Answer:Aluminium
Explanation:
Given
Aluminium ball sinks after placing inside the water while wooden block floats.
Buoyancy on a body is given by
[tex]volume\ inside\ the\ water \times density\ of\ fluid\times g[/tex]
here [tex]\rho _{Al}>\rho _{water}[/tex]
thus aluminium ball sinks
but for
[tex]\rho _{wood}<\rho _{water}[/tex]
only a part of wood is under the water i.e. not whole volume is not under water therefore net buoyancy is less in case of wood
A vertical solid steel post of diameter d = 26cmand length L = 2.40m is required to support a load of mass m = 7800kg . You can ignore the weight of the post. Take free fall acceleration to be g=9.8m/s2.Part AWhat is the stress in the post?Express your answer using two significant figures.Part BWhat is the strain in the post?Express your answer using two significant figures.Part CWhat is the change in the post's length when the load is applied?Express your answer using two significant figures.
Answer
given,
diameter of steel = d = 26 c m
radius = 13 cm = 0.13 m
length = L = 2.4 m
mass = 7800 Kg
g = 9.8 m/s²
a) stress = F/A
stress = [tex]\dfrac{mg}{\pi\ r^2}[/tex]
stress = [tex]\dfrac{7800 \times 9.8}{\pi\ 0.13^2}[/tex]
stress = 1.44 x 10⁶ N/m²
b) Young's modulus x strain = stress
Young's modulus for steel = 200 x 10⁹ N/m²
200 x 10⁹ x strain = 1.44 x 10⁶
strain = 7.2 x 10⁻⁶ m
c) change in length
[tex]Strain= \dfrac{\Delta L}{L}[/tex]
[tex]7.2 \times 10^{-6} = \dfrac{\Delta L}{2.4}[/tex]
[tex]\Delta L= 17.28\times 10^{-6}\ m[/tex]
A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 626 nm (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 290 nm and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?
nfilm = 1Your answer is incorrect.I did t=(m)(wavelengthfilm)/(2) solving for the wavelegth of filmThenI did: wavelength film = wavelength vaccum / n and my n comes out as 1.18 which is the wrong answer can anyone help??
Answer:
1.08
Explanation:
This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then
path difference created will be 2μ t.
For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference
path diff = nλ , for minimum t , n =1
path diff = λ
2μ t. = λ
μ = λ / 2t
= 626 / 2 x 290
= 1.08
A 20-cm-long spring is attached to a wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 22 cm. The same spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end. How long is the stretched spring in cm? (Do not include unit in answer)
Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm
A horizontal pipe has an abrupt expansion from D1 = 5 cm to D2 = 10 cm. The water velocity in the smaller section is 8 m/s and the flow is turbulent. The pressure in the smaller section is P1 = 380 kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure P2, and estimate the error that would have occurred if Bernoulli’s equation had been used. Take the density of water to be rho = 1000 kg/m3.
Answer:
P₂ = 392720.38 Pa = 392.72 kPa
Explanation:
Given
D₁ = 5 cm = 0.05 m
D₂ = 10 cm = 0.10 m
v₁ = 8 m/s
P₁ = 380 kPa = 380000 Pa
α = 1.06
ρ = 1000 kg/m³
g = 9.8 m/s²
We can use the following formula
(P₁ / (ρg)) + α*(V₁² / (2g)) + z₁ = (P₂ / (ρg)) + α*(V₂² / (2g)) + z₂ + +hL
knowing that z₁ = z₂ we have
(P₁ / (ρg)) + α*(V₁² / (2g)) = (P₂ / (ρg)) + α*(V₂² / (2g)) + +hL (I)
Where
V₂ can be obtained as follows
V₁*A₁ = V₂*A₂ ⇒ V₁*( π* D₁² / 4) = V₂*( π* D₂² / 4)
⇒ V₂ = V₁*(D₁² / D₂²) = (8 m/s)* ((0.05 m)² / (0.10 m)²)
⇒ V₂ = 2 m/s
and
hL is a head loss factor: hL = α*(1 - (D₁² / D₂²))²*v₁² / (2*g)
⇒ hL = (1.06)*(1 – ((0.05 m) ² / (0.10 m)²))²*(8 m/s)² / (2*9.8 m/s²)
⇒ hL = 1.9469 m
Finally we get P₂ using the equation (I)
⇒ P₂ = P₁ - ((V₂² - V₁²)* α*ρ / 2) – (ρ*g* hL)
⇒ P₂ = 380000 Pa - (((2 m/s)² - (8 m/s)²)*(1.06)*(1000 kg/m³) / 2) – (1000 kg/m³*9.8 m/s²*1.9469 m)
⇒ P₂ = 392720.38 Pa = 392.72 kPa
The velocity of the water in the larger section of the pipe is 3.2 m/s and the pressure is 412 kPa. The downstream pressure P2 is 356.7 kPa and the error if Bernoulli's equation had been used is approximately 55.3 kPa.
