
- The die diameter for the blanking operation should match the outside diameter of the washer, which is 25.0 mm. This is the size of the hole in the die through which the punch will pass to cut the washer's outer edge.
- The punch diameter for the blanking operation should match the inside diameter of the washer, which is 12.0 mm. This is the size of the punch that will push through the sheet stock to create the inner hole of the washer.
(b) Punching Operation:
The punching operation involves cutting the inside hole of the washer after the blanking operation has been completed. The punch and die sizes for this operation are determined by the inside diameter of the washer and the required clearance.
- The punch diameter for the punching operation should also be 12.0 mm, the same as the inside diameter of the washer. This ensures that the punch cuts the material exactly at the desired inner diameter.
- The die diameter for the punching operation should be slightly larger than the punch diameter to allow for clearance. Clearance is necessary to prevent the punch and die from binding together and to allow for the removal of the slug (the piece of material that is punched out). The typical clearance ranges from 0.05 mm to 0.15 mm per side. Therefore, the die diameter would be 12.0 mm plus two times the clearance. If we take the middle value of the clearance range, which is 0.10 mm per side, the die diameter would be 12.0 mm + 2 * 0.10 mm = 12.20 mm.
A. Situation A: A 20-kg block is pressed against a relaxed spring on a frictionless surface. The spring is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring uncompresses. The speed of the block at x = x 0 e is va,f = 5.0 m/s.
B. Situation B: The same 20-kg block is pressed against two relaxed springs that are attached in series on a frictionless surface. (These two springs are both identical to the spring in Situation A). The entire spring assembly is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring assembly uncompresses. The speed of the block at x = x 0 e is vb,f Question:
What is the value of vb,f ?
(A) 3.5 m/s
(B) 2.5 m/s/s
(C) 5.0 m/s
(D) 6.4 m/s
(E) 7.0 m/s.
Final answer:
In Situation B, the speed of the block at x = x0e is 3.5 m/s.
Explanation:
In Situation A, a 20-kg block is pressed against a single spring and released. The speed of the block at x = x0e is va,f = 5.0 m/s. In Situation B, the same 20-kg block is pressed against two identical springs in series. The entire spring assembly is compressed by Δx = 10 cm. The question asks for the speed of the block at x = x0e, which we can find using the principle of conservation of mechanical energy.
The speed of the block in Situation B, vb,f, can be found as follows:
Calculate the potential energy stored in the spring assembly when it is compressed by Δx = 10 cm. This is done using the formula U = (1/2)k(Δx)^2, where k is the spring constant.Use the principle of conservation of mechanical energy to relate the potential energy of the spring assembly to the kinetic energy of the block at x = x0e. This can be written as (1/2)mvb,f^2 = U.Solve for vb,f to find the speed of the block at x = x0e.Solving this equation gives us vb,f = 3.5 m/s.
A high-speed drill rotating counterclockwise takes 2.5 s to speed up to 2400 rpm. (a) What is the drill’s angular acceleration?
(b) How many revolutions does it make as it reaches top speed?
Answer:
(a) Angular acceleration will be [tex]100.53rad/sec^2[/tex]
(b) 50 revolution
Explanation:
We have given time t = 2.5 sec
Initial speed of the drill [tex]\omega _0=0rad/sec[/tex]
Speed after 2.5 sec [tex]=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec[/tex]
From first equation of motion we know that
[tex]\omega =\omega _0+\alpha t[/tex]
[tex]251.327 =0+\alpha \times 2.5[/tex]
[tex]\alpha =100.53rad/sec^2[/tex]
(b) From second equation of motion we know that
[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2[/tex]
[tex]\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times 2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution[/tex]
An audio engineer takes decibel readings at distances of r1 = 13 m and r2 = 24 m from a concert stage speaker during a sound check. When he is r1 from the speaker, the engineer registers a decibel level of β1 = 101 dB on his loudness meter.
Answer
given,
Distance for decibel reading
r₁ = 13 m
r₂ = 24 m
When the engineer is at r₁ reading is β₁ = 101 dB
now, Calculating the Intensity at r₁
Using formula
[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]
I₀ = 10⁻¹² W/m²
now inserting the given values
[tex]101= 10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]10.1= log(\dfrac{I}{10^{-12}})[/tex]
[tex]\dfrac{I}{10^{-12}}= 10^{10.1}[/tex]
[tex]I =10^{10.1} \times 10^{-12}[/tex]
I = 0.01258 W/m²
now, calculating power at r₁
P₁ = I₁ A₁
P₁ = 0.01258 x 4 π r²
P₁ = 0.01258 x 4 π x 13²
P₁ = 26.72 W
Final answer:
The question deals with acoustics in physics, where an audio engineer uses the decibel scale to measure sound pressure levels and requires an understanding of the inverse square law and logarithmic measurements of sound intensity.