Explanation:First, we can calculate the velocity of the water in the larger section of the pipe using the continuity equation, which states that the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region. Using this equation, we can find that the velocity in the larger section is 3.2 m/s.
Next, we can use Bernoulli's equation to solve for the pressure in the larger section of the pipe. Bernoulli's equation states that the pressure in a fluid decreases as the velocity increases. Plugging in the given values and the calculated velocity of 3.2 m/s, we can find that the pressure in the larger section is 412 kPa.
Finally, we can use the pressure drop equation, P2 - P1 = RQ, to solve for the downstream pressure P2. The resistance R can be calculated using the kinetic energy correction factor and the given values. Plugging in all the values, we can find that the downstream pressure P2 is 356.7 kPa. To estimate the error that would have occurred if Bernoulli's equation had been used, we can calculate the difference between the actual downstream pressure and the pressure calculated using Bernoulli's equation. This error is approximately 55.3 kPa.
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What trend is visible in the sandstone beds as they are traced from east to west?
a There is no information on how the sandstone beds change from east to west.
b. The sandstone beds grade into conglomerate from east to west.
c. The sandstone beds become thinner to the west.
d. The sandstone beds become thicker to the west.
Answer: d.
Explanation:
The sandstone beds become thicker to the west.
Based on the information provided, there is no explicit trend mentioned in the sandstone beds from east to west. To find a trend, one would need to look for changes in grain size, sediment type, or sedimentary structures, none of which are detailed in the given data.
Explanation:The trend visible in the sandstone beds as they are traced from east to west is not explicitly listed in the provided information. However, looking at the descriptions, it seems that no direct trend such as grading into conglomerate, becoming thinner, or becoming thicker is mentioned.
To determine such a trend, geologists might look for changes in the grain size, sediment type, or sedimentary structures within the sandstone layers. For example, if the beds were observed to contain increasingly larger grain sizes, or a transition to a different rock type such as conglomerate, that might indicate a trend in energy conditions and depositional environments.
Since no such trend is described in the information given, the most appropriate answer based on the provided options would be There is no information on how the sandstone beds change from east to west, which corresponds to option a.
You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the ball or deflect the ball back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to minimize your speed on the skateboard? Choose one of the following three answers.
(A) Your final speed on the skateboard will be the same regardless whether you catch the ball or deflect the ball.
(B) You should catch the hall.
(C) You should deflect the ball
Answer:
(B) You should catch the hall
Explanation:
B) You should catch the ball.
case 1 : when ball is catched.
Mass of person catching = M
mass of ball = m
velocity of the ball just before catching = v
velocity of ball just after catching = 0
[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (0 - v) = - mv
since,
[tex]\Delta P_{person} =-\Delta P_{ball}[/tex]
[tex]\Delta P_{person} =mv[/tex]
case 2 : when the ball is deflected back
velocity of the ball just before catching = v
velocity of ball just after catching = - v
[tex]\Delta P_{ball}[/tex] = change in momentum of the ball = m (- v - v) = - 2mv
since,
[tex]\Delta P_{person} =- \Delta P_{ball}[/tex]
[tex]\Delta P_{person} = 2mv[/tex]
clearly we can see that the change in momentum is minimum in case when the ball is catched and hence keeps the speed of skateboard minimum
Super Invar, an alloy of iron and nickel, is a strong material with a very low coefficient of thermal expansion (0.20× 10−6 C∘). A 1.9-m-long tabletop made of this alloy is used for sensitive laser measurements where extremely high tolerances are required.Part AHow much will this alloy table expand along its length if the temperature increases 5.5 C∘?Express your answer to two significant figures and include the appropriate units.Part BHow much will a 1.8-m-long table made of steel expand along its length if the temperature increases5.5 C∘?Express your answer to two significant figures and include the appropriate units.
Answer:
a) 2.1 × 10^-6 m b) 1.3 × 10^-4 m
Explanation:
The thermal expansion of super Invar 0.20 ×10^-6oC^-1 and the length of Super is 1.9 m
using the linear expansivity formula
ΔL = αLoΔt where ΔL is the change in length meters, α is the linear expansivity of Super Invar in oC^-1 and Δt is the change in temperature in oC. Substituting the value into the formula gives
ΔL = 0.2 × 10 ^-6 × 1.9 m ×5.5 = 2.1 × 10^-6 meters to two significant figure
b) the thermal expansion of steel is 1.3 × 10^-5 oC^-1 and the length of steel is 1.8 m
using the formula
ΔL of steel = 1.3 × 10^-5 × 1.8 × 5.5 = 1.3 × 10^-4 m to two significant figure.