Explanation:
The student's question involves the concept of sound intensity levels and decibels, which are part of the field of acoustics, a topic in physics. Specifically, the student is dealing with the measurement of sound pressure levels at various distances from a sound source, which is an application of the inverse square law in acoustics. The decibel (dB) scale is a logarithmic unit used to measure sound intensity, with 0 dB representing the threshold of human hearing at an intensity of 10-12 W/m².
The loudness experienced by the audio engineer at various distances can be computed using the formula for sound intensity level (SIL), which is SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10-12 W/m².
A soap bubble is 103 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength (in nm) and color of visible light is most constructively reflected, assuming the same index of refraction as water (nw = 1.33)?
Answer:
λ = 547.96 nm
Explanation:
given,
thickness of soap bubble = 103 nm
refractive index of thin film = 1.33
using formula of constructive interference
[tex]2 n t = (m + \dfrac{1}{2})\times \lambda[/tex]
t is thickness of the medium
n is refractive index
m = 0,1,2...
now,
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
at m = 0
[tex]2\times 1.33 \times 103 =0.5\times \lambda[/tex]
λ = 547.96 nm
at m = 1
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
[tex]2\times 1.33 \times 103 =1.5\times \lambda[/tex]
λ = 182.65 nm
the only visible light enhanced by the thin film is of wavelength
λ = 547.96 nm
The Bullet Train The Shinkansen, the Japanese "bullet" train, runs at high speed from Tokyo to Nagoya. Riding on the Shinkansen, you notice that the frequency of a crossing signal changes markedly as you pass the crossing. As you approach the crossing, the frequency you hear is f; as you recede from the crossing the frequency you hear is 2f/3.What is the speed of the train?
Answer:
68.6 m/s
Explanation:
v = Speed of sound in air = 343 m/s
u = Speed of train
[tex]f_1[/tex] = Actual frequency
From the Doppler effect we have the observed frequency as
When the train is approaching
[tex]f=f_1\frac{v+u}{v}[/tex]
When the train is receeding
[tex]\frac{2f}{3}=f_1\frac{v-u}{v}[/tex]
Dividing the above equations we have
[tex]\frac{f}{\frac{2f}{3}}=\frac{f_1\frac{v+u}{v}}{f_1\frac{v-u}{v}}\\\Rightarrow \frac{3}{2}=\frac{v+u}{v-u}\\\Rightarrow 3v-3u=2v+2u\\\Rightarrow v=5u\\\Rightarrow u=\frac{v}{5}\\\Rightarrow u=\frac{343}{5}\\\Rightarrow u=68.6\ m/s[/tex]
The speed of the train is 68.6 m/s
Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. Take the density of water to be 1000 kg/m3.
According to the information presented, it is necessary to take into account the concepts related to mass flow, specific potential energy and the power that will determine the total work done in the system.
By definition we know that the change in mass flow is given by
[tex]\dot{m} = \rho AV[/tex]
[tex]\dot {m} = \rho Q[/tex]
Remember that the Discharge is defined as Q = AV, where A is the Area and V is the speed.
Substituting with the values we have we know that the mass flow is defined by
[tex]\dot{m} = 1000*0.03[/tex]
[tex]\dot{m} = 30kg/s[/tex]
To calculate the power we need to obtain the specific potential energy, which is given by
[tex]\Delta pe = gh[/tex]
[tex]\Delta pe = 9.8*45[/tex]
[tex]\Delta pe = 441m^2/s^2[/tex]
So the power needed to deliver the water into the storage tank would be
[tex]\dot {E} = \dot{m}\Delta pe[/tex]
[tex]\dot {E} = 30*441[/tex]
[tex]\dot{E} = 13230W = 13.23kW[/tex]
Finally the mechanical power that is converted to thermal energy due to friction effects is:
[tex]\dot{W}_f = \dot{W}_s - \dot{E}[/tex]
[tex]\dot{W}_f 20-13.23[/tex]
[tex]\dot{W} = 6.77kW[/tex]
Therefore the mechanical power due to friction effect is 6.77kW
The irreversible head loss in this pump system is 23 meters and the lost mechanical power during the pumping process is 6.8 kW.
Explanation:In this system, the pump supplies energy to the water and drives it up to the higher reservoir. Out of this energy provided by the pump, part of it is used for lifting the water while some is lost as head loss due to friction and other losses in the system. Using the principle of conservation of energy and the Bernoulli's equation, one can calculate the irreversible head loss and the mechanical power loss.