Two-out-of-tune flutes play the same note. One produces a tone that has a frequency of 250 Hz, while the other produces 266 Hz. When a tuning fork is sounded together with the 250-Hz tone, a beat frequency of 9 Hz is produced. When the same tuning fork is sounded together with the 266-Hz tone, a beat frequency of 7 Hz is produced. What is the frequency of the tuning fork?
Answer:
x = 259 Hz
Explanation:
given,
frequency of one tuning fork = 250 Hz
frequency of another tuning fork = 266 Hz
when a tuning fork is sounded together beat frequency heard = 9
let x be the frequency of unknown
x - 250 = 9 Hz..............(1)
x = 259 Hz
when a another tuning fork is sounded together beat frequency heard = 7
266 - x = 7 Hz..............(2)
x = 259 Hz
now, on solving both the equation the frequency comes out to be 259 Hz.
so, The frequency of the tuning fork is equal to 259 Hz
Joe the house painter stands on a uniform oak board weighing 600 N and held up by vertical ropes at each end. Joe has been dieting, and now weighs 844 N. The length of the board is 4.00 m, and Joe is standing 1.00 m from the left end of the board. What is the tension in each of the supporting ropes?
Answer
given,
weight of the oak board = 600 N
Weight of Joe = 844 N
length of board = 4 m
Joe is standing at 1 m from left side
vertical wire is supporting at the end.
Assuming the system is in equilibrium
T₁ and T₂ be the tension at the ends of the wire
equating all the vertical force
T₁ + T₂ = 600 + 844
T₁ + T₂ = 1444...........(1)
taking moment about T₂
T₁ x 4 - 844 x 3 - 600 x 2 = 0
T₁ x 4 = 3732
T₁ = 933 N
from equation (1)
T₂ = 1444 - 933
T₂ = 511 N
Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first law of thermodynamics, What is the value (including algebraic sign) of W, the work done by the system?
To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.
[tex]dQ = dU + dW[/tex]
Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
[tex]W = -168J \righarrow[/tex] Work is done ON the system
[tex]Q = -305.6J \rightarrow[/tex] Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a mass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?
Answer:
Velocity of the first car after the collision, [tex]v_1=8.93\ m/s[/tex]
Explanation:
It is given that,
Mass of the car, [tex]m_1 = 480\ kg[/tex]
Initial speed of the car, [tex]u_1 = 14.4\ m/s[/tex]
Mass of another car, [tex]m_2 = 570\ kg[/tex]
Initial speed of the second car, [tex]u_2 = 13.3\ m/s[/tex]
New speed of the second car, [tex]v_2 = 17.9\ m/s[/tex]
Let [tex]v_1[/tex] is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]m_1u_1+m_2u_2-m_2v_2=m_1v_1[/tex]
[tex]480\times 14.4+570\times 13.3-570\times 17.9=480v_1[/tex]
[tex]v_1=8.93\ m/s[/tex]
So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine the gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon. (b) Repeat the calculation when Earth has rotated so that the Pacific Ocean faces toward the Moon. (c) Estimate the maximum speed of the water in the Pacific Ocean due to the tidal influence of the Moon. For the sake of the calculations, treat the Pacific Ocean as a pointlike object (obviously a very rough approximation)
To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.
Potential Energy because of a planet would be given by the equation,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Ocean
M = Mass of Moon
r = Radius
From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.
When it is far away, it will be the total diameter from the center of the earth to the center of the moon.
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
PART A) Potential energy when the ocean is at its furthest point to the moon,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) Potential energy when the ocean is at its closest point to the moon
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]
PART C) The maximum speed. This can be calculated through the conservation of energy, where,
[tex]\Delta KE = \Delta PE[/tex]
[tex]\frac{1}{2}mv^2 = PE_2-PE_1[/tex]
[tex]v=\sqrt{2(PE_2-PE_1)/m}[/tex]
[tex]v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}[/tex]
[tex]v = 29.4m/s[/tex]
A spherical balloon is made from a material whose mass is 2.70 kg. The thickness of the material is negligible compared to the 1.55 m radius of the balloon. The balloon is filled with helium (He) at a temperature of 290 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m³ and the molar mass of helium is 4.0026×10-3 kg/mol. Find the absolute pressure of the helium gas.
To develop this problem it is necessary to apply the concepts related to the calculation of the Force through density and volume as well as the ideal gas law.