Given the mechanical power provided by the pump, P = 20 kW, the height difference h = 45 m, the flow rate Q = 0.03 m3/s, and the density of water ρ = 1000 kg/m3, the useful power for lifting the water equals ρghQ=1000x 9.81x 45 x 0.03=13.2 kW. Thus, the lost power due to irreversible head loss equals the total power minus the useful power, or 20 kW - 13.2 kW = 6.8 kW. The corresponding head loss equals the lost power divided by the power used for lifting the water, or h_loss = (6.8 kW)/(ρgQ) = 23 m.
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Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?
Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
A bullet is shot straight up in the air with an initial velocity of v=1000 feet per second. If the equation describing the bullets height from the ground is given by f(t)=−16t2+1000t+7, over what interval is the bullet speeding up, and over what interval is it slowing down?
Answer:
It slows down from 0-31.25s
It speeds up from 31.25-62.507s
Explanation:
If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.
[tex]Y = -16t^2+1000t+7[/tex]
[tex]Y' = -32t+1000 = 0[/tex] Solving for t:
t = 31.25s
The bullet will slow down in the interval 0-31.25s
Let's now find the moment when it hits ground:
[tex]Y = 0 = -16t^2+1000t+7[/tex] Solving for t:
t1 = -0.0069s t2 = 62.507s
The bullet will speed up in the interval 31.25-62.507s
9. A baseball pitcher hurls a 400 g baseball toward a batter with a speed of 30 m/s. The batter hits the pitch directly back toward the pitcher and it travels at 38 m/s. If the collision with the bat required 80 ms (milliseconds), find the force the bat applies to the ball. (Watch signs.)
Answer:
The force the bat applies to the ball is 340 N
Explanation:
given information:
mass of the ball, m = 400 g = 0.4 kg
initial velocity, [tex]v_{i}[/tex] = 30 m/s
final velocity, [tex]v_{f}[/tex] = 38 m/s
time, t = 80 ms = 0.08 s
the correlation between the change in momentum and the force is shown by the following equation
ΔP = FΔt
F = ΔP/Δt
where,
Δp = mΔv
= m([tex]v_{f} - v_{i}[/tex])
= m([tex]v_{f} + v_{i}[/tex])
[tex]v_{i}[/tex] = - because in opposite direction, thus
F = ΔP/Δt
= m([tex]v_{f} + v_{i}[/tex])/Δt
= (0.4)(38+30)/0.08
= 340 N
What is a particulate ? Name a couple of examples.
Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles
Explanation:
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave is 0.4. What is the largest ratio of tensile forces between the two ends of the rope before the rope starts to slide over the sheave?
Answer:3.51
Explanation:
Given
Coefficient of Friction [tex]\mu =0.4 [/tex]
Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]
Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]
[tex]N=T\cdot d\theta [/tex]
Friction [tex]f=\mu \times Normal\ Reaction[/tex]
[tex]f=\mu \cdot N[/tex]
and [tex]T+dT-T=f=\mu Td\theta [/tex]
[tex]dT=\mu Td\theta [/tex]
[tex]\frac{dT}{T}=\mu d\theta [/tex]
[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]
[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]
[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]
[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]
[tex]\frac{T_2}{T_1}=3.51[/tex]
A rectangular settling tank is being designed to settle sand particles that have a settling velocity of 2.0 cm/s. The water flow is 10,000 m3/d. The tank length should be four times the width and the width is equal to the depth. Please find the dimensions for the tank in meter.
Answer:
b = 1.2 m
L = 4.8 m
d= 1.2 m
Explanation:
Lets take
Width of the tank = b
Depth = d
length = L
Given that L = 4 b
v= 2 cm/s
Flow rate Q= 10,000 m³/d
We know that 1 d = 24 hr = 24 x 3600 s
1 d= 86400 s
Q= 0.115 m³/s
Flow rate Q
Q= Area x velocity
Q= L . b .v
0.115 = 4 b . b . 0.02
4 b² = 5.78
b = 1.2 m
L = 4.8 m
d= 1.2 m
This is the dimension of the tank.
To design the settling tank with the given water flow and settling velocity, we use conservation of mass and settling speed to calculate the dimensions. We convert the flow rate to m3/s, relate it to the tank's cross-sectional area, and given settling characteristics to find the tank's width, length, and depth.
Explanation:To design a rectangular settling tank for settling sand particles with a settling velocity of 2.0 cm/s and a water flow of 10,000 m3/d, we will apply the principles of conservation of mass and the given settling velocity to determine the dimensions of the tank. The problem states that the tank length (L) should be four times the width (W), and the width should be equal to the depth (D), therefore L = 4W and W = D.