By definition, force can be expelled as
F = ma
Where,
m = mass
a = Acceleration
At the same time the mass can be defined as function of density and Volume
[tex]m = \rho V[/tex]
Therefore if we do a sum in the spherical balloon we have,
[tex]\sum F = 0[/tex]
[tex]F_w +F_h-F_b=0[/tex]
Where,
[tex]F_W[/tex]= Force by weight of balloon
[tex]F_h[/tex]= Force by weight of helium gas
[tex]F_b[/tex]= Buoyant force
[tex]mg + V \rho g - V\rho_a g = 0[/tex]
Re-arrange to find [tex]\rho,[/tex]
[tex]\rho = \rho_a - \frac{m}{V}[/tex]
Our values are given as,
[tex]r= 1.55m[/tex]
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]V = \frac{4}{3} \pi (1.55)^3[/tex]
[tex]V = 15.59m^3[/tex]
Replacing the values we have,
[tex]\rho = 1.19kg/m^3 - \frac{2.7}{15.59}[/tex]
[tex]\rho = 1.0168kg/m^3[/tex]
Applying the ideal gas law we have finally that
[tex]P = \frac{\rho}{M_0} RT[/tex]
Where,
P = Pressure
[tex]\rho =[/tex] Density
M_0 Molar mass (0.004Kg/mol for helium)
R= Gas constant
T = Temperature
Substituting
[tex]P = \frac{1.0168}{0.004} *8.314*290[/tex]
[tex]P = 612891.452Pa[/tex]
[tex]P = 0.613Mpa[/tex]
Therefore the absolute pressure of the helium gas is [tex]0.613Mpa[/tex]
There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the direction of the electric field, E→ , if the magnetic field, B→ , is pointed in the +x direction?
Answer: The direction of the electric field, E→, is pointed in the +y direction.
Explanation:
One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.
The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (in this case the -z direction) and the curling of the fingers point in the direction of the magnetic field B→ (in this case the +x direction), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.
Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What is the least possible uncertainty within which we can determine the position of this electron? (melectron = 9.11 × 10-31 kg, h = 6.626 × 10-34 J ∙ s)
Answer: [tex]2.89(10)^{-3} m[/tex]
Explanation:
The Heisenberg uncertainty principle postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.
In other words:
It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.
Mathematically this principle is written as:
[tex]\Delta x \geq \frac{h}{4 \pi m \Delta V}[/tex] (1)
Where:
[tex]\Delta x[/tex] is the uncertainty in the position of the electron
[tex]h=6.626(10)^{-34}J.s[/tex] is the Planck constant
[tex]m=9.11(10)^{-31}kg[/tex] is the mass of the electron
[tex]\Delta V[/tex] is the uncertainty in the velocity of the electron.
If we know the accuracy of the velocity is [tex]0.001\%[/tex] of the velocity of the electron [tex]V=2 km/s=2000 m/s[/tex], then [tex]\Delta V[/tex] is:
[tex]\Delta V=2000 m/s(0.001\%)[/tex]
[tex]\Delta V=2000 m/s(\frac{0.001}{100})[/tex]
[tex]\Delta V=2(10)^{-2} m/s[/tex] (2)
Now, the least possible uncertainty in position [tex]\Delta x_{min}[/tex] is:
[tex]\Delta x_{min}=\frac{h}{4 \pi m \Delta V}[/tex] (3)
[tex]\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}[/tex] (4)
Finally:
[tex]\Delta x_{min}=2.89(10)^{-3} m[/tex]
The least possible uncertainty within which we can determine the position of this electron is 0.29 micrometers. This value is derived using the Heisenberg's Uncertainty Principle, which states that the more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.
Explanation:The subject question is dealing with the principle of uncertainty in quantum mechanics. According to Heisenberg's Uncertainty Principle, the position and momentum of a particle cannot both be accurately measured at the same time. The more precisely one measures the position of a particle, the less precisely one can measure its speed, and vice versa.
The accuracy is given as 1 part in 105, that is 0.001% or 1e-5. The speed of the electron is known to be 2.0 km/s so the uncertainty in velocity (Δv) would be 2.0 km/s * 1e-5 = 2e-8 m/s. The mass of the electron (melectron) is given to be 9.11 × 10-31 kg.
Momentum is the product of mass and velocity, so Δp = melectron x Δv = (9.11 × 10-31 kg) * (2e-8 m/s) = 1.822e-38 kg m/s. According to Heisenberg's Uncertainty Principle (ΔxΔp ≥ h/4π) the least uncertainty in position (Δx) = h / (4πΔp), where h is Planck's constant 6.626 × 10-34 J.s. Substituting the values, Δx = 6.626 × 10-34 J.s / (4π * 1.822e-38 kg m/s) = 2.9e-7 m or 0.29 μm.
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A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two pieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.35 c relative to the original nucleus. Part A Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.
Final answer:
To find the velocities of pieces A and B as measured by an observer in the laboratory, use the relativistic velocity addition formula.
Explanation:
To find the velocities of pieces A and B as measured by an observer in the laboratory, we need to use the relativistic velocity addition formula. Let's call the initial velocity of the uranium nucleus as v. Piece A is moving forward with a speed of 0.43c relative to the original nucleus and piece B is moving backward at 0.35c relative to the original nucleus.
The velocity of piece A as measured by an observer in the laboratory is given by vA = (v + vA') / (1 + v*vA'/c^2), where vA' is the velocity of piece A relative to the original nucleus. Plugging in the values, we get vA = (v + 0.43c) / (1 + v*0.43c/c^2).
The velocity of piece B as measured by an observer in the laboratory is given by vB = (v - vB') / (1 - v*vB'/c^2), where vB' is the velocity of piece B relative to the original nucleus. Plugging in the values, we get vB = (v - 0.35c) / (1 - v*0.35c/c^2).
(Serway 9th ed., 6-27) The mass of a sports car is 1200 kg. The shape of the body is such that the aerodynamic drag coefficient is 0.250 and the frontal area is 2.20 m2. Ignoring all other sources of friction, calculate the initial acceleration the car has if it has been traveling at 100 km/hr and is now shifted into neutral and allowed to coast. (Ans. 0.212 m/s2, opposite the velocity vector)
Answer:
a = - 0.248 m/s²
Explanation:
Frictional drag force
F = ½ *(ρ* v² * A * α)
ρ = density of air , ρ = 1.295 kg/m^3
α = drag coef , α = 0.250
v = 100 km/h x 1000m / 3600s
v = 27.77 m/s
A = 2.20m^2
So replacing numeric in the initial equation
F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)
F = 298.6 N
Now knowing the force can find the acceleration
a = - F / m
a = - 298.6 N / 1200 kg
a = - 0.248 m/s²
Final answer:
The question involves using aerodynamic drag force and Newton's second law to calculate the deceleration of a sports car shifted into neutral.
Explanation:
The question revolves around calculating the initial acceleration of a sports car weighing 1200 kg that has been shifted into neutral and allowed to coast, assuming the car was traveling at 100 km/hr, and given an aerodynamic drag coefficient of 0.250 and a frontal area of 2.20 m2. To calculate acceleration due to the forces acting on the car, we use the formula for aerodynamic drag force which is Fd = 0.5 × Cd × ρ × A × v2 where Cd is the drag coefficient, ρ is the air density (approximately 1.225 kg/m3 at sea level), A is the frontal area, and v is the velocity in meters per second. To find the acceleration, we can then apply Newton's second law, F = m × a, where F is the net force acting on the car, m is the mass, and a is the acceleration. In this case, the net force is opposite to the direction of the velocity vector due to drag force, thus providing the deceleration or negative acceleration of the car.
Suppose one of the Global Positioning System satellites has a speed of 4.46 km/s at perigee and a speed of 3.54 km/s at apogee. If the distance from the center of the Earth to the satellite at perigee is 2.23×104 km , what is the corresponding distance at apogee?
To solve this problem it is necessary to apply the concepts related to Kepler's second law and the conservation of angular momentum.
Kepler's second law tells us that the vector radius that unites a planet and the sun sweeps equal areas at equal times, that is, when the planet is farther from the sun, the speed at which it travels is less than when it is close to the sun.
The angular momentum is defined as
[tex]L = m*r*v[/tex]
Where,
m= mass
r = Radius
v = Velocity
For conservation of angular momentum
[tex]L_{apogee}=L_{perigee}[/tex]
[tex]mv_a*r_a = mv_p*r_p[/tex]
[tex]v_a*r_a= v_p*r_p[/tex]
[tex]r_a = \frac{v_p*r_p}{v_a}[/tex]
[tex]r_a = \frac{(4.46)(2.23*10^4)}{(3.54)}[/tex]
[tex]r_a = 2.81*10^4km[/tex]
Therefore the corresponding distance at apogee is [tex]2.81*10^4km[/tex]
A long, straight wire carrying a current of 380 A is placed in a uniform magnetic field that has a magnitude of 6.59 × 10-3 T. The wire is perpendicular to the field. Find a point in space where the net magnetic field is zero. Locate this point by specifying its perpendicular distance from the wire.