To find the dimensions, we first convert the water flow rate into m3/s: flow rate = 10,000 m3/d × (1 day/86400 s) ≈ 0.1157 m3/s. Since the sand particles settle at 2 cm/s, for a particle to settle before reaching the end of the tank, the residence time should be at least the depth divided by the settling velocity. From here, we use the flow rate (Q), which is also equal to the cross-sectional area (A) times the water velocity (V), and we know that A = W² for this tank because W = D.
Therefore, Q = W²V. Using V = L / (residence time) and substituting for L = 4W, we find that Q = W² × (4W / (W/0.02 m/s)) which simplifies to Q = 0.08 m/s × W³. Solving for W gives us W = ∛(Q / 0.08 m/s) and subsequently we find L = 4W and D = W. Finally, we can plug in the flow rate and solve for the dimensions in meters.
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A 20-cm-diameter vertical cylinder is sealed at the top by a frictionless 23 kg piston. The piston is 86 cm above the bottom when the gas temperature is 307°C. The air above the piston is at 1.00 atm pressure.
A) What is the gas pressure inside the cylinder?
B) What will the height of the piston be if the temperature is lowered to 13°C?
Answer:
108507.02596 Pa
42.42 cm
Explanation:
F = Force
m = Mass of piston = 23 kg
[tex]T_1[/tex] = Initial temperature = 307°C
[tex]T_2[/tex] = Final temperature = 13°C
[tex]P_o[/tex] = Outside pressure
[tex]P_i[/tex] = Pressure inside cylinder
A = Area of pistion
h = Height of piston
Change in pressure is given by
[tex]\Delta P=\frac{F}{A}\\\Rightarrow \Delta P=\frac{mg}{\pi r^2}\\\Rightarrow \Delta P=\frac{23\times 9.81}{\pi 0.1^2}\\\Rightarrow \Delta P=7182.02596\ Pa[/tex]
[tex]P_i-P_o=\Delta P\\\Rightarrow P_i=\Delta P+P_0\\\Rightarrow P_i=7182.02596+101325\\\Rightarrow P_i=108507.02596\ Pa[/tex]
Gas pressure inside the cylinder is 108507.02596 Pa
From ideal gas law we have the relation
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow \frac{Ah_1}{T_1}=\frac{Ah_2}{T_2}\\\Rightarrow \frac{h_1}{T_1}=\frac{h_2}{T_2}\\\Rightarrow h_2=\frac{h_1}{T_1}\times T_2\\\Rightarrow h_2=\frac{0.86}{307+273.15}\times (13+273.15)\\\Rightarrow h_2=0.42418\ m=42.42\ cm[/tex]
The height of the piston at 13°C will by 42.42 cm
(a) The gauge pressure inside the cylinder is 108,500 Pa.
(b) The height of the piston when the temperature is lowered to 13°C is 42.4 cm.
Change in gas pressure inside the cylinderThe change in the pressure of the gas inside the cylinder is calculated as follows;
ΔP = F/A
ΔP = mg/πr²
ΔP = (23 x 9.8)/(π x0.1²)
ΔP = 7,174.7 Pa
Gauge pressure inside the cylinderThe gauge pressure inside the cylinder is calculated as follows;
Pi = ΔP + Po
Pi = 7,174.7 + 101325
Pi = 108,500 Pa
Height of the pistonThe height of the piston is calculated as follows;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\\frac{Ah_1}{T_1} = \frac{Ah_2}{T_2} \\\\h_2 = \frac{T_2h_1}{T_1} \\\\h_2 = \frac{(13 + 273) \times 0.86}{(307 + 273)} \\\\h_2 = 0.424 \ m[/tex]
h₂ = 42.4 cm
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What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a final velocity of 7.3 m/s? Express the moment of inertia as a multiple of MR2, where M is the mass of the object and R is its radius. (Hint: Use total conservation of mechanical energy)
Answer:
I = 0.287 MR²
Explanation:
given,
height of the object = 3.5 m
initial velocity = 0 m/s
final velocity = 7.3 m/s
moment of inertia = ?
Using total conservation of mechanical energy
change in potential energy will be equal to change in KE (rotational) and KE(transnational)
PE = KE(transnational) + KE (rotational)
[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2[/tex]
v = r ω
[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}[/tex]
[tex]I = \dfrac{m(2gh - v^2)r^2}{v^2}[/tex]
[tex]I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}[/tex]
[tex]I =mr^2(0.287)[/tex]
I = 0.287 MR²
An electromagnetic wave is propagating in the positive x direction. At a given moment in time, the magnetic field at the origin points in the positive y direction. In what direction does the electric field at the origin point at that same moment?a. Positive xb. Negative xc. Positive yd. Negative ye. Positive zf. Negative z
To solve this exercise it is necessary to take into account the concepts related to the magnetic field and its vector representation through the cross product or vector product.