Answer:
The point in space where the net magnetic field is zero lies by specifying its perpendicular distance from the wire is 0.01153 m.
Explanation:
Given that,
Current = 380 A
Magnetic field [tex]B=6.59\times10^{-3}\ T[/tex]
We need to calculate the distance
Using formula of magnetic field
[tex]B = \dfrac{\mu_{0}I}{2\pi r}[/tex]
[tex]r=\dfrac{\mu_{0}I}{2\pi B}[/tex]
Where, B = magnetic field
I = current
Put the value into the formula
[tex]r=\dfrac{4\pi\times10^{-7}\times380}{6.59\times10^{-3}\times2\pi}[/tex]
[tex]r=0.01153\ m[/tex]
Hence, The point in space where the net magnetic field is zero lies by specifying its perpendicular distance from the wire is 0.01153 m.
The vapor pressure of the liquid HF is measured at different temperatures. The following vapor pressure data are obtained: Temperature 270.6K and 287.5K, Pressure 324.5 mmHg and 626.9 mmHG. Calculate the enthapy of vaporization ( delta H vap ) in kJ/mol for this liquid.
Answer:
Enthalpy is 44.95 kJ/mol
Solution:
As per the question:
Temperature, T = 270.6 K
Temperature, T' = 287.5 K
Pressure, P = 324.5 mmHg
Pressure, P' = 626.9 mmHg
Now,
To calculate the enthalpy, we make use of the Clausius-Clapeyron eqn:
[tex]ln\frac{P}{P'} = \frac{\Delta H}{R}(\frac{1}{T'} - \frac{1}{T})[/tex]
where
[tex]\Delta H = Enthalpy[/tex]
R = Rydberg's constant
Substituting suitable values in the above eqn:
[tex]ln\frac{324.5}{626.9} = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]
[tex]- 0.658 = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]
[tex]\Delta H = 44.95\ kJ/mol[/tex]
Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between observations is the period of the comet and take its eccentricity as 0.997. What are (a) the semimajor axis of the comet's orbit and (b) its greatest distance from the Sun?
To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that
[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]
Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by
[tex]T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s[/tex]
PART A) Replacing the values to find a, we have
[tex]a^3= \frac{T^2 GM}{4\pi^2}[/tex]
[tex]a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}[/tex]
[tex]a^3 = 6.46632*10^{39}[/tex]
[tex]a = 1.86303*10^{13}m[/tex]
Therefore the semimajor axis is [tex]1.86303*10^{13}m[/tex]
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by
[tex]R = a(1-e)[/tex]
[tex]R = 1.86303*10^{13}(1-0.997)[/tex]
[tex]R= 5.58*10^{10}m[/tex]
A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the fin is 75oC and it is exposed to flowing air at 20oC with h = 20 W/m2-K. That is the rate of heat transfer from this pin?
This question asks for the rate of heat transfer from a copper fin, which requires the application of heat transfer theory, including the use of specific equations for fin efficiency and heat loss. However, a sample calculation provided deals with finding the initial temperature of a copper piece mixed with water, based on the conservation of energy and specific heat capacities.
Explanation:The subject of this question involves the calculation of heat transfer from a fin made of copper. Given the material's thermal conductivity (k), the fin's dimensions, and the conditions surrounding it (base temperature, air temperature, and convection heat transfer coefficient h), we can find the rate of heat transfer. However, the actual calculation for the rate of heat transfer from this fin isn't provided directly in the context of the question. For such a problem, one would typically employ the fin equations from heat transfer theory, which might involve the use of Biot number, fin efficiency, and heat loss calculations. Instead, to answer a closely related question with given data, we can calculate the initial temperature of a 248-g piece of copper when dropped into 390 mL of water at 22.6°C, where the final temperature is measured as 39.9°C, assuming all heat transfer occurs between the copper and the water. This problem relies on the conservation of energy principle, requiring the application of the specific heat capacities of copper and water
A nearsighted person cannot see objects beyond 80 cm from his eyes. Which one of the followingcombinations represents the correct focal length and the refractive power of the contact lenses thatwill enable him to see the distant objects clearly?A) -80 cm, -1.3 dioptersB) -1.3 cm, +1.3 dioptersC) -80 cm, +1.3 dioptersD) +80 cm, +1.3 dioptersE) +80 cm, -1.3 diopters
Answer:
Option (A) is correct.
Explanation:
for a near sighted person, distance of object from the lens = u = ∞
distance of image from the lens, v = - 80 cm
Use lens formula
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]\frac{1}{f}=\frac{1}{-80}-\frac{1}{\infty }[/tex]
So, f = - 80 cm
Power of the lens is the reciprocal of the focal length of the lens.
P = 100/f
where, f is the focal length when it is measured in the units of cm.
P = - 100 / 80 = - 1.3 Dioptre
Thus, option (a) is correct.
While taking a shower, you notice that the shower head is made up of 44 small round openings, each with a radius of 2.00 mm. You also determine that it takes 3.00 s for the shower to completely fill a 1.00-liter container you hold in the water stream. The water for the shower is pumped by a pump that is 5.70 m below the level of the shower head. The pump maintains an absolute pressure of 1.50 atm. Use g = 10 m/s2, and assume that 1 atmosphere is 1.0 105 Pa. (a) At what speed does the water emerge from the shower head? (b) What is the speed of the water in the pipe connected to the pump? (c) What is the cross-sectional area of the pipe connected to the pump?
To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.
PART A) For the calculation of the velocity we define the area and the flow, thus
[tex]A = \pi r^2[/tex]
[tex]A = pi (2*10^{-3})^2[/tex]
[tex]A = 12.56*10^{-6}m^2[/tex]
At the same time the rate of flow would be
[tex]Q = \frac{1L}{2s}[/tex]
[tex]Q = 0.5L/s = 0.5*10^{-3}m^3/s[/tex]
By definition the discharge is expressed as
[tex]Q = NAv[/tex]
Where,
A= Area
v = velocity
N = Number of exits
Q = NAv
Re-arrange to find v,
[tex]v = \frac{Q}{NA}[/tex]
[tex]v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}[/tex]
[tex]v = 0.9047m/s[/tex]
PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe
[tex]P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2[/tex]
Replacing with our values we have that
[tex]1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)[/tex]
[tex]v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)[/tex]
[tex]v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}[/tex]
[tex]v_1 = 3.54097m/s[/tex]
PART C) Assuming that water is an incomprehensible fluid we have to,
[tex]Q_{pipe} = Q_{shower}[/tex]
[tex]v_{pipe}A_{pipe}=v_{shower}A_{shower}[/tex]
[tex]3.54097*A_{pipe}=0.9047*12.56*10^{-6}[/tex]
[tex]A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}[/tex]
[tex]A_{pipe = 3.209*10^{-6}m^2[/tex]
The answer provides calculations for the speed of water from the shower head, the speed in the connected pipe, and the cross-sectional area of the pipe based on the given parameters.
Speed from the shower head: The speed at which water emerges from the shower head can be calculated using Bernoulli's equation. Using the given data, we can determine that the speed of the water coming out of the shower head is approximately 9.06 m/s.Speed in the connected pipe: To find the speed of the water in the pipe connected to the pump, we need to consider the change in height and convert the potential energy into kinetic energy. The speed in the connected pipe would be about 10.92 m/s.Cross-sectional area of the pipe: The cross-sectional area of the pipe can be calculated using the formula A = πr². For the pipe connected to the pump, the cross-sectional area is approximately 9.16 × 10⁻⁴ m².A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes the turntable to accelerate from 0 to 24 revolutions per second in 8.0 s.Calculate the rotational acceleration.
Answer:
Angular acceleration will be [tex]18.84rad/sec^2[/tex]
Explanation:
We have given that mass m = 0.18 kg
Radius r = 0.32 m
Initial angular velocity [tex]\omega _i=0rev/sec[/tex]
And final angular velocity [tex]\omega _f=24rev/sec[/tex]
Time is given as t = 8 sec
From equation of motion
We know that [tex]\omega _f=\omega _i+\alpha t[/tex]
[tex]24=0+\alpha \times 8[/tex]
[tex]\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2[/tex]
So angular acceleration will be [tex]18.84rad/sec^2[/tex]
Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction. The impulse on the ball caused by the bat will bo in the negative x direction. Part E Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of -8.4 N s to the baseball? Enter your answer numerically in meters per second using two significant figures.
The ball's velocity just after leaving the bat is -25.93 m/s.
Explanation:To find the ball's velocity just after leaving the bat, we can use the principle of conservation of momentum. The impulse on the ball caused by the bat is equal to the change in momentum of the ball. Since impulse is defined as force multiplied by time, we can use the given impulse of -8.4 N s and the mass of the ball (0.145 kg) to find the change in velocity of the ball.
The formula for impulse is impulse = change in momentum = mass * change in velocity. Rearranging the formula, we can solve for the change in velocity: change in velocity = impulse/mass = -8.4 N s / 0.145 kg = -57.93 m/s.
Since the initial velocity of the ball was 32 m/s in the +x direction, the final velocity of the ball can be found by adding the change in velocity to the initial velocity: final velocity = initial velocity + change in velocity = 32 m/s + (-57.93 m/s) = -25.93 m/s.