According to the definition the direction of the electromagnetic wave propagation is given by
[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]
Where,
E = Electric Field
B = Magnetic Field
According to the information provided, the direction of propagation of the electromagnetic wave is on the X axis, which for practical purposes we will denote as [tex]\hat {i}[/tex], on the other hand it is also indicated that the magnetic field is in the Y direction, that for practical purposes we will denote it as [tex]\hat {j}[/tex]. In this way using the previous equation we would have to,
[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]
[tex]\hat{i} = \vec{E} \times \hat{j}[/tex]
The cross product identity is
[tex]\hat{i}=-\hat{k} \times \hat{j}[/tex]
From the equation we can notice that the electric field would be given by,
[tex]\vec{E} = -\hat{k}[/tex]
Therefore the direction of electric field is negative z-axis.
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationary nitrogen nucleus of mass 14m. Find the velocities of the proton and the nitrogen nucleus after the collision.
Answer:
[tex]V_p = 267.258 m/s[/tex]
[tex]V_n = 38.375 m/s[/tex]
Explanation:
using the law of the conservation of the linear momentum:
[tex]P_i = P_f[/tex]
where [tex]P_i[/tex] is the inicial momemtum and [tex]P_f[/tex] is the final momentum
the linear momentum is calculated by the next equation
P = MV
where M is the mass and V is the velocity.
so:
[tex]P_i = m(270 m/s)[/tex]
[tex]P_f = mV_P + M_nV_n[/tex]
where m is the mass of the proton and [tex]V_p[/tex] is the velocity of the proton after the collision, [tex]M_n[/tex] is the mass of the nucleus and [tex]V_n[/tex] is the velocity of the nucleus after the collision.
therefore, we can formulate the following equation:
m(270 m/s) = m[tex]V_p[/tex] + 14m[tex]V_n[/tex]
then, m is cancelated and we have:
270 = [tex]V_p[/tex] + [tex]14V_n[/tex]
This is a elastic collision, so the kinetic energy K is conservated. Then:
[tex]K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2[/tex]
and
Kf = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]
then,
[tex]\frac{1}{2}m(270)^2[/tex] = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]
here we can cancel the m and get:
[tex]\frac{1}{2}(270)^2[/tex] = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]
now, we have two equations and two incognites:
270 = [tex]V_p[/tex] + [tex]14V_n[/tex] (eq. 1)
[tex]\frac{1}{2}(270)^2[/tex] = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]
in the second equation, we have:
36450 = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex] (eq. 2)
from this last equation we solve for [tex]V_n[/tex] as:
[tex]V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]
and replace in the other equation as:
270 = [tex]V_p +[/tex] 14[tex]\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]
so,
[tex]V_p = -267.258 m/s[/tex]
Vp is negative because the proton go in the -i hat direction.
Finally, replacing this value on eq. 1 we get:
[tex]V_n = \frac{270+267.258}{14}[/tex]
[tex]V_n = 38.375 m/s[/tex]
The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.Express your answer using one significant figure.P = ? Pa
Answer:
[tex]1\times 10^{8}\ Pa[/tex]
Explanation:
k = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]
r = Radius of gas molecule = [tex]2\times 10^{-10}\ m[/tex]
t = Temperature = 300 K
P = Pressure
Volume of gas per molecule is given by
[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi (2\times 10^{-10})^3\\\Rightarrow V=3.35103\times 10^{-29}\ m^3[/tex]
From the ideal gas law we have
[tex]PV=kt\\\Rightarrow P=\frac{kt}{V}\\\Rightarrow P=\frac{1.38\times 10^{-23}\times 300}{3.35103\times 10^{-29}}\\\Rightarrow P=123544104.35\ Pa=1\times 10^{8}\ Pa[/tex]
The pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior is [tex]1\times 10^{8}\ Pa[/tex]
The pressure at which the finite volume of oxygen molecules causes deviations from ideal-gas behavior can be estimated using the ideal gas law and considering the volume of a molecule. At ordinary temperatures, the approximate pressure is 2.04 × 10^11 Pa.
Explanation:To estimate the pressure at which the finite volume of oxygen molecules causes noticeable deviations from ideal-gas behavior, we need to equate the volume of gas per molecule to the volume of a molecule. The volume of the gas per molecule is equivalent to the volume of a sphere with a radius of 2.0 × 10-10 m, which is approximately 4.19 × 10-30 m3. Using the ideal gas law, we can calculate the pressure using the formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Assuming 1 mole of oxygen gas (which contains 6.022 × 1023 molecules), the number of moles can be calculated by dividing the volume of the gas per molecule by Avogadro's number. The approximate pressure at which deviations from ideal-gas behavior become noticeable is around 2.04 × 1011 Pa (Pascals).
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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction µ between the object and the surface. The object has speed v when it reaches x=0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x=0 on its return trip, it stops.
1. Find k, the spring constant.
2. Express k in terms of µ, m, g, and v.
Answer:
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Explanation:
The mass of object is m
The coefficient of friction is μ
the speed of object at x= 0 is v
when the object compress the springs stops so,
The kinetic energy of the spring is equal to and friction work is done
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x[/tex]...........
after that it returns from the rest and finally comes to rest after coming to x=0
position.
So, now the potential energy of the spring is equal to the work done by the friction force.
[tex]\frac{1}{2}kx^2= \mu mgx[/tex]
[tex]\frac{1}{2}kx= \mu mg[/tex]
[tex]x= \frac{2\mu mg}{x}[/tex]
substitute the x value in equation 1
mv^2 = kx^2+2\mu (mg)x
[tex]mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}[/tex]
solving this we get
[tex]mv^2= \frac{8m^2\mu^2g^2}{k}[/tex]
therefore the force constant
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
What is the spring constant?Spring Constant is the measure of a spring's stiffness.
Given to us
Mass of object = m
The coefficient of friction = μ
The speed of the object at (x= 0) = v
When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,
The kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x[/tex]
The kinetic energy of the object after complete compression,
[tex]\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}[/tex]
Substitute the value of x in the kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }[/tex]
Hence, the value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
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A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)
Answer:
-252.52
Explanation:
L = Distance between lenses = 10 cm
D = Near point = 25 cm
[tex]f_o[/tex] = Focal length of objective = 0.9 cm
[tex]f_e[/tex] = Focal length of eyepiece = 1.1 cm
Magnification of a compound microscope is given by
[tex]m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52[/tex]
The angular magnification of the compound microscope is -252.52
The wavelength range for infrared radiation is 10^-5 m, while that of ultraviolet radiation is 10^-8 m. Which type of radiation has more energy and why?A.) Ultraviolet because it has a higher frequencyB.) Ultraviolet because it has a longer wavelengthC.) Infrared because it has a lower frequencyD.) Infrared because it has a shorter wavelength.
Answer:
A,
Explanation:
To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which
[tex]E = \frac{hc}{\lambda}[/tex]
Where,
h = Plank's constant [tex](6.626*10^{-34}J.s)[/tex]
c = Speed of light [tex](3*10^8m/s)[/tex]
[tex]\lambda[/tex] =Wavelength
If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.
The larger the amplitude of the wave, the smaller the energy.
On the other hand we have the frequency value defined as
[tex]f = \frac{c}{\lambda}[/tex]
In this case the frequency is also inversely proportional to the wavelength.
In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.
In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.
Please help I have marked the circles with numbers which one is it anybody knows?
Where is the natural light display called aurora borealis located?
Answer:
hi sandra!!
Explanation:
the number 3 is the correct!!
The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth’s atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.
Auroras that occur in the northern hemisphere are called ‘Aurora Borealis’ or ‘northern lights’ and auroras that occur in the southern hemisphere are called ‘Aurora Australis’ or ‘southern lights’.
Auroral displays can appear in many differents colours, but green is the most common. Colours such as red, yellow, green, blue and violet are also seen occasionally. The auroras can appear in many forms, from small patches of light that appear out of nowhere to streamers, arcs, rippling curtains or shooting rays that light up the sky with an incredible glow.
Answer:
Option-(3): The aurora borealis are the formation of the colored patterns on the pole of the planet, but other celestial bodies also experience such effects due to the heat flares or the sun flares when exposed to them.
Explanation:
Aurora borealis:
The natural phenomenon which is caused in most of the celestial bodies, but when our planet earth is exposed to such electromagnetic rays or flares released from the Sun, when the energy level on the surface increases and the flare is cut off from the environment of the sun. As, starts motion towards the different celestial bodies present in the solar system.
And when it reaches earth, the earth surface reflects the high in energy atoms coming towards its atmosphere. As the magnetic field present on the earth surface interacts with the Sun's flare and then it cut off's into two moving towards the two poles of the planet.
And when the molecules cools down, they are seen are different patterns on the poles.Which are also called as the aurora borealis in scientific terms.a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially at rest. Assuming the collision takes place in nearly empty space (that is no external forces are involved so momentum will be conserved), what are the velocities of the proton and helium after the collision?
Answer:
Velocity of the helium nuleus = 1.44x10⁴m/s
Velocity of the proton = 2.16x10⁴m/s
Explanation:
From the conservation of linear momentum of the proton collision with the He nucleus:
[tex] P_{1i} + P_{2i} = P_{1f} + P_{2f] [/tex] (1)
where [tex]P_{1i}[/tex]: is the proton linear momentum initial, [tex]P_{2i}[/tex]: is the helium nucleus linear momentum initial, [tex]P_{1f}[/tex]: is the proton linear momentum final, [tex]P_{2f}[/tex]: is the helium nucleus linear momentum final
From (1):
[tex] m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex] (2)
where m₁ and m₂: are the proton and helium mass, respectively, [tex]v_{1i}[/tex] and [tex]v_{2i}[/tex]: are the proton and helium nucleus velocities, respectively, before the collision, and [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex]: are the proton and helium nucleus velocities, respectively, after the collision
By conservation of energy, we have:
[tex] K_{1i} + K_{2i} = K_{1f} + K_{2f} [/tex] (3)
where [tex]K_{1i}[/tex] and [tex]K_{2i}[/tex]: are the kinetic energy for the proton and helium, respectively, before the colission, and [tex]K_{1f}[/tex] and [tex]K_{2f}[/tex]: are the kinetic energy for the proton and helium, respectively, after the colission
From (3):
[tex] \frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] (4)
Now we have two equations: (2) ad (4), and two incognits: [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex].
Solving equation (2) for [tex]v_{1f}[/tex], we have:
[tex] v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} [/tex] (5)
From getting (5) into (4) we can obtain the [tex]v_{2f}[/tex]:
[tex] v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0 [/tex]
[tex] v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0 [/tex]
From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:
[tex] v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} [/tex] (6)
Now, by introducing (6) into (5) we get the proton velocity after the collision:
[tex] v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4} [/tex]
[tex] v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s} [/tex]
The negative sign means that the proton is moving in the opposite direction after the collision.
I hope it helps you!
Final answer:
In an elastic collision between a proton and a helium nucleus, their final velocities can be found using the principles of conservation of momentum and kinetic energy.
Explanation:
In an elastic collision, both momentum and kinetic energy are conserved. To find the final velocities of the proton and helium nucleus after the collision, we can use the principles of conservation of momentum and kinetic energy.
Since the helium nucleus is initially at rest, its momentum before the collision is zero. The momentum before the collision can be calculated as:
Initial momentum proton = proton mass × proton velocity
Final momentum proton + Final momentum helium nucleus = proton mass × proton final velocity + helium nucleus mass × helium nucleus final velocity
Using the conservation of momentum, we can solve for the final velocities of the proton and helium nucleus. The final kinetic energy of the helium nucleus can also be calculated using the equation:
Final kinetic energy helium nucleus = 0.5 × helium nucleus mass × helium nucleus final velocity^2
A rotating wheel accelerates at a constant rate from an angular speed of 25 rad/s to 37 rad/s in a time interval of 8 s. (a) What is the angular acceleration in rad/s/s? rad/s/s (b) What is the average angular speed? rad/s (c) What is the angle in radians through which the wheel rotates? rad (d) What is this angle in degrees?
The value of all options are mathematically given as
a) a=1.5 rad/s^2
b)theta=248 rad
c) theta '=248 radians
d) angles in degree=14207.511
What is the angular acceleration in rad/s/s?Question Parameter(s):
Generally, the equation for the final angular speed is mathematically given as
w2=w+at
Therefore
37=25+a 8
a=1.5 rad/s^2
What is the average angular speed?b)
Average angular speed
[tex]\theta =\omega _1t+\frac{1}{2}\alpha t^2[/tex]
[tex]\theta =25\times 8+\frac{1}{2}\times 1.5\times 8^2[/tex]
theta=248 rad
Hence
theta =200+48
theta=248 rad
What is the angle in radians through which the wheel rotates, in rad and degree?In conclusion,
c) theta '=248 radians
d) angles in degree=14207.511
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Final answer:
The angular acceleration is 1.5 rad/s², the average angular speed is 31 rad/s, the angle through which the wheel rotates is 320 rad, and the angle in degrees is 18333 degrees.
Explanation:
(a) To find the angular acceleration, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time interval. Plugging in the given values, we get: angular acceleration = (37 rad/s - 25 rad/s) / 8 s = 1.5 rad/s².
(b) The average angular speed can be found by taking the average of the initial and final angular velocities. So, the average angular speed = (25 rad/s + 37 rad/s) / 2 = 31 rad/s.
(c) The angle through which the wheel rotates can be found using the formula: angle = initial angular velocity * time + (1/2) * angular acceleration * time². Plugging in the given values, we get: angle = 25 rad/s * 8 s + (1/2) * 1.5 rad/s² * (8 s)² = 320 rad.
(d) To convert radians to degrees, we use the conversion factor: 1 radian = 180 degrees / pi. So, the angle in degrees = 320 rad * 180 degrees / pi = 18333 degrees (rounded to the nearest whole number).
Four point masses, each of mass 1.3 kg are placed at the corners of a rigid massles square of side 1.1 m. Find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.
Answer:
I= 6.292 kg.m²
Explanation:
Given that
m = 1.3 kg
Side of square a= 1.1 m
The distance r
[tex]r=\sqrt{{a^2}+{a^2}}[/tex]
[tex]r={a}{\sqrt 2}[/tex]
[tex]r={1.1}{\sqrt 2}[/tex]
The moment of inertia I
The axis passes through one of the mass then the distance of the that mass from the axis will be zero.
I = m a² + m a² + m r²
By putting the values
I = m a² + m a² + m r²
I =m( 2 a² + r²)
I =1.3( 2 x 1.1² + 2 x 1.1²)
I = 1.3 x 4 x 1.1² kg.m²
I= 6.292 kg.m²
5.00 g of molten NaCl is at its melting temperature. The salt is added to 25.00 mL water at 25.0 °C in a coffee cup calorimeter. The salt becomes solid and the temperature of the water increases to 47.9 °C. a) Calculate the amount of heat transferred.
Answer:
heat absorbed by water = 2.39 KJ
Explanation:
given,
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Temperature increased to = 47.9 °C
heat absorbed by water = m S ΔT
S is specific heat in J/gm °C
Specific heat of water = 4.184 J/gm °C
m is mass of water
heat absorbed by water = 25 x 4.184 x (47.9-25)
= 25 x 4.184 x 22.9
= 2395.34 J
heat absorbed by water = 2.39 KJ
The amount of heat transferred is 2.39 KJ.
What is Heat?
This is the form of energy which is transferred from one body to another as the result of a difference in temperature.
Parameters
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Final Temperature of water = 47.9 °C
Specific heat of water = 4.184 J/gm °C
Heat = m S ΔT
Heat absorbed by water = 25 x 4.184 x (47.9 -25)
= 25 x 4.184 x 22.9 = 2395.34 J
= 2.39 KJ.
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A boy exerts a force of 11.8 N at 28.0° above the horizontal on a 6.15 kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.75 m, assuming the sled starts with an initial speed of 0.370 m/s and slides horizontally without friction.
Explanation:
It is given that,
Force exerted by a boy, F = 11.8 N
Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]
Mass of the sled, m = 6.15 kg
Distance moved, d = 2.75 m
Initial speed, u = 0.37 m/s
Let W is the work done by the boy. Using the expression for the work done to find it as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=11.8\times 2.75\ cos(28)[/tex]
W = 28.65 joules
Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]
[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]
v = 3.07 m/s
So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.
The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.
To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.
First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:
[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]
[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]
Now, we can calculate the work done:
[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]
[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]
\[ W \approx 28.64 \, \text{J} \]
Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):
[tex]\[ W = \Delta KE \][/tex]
[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]
Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:
[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]
[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]
[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]
Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of a bucket of water is 20 kg. Someone cranks the bucket up and then lets go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?
To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.
Explanation:The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.
First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.
The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.
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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².
To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:
I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²
where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:
I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²
The torque (τ) caused by the falling bucket is:
τ = Force[tex]\times[/tex] RadiusThe force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):
F = 20 kg [tex]\times[/tex] 9.8 m/s² = 196 NSo,
τ = 196 N [tex]\times[/tex] 0.125 m = 24.5 N mUsing Newton's second law for rotation:
τ = I[tex]\times[/tex] αwhere α is the angular acceleration:
24.5 N m = 0.390625 kg m²[tex]\times[/tex] αSolving for α:
α = 24.5 N m / 0.390625 kg m² ≈ 62.7 rad/s²Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical surface with radius 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the concave surface.A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface. (answer is s1 in units cm)B) Find the height of the image formed by paraxial rays incident on the convex surface. (answer is y1 in units mm)C) Is the image erect or inverted?
Answer:
a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual
Explanation:
This is an optical exercise, where the constructor equation should be used
1 / f = 1 / p + 1 / q
Where f is the focal length, p the distance to the object and q the distance to the image
A) The cocal distance is framed with the relationship
1 / f = (n₂-1) (1 /R₁ -1 /R₂)
In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave
1 / f = (1.60 -1) (1 /∞ - 1 / (-4))
1 / f = 0.6 / 4 = 0.15
f = 6.67 cm
We have the distance to the object p = 24.0 cm, let's calculate
1 / q = 1 / f - 1 / p
1 / q = 1 / 6.67 - 1/24
1 / q = 0.15 - 0.04167 = 0.10833
q = -9.23 cm
distance to the negative image is before the lens
B) the magnification of the lenses is given by
M = h ’/ h = - q / p
h’= - q / p h
h’= - (-9.23) / 24.0 0.150
h’= 0.05759 cm
h’= 0.577 mm
C) the object is after the focal length, therefore, the image is right and virtual