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The eccentricity of an asteroid's orbit is 0.0442, and the semimajor axis is 1.12 x 1011 m. The Sun's center is at one focus of the asteroid's orbit. (a) How far from this focus is the other focus in meters? (b) What is the ratio of this distance to the solar radius, 6.96 x 108 m?
Answer:
a. [tex]d=99x10^{8}m[/tex]
b. [tex]r=14.22 R_s[/tex]
Explanation:
The eccentricity of an asteroid's is 0.0442 so
a.
to find the focus distance between both focus is
[tex]d=2*e*a[/tex]
[tex]e=1.12x10^{11}m\\a=0.0442[/tex]
So replacing numeric
[tex]d=2*1.12x10^{11}m*0.0442=9900800000[/tex]
[tex]d=99x10^{8}m[/tex]
b.
Now to find the ratio of that distance between the solar radius and the distance
[tex]r=\frac{d_1}{d_s}[/tex]
[tex]r=R_s*\frac{99x10^8m}{6.96x10^8m}[/tex]
[tex]r=14.22 R_s[/tex]
When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ice cube in a glass of some liquid whose density is less than that of water. How much of the ice cube will be submerged below the surface of this liquid? a. More than 90% b. 90% c. Less than 90%
When an ice cube, which floats with 90% submerged in water, is placed in a liquid that is less dense than water, less than 90% of the cube will be submerged. This is due to the interplay between the densities of the ice cube and the liquid.
Explanation:The extent to which an object is submerged in a fluid depends on the density of both the object and the fluid. In the case of an ice cube in water, the ice cube is less dense than the water, causing about 90% of it to be submerged under the surface. Now if you place the same ice cube in a liquid that is less dense than water, less than 90% of the ice cube will be submerged. This is because the ice cube is denser than this new liquid, and it displaces less liquid to balance its weight.
Learn more about Density and Buoyancy here:https://brainly.com/question/7103480
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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4 cm for 4 seconds, what linear speed does the wheel achieve, assuming it rolls without slipping?
a) 3 m/s
b) 24 m/s
c) 5.9 m/s
d) 7.1 m/s
To solve the problem it is necessary to apply the Torque equations and their respective definitions.
The Torque is defined as,
[tex]\tau = I \alpha[/tex]
Where,
I=Inertial Moment
[tex]\alpha =[/tex] Angular acceleration
Also Torque with linear equation is defined as,
[tex]\tau = F*d[/tex]
Where,
F = Force
d= distance
Our dates are given as,
R = 30 cm = 0.3m
m = 1.5 kg
F = 20 N
r = 4.0 cm = 0.04 m
t = 4.0s
Therefore matching two equation we have that,
[tex]d*F = I\alpha[/tex]
For a wheel the moment inertia is defined as,
I= mR2, replacing we have
[tex]d*F= \frac{mR^2a}{R}[/tex]
[tex]d*F= mRa[/tex]
[tex]a = \frac{rF}{ mR}[/tex]
[tex]a = \frac{0.04*20}{1.5*0.3}[/tex]
[tex]a=1.77 m/s^2[/tex]
Then the velocity of the wheel is
[tex]V = a *t \\V=1.77*4 \\V=7.11 m/s[/tex]
Therefore the correct answer is D.
A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were traveling at a velocity of 2v?
a. 4d
b. 2d
c. 8d
d. √2 d
e. d
Answer:
a. 4d.
If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.
Explanation:
Hi there!
The equations of distance and velocity of the car are the following:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the car at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity of the car at time t.
Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:
velocity = v0 + a · t v0 = v
0 = v + a · t
Solving for t:
-v/a = t
Since the acceleration is negative because the car is stopping:
v/a = t
Now replacing t = v/a in the equation of position:
x = x0 + v0 · t + 1/2 · a · t² (let´s consider x0 = 0)
x = v · (v/a) + 1/2 · (-a) (v/a)²
x = v²/a - 1/2 · v²/a
x = 1/2 v²/a
At a velocity of v, the stopping distance is 1/2 v²/a = d
Now, let´s do the same calculations with an initial velocity v0 = 2v:
Using the equation of velocity:
velocity = v0 + a · t
0 = 2v - a · t
-2v/-a = t
t = 2v/a
Replacing in the equation of position:
x1 = x0 + v0 · t + 1/2 · a · t²
x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²
x1 = 4v²/a - 2v²/a
x1 = 2v²/a
x1 = 4(1/2 v²/a)
x1 = 4x
x1 = 4d
If the car travels at a velocity 2v, the minimum stopping distance will be 4d.
Answer:
4d
Explanation: