Answer:
1556.906J/K=ΔS
Explanation:
Entropy is a thermodynamic property that measures the level of molecular disorder in a substance.
This property is already calculated for all pressure and temperature values.
Therefore, to solve this problem we must use thermodynamic tables for water and calculate the specific entropy in the two states, finally multiply by the mass to find the entropy change.
the entropy change is given by the following equation
ΔS=m(s2-s1)
where
ΔS=change in the entropy
s= especific entropy
m=mass=257g=0.257Kg
for the state 1: entropy for liquid water at 99°C
s1=1296J/kgK
for the state 2: entropy for steam at 100°C
s2=7354J/kgK
solving
ΔS=0.257(7354-1296)=1556.906J/K
A golfer takes three strokes to putt a golf ball into a hole. On the first stroke, the ball moves 4.8 m due east. On the second, it moves 2.7 m at an angle 20° north of east. On the third, it moves 0.50 m due north. If the golfer had instead hit the ball directly into the hole on the first stroke, what would have been the magnitude (in m) and direction of the ball's displacement? Give the direction as a positive angle measured counterclockwise from due east.
The magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.
1. First stroke:
[tex]Displacement \( \vec{d}_1 = 4.8 \, \text{m} \) due east.[/tex]
2. Second stroke:
[tex]Eastward component: \( 2.7 \, \text{m} \times \cos(20^\circ) \approx 2.564 \, \text{m} \) due east.\\ Northward component: \( 2.7 \, \text{m} \times \sin(20^\circ) \approx 0.924 \, \text{m} \) due north.\\\\ Displacement \( \vec{d}_2 = 2.564 \, \text{m} \) due east and \( 0.924 \, \text{m} \) due north.[/tex]
3. Third stroke:
Displacement[tex]\( \vec{d}_3 = 0.50 \, \text{m} \)[/tex] due north.
Now let's calculate the total eastward and northward displacements:
Total eastward displacement:[tex]\( 4.8 \, \text{m} + 2.564 \, \text{m} = 7.364 \, \text{m} \)[/tex]
Total northward displacement: [tex]\( 0.924 \, \text{m} + 0.50 \, \text{m} = 1.424 \, \text{m} \)[/tex]
Using the Pythagorean theorem, the magnitude of the total displacement is:
[tex]Magnitude = \( \sqrt{(\text{Total eastward displacement})^2 + (\text{Total northward displacement})^2} \)Magnitude = \( \sqrt{7.364^2 + 1.424^2} \approx 7.579 \, \text{m} \)[/tex]
The direction of the total displacement is given by:
[tex]Direction = \( \tan^{-1}\left(\frac{\text{Total northward displacement}}{\text{Total eastward displacement}}\right) \)\\Direction = \( \tan^{-1}\left(\frac{1.424}{7.364}\right) \approx 11.08^\circ \) counterclockwise from due east.[/tex]
So, if the golfer had hit the ball directly into the hole on the first stroke, the magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.
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Final answer:
The golfer's ball would have a direct displacement of approximately 7.434 meters in a direction of roughly 11° north of east if it were hit directly into the hole.
Explanation:
To determine the direct displacement and direction of the ball if the golfer had hit it directly into the hole, we need to sum up the three given vectors. The first is 4.8 m due east, the second is 2.7 m at a 20° angle north of east, and the third is 0.50 m due north.
First, decompose the second stroke into its north and east components. The east component is 2.7 cos(20°) m and the north component is 2.7 sin(20°) m.
East component of the second stroke: 2.7 cos(20°) ≈ 2.536 m
North component of the second stroke: 2.7 sin(20°) ≈ 0.923 m
Now, sum the components in each direction:
Total east displacement: 4.8 m + 2.536 m = 7.336 m
Total north displacement: 0.923 m + 0.50 m = 1.423 m
Finally, calculate the magnitude and direction of the resultant displacement vector:
Magnitude (√(x² + y²)): √(7.336² + 1.423²) ≈ 7.434 m
Direction (tan⁻¹(y/x)): tan⁻¹(1.423/7.336) ≈ 11° north of east
A crate is resting on a ramp that is inclined at an angleabove
the horizontal. As the angle is incrased the crateremains in place
until the angle reaches a value of 38degrees. Then the crate begins
to slide down the slope. (a) Determine the coefficient of static
friction between the crateand the ramp surface. (b) The coefficient
of kinetic frictionbetween the crate and the ramps surface is
0.600. Find theacceleration of the moving crate.
Answer:
1.4 m/s^2
Explanation:
The angle of repose or the angle of sliding is defined as the angle of inclined plane with the horizontal at which an object placed on the plane is just start to slide down.
The relation between the angle of friction and the coefficient of static friction is given by
[tex]\mu _{s}=tan\theta[/tex]
[tex]\mu _{s}=tan38=0.78[/tex]
Thus, the coefficient of static friction is 0.78.
(b) As the crate is moving down, then the friction force is kinetic friction.
The force acting along the plane downward = mg Sinθ
The normal reaction, N = mg cosθ
The friction force acting upward along the plane, f = μk N = μk mg Cosθ
The net force acting along the plane downward
Fnet = mg Sinθ - μk mg Cosθ
According to the newton's second law, Fnet = mass x acceleration
so, m x a = mg Sinθ - μk mg Cosθ
a = g Sinθ - μk g Cosθ
here, μk = 0.6 and θ = 38°, g = 9.8 m/s^2
By substituting the value, we get
a = 9.8 ( Sin38 - 0.6 x Cos 38)
a = 9.8 (0.6156 - 0.6 x 0.788)
a = 1.4 m/s^2
The coefficient of static friction between the crate and the ramp surface is 0.781. The acceleration of the moving crate is 5.88 m/s².
Explanation:To find the coefficient of static friction, we need to calculate the tangent of the angle at which the crate begins to slide. The tangent of 38 degrees is approximately 0.781. Since the crate remains in place until this angle, the coefficient of static friction must be equal to or greater than this value. Therefore, the coefficient of static friction between the crate and the ramp surface is 0.781.
To find the acceleration of the moving crate, we can use the formula a = μk * g, where a is the acceleration, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity. Given that the coefficient of kinetic friction is 0.600, and the acceleration due to gravity is approximately 9.8 m/s², we can substitute these values into the formula to find the acceleration.
Substituting the values, we get a = (0.600) * (9.8) = 5.88 m/s². Therefore, the acceleration of the moving crate is 5.88 m/s².
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Two point charges are located on the x-axis. One has a charge of 1.77 μC and is located at x = 0.0 m, and the other has a charge of -4.09 μC and is located at x = 15.1 m. At what location on the x-axis (other than at infinity) would the electric force on a third point charge of 3.32 μC be zero?
Answer:
x = -29.032m
Explanation:
Since the third charge is positive, it cannot be between the other two charges, because it would be repelled by the positive one and attracted by the negative one, so the electric force would never be zero.
This leaves only two options: To the left of the positive one or to the right of the negative one.
If it was located on the right of the negative charge, the force of the positive charge would be weaker because of both the distance is larger and its charge is smaller than the negative charge. So, there is no point the would make the result force equal zero.
This means that the third charge has to be at the left of the positive charge. With this in mind, we make the calculations:
[tex]F_{13}=K*\frac{Q_{1}*Q_{3}}{d^{2}} =F_{23}=K*\frac{Q_{2}*Q_{3}}{(x_{2}+d)^{2}}[/tex]
Replacing the values of Q1=1.77, Q2=4.09, X2=15.1, we solve for d and get two possible results:
d1 = 29.032m and d2 = -5.99m
Since we assumed in our formula that the third charge was on the left of the positive charge, the distance d has to be positive so that our final result can be a negative position. This is X = -d
This way, we get:
X = -29.032m
An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following statements best describes the electric field at a distance r from the electron? The electric field is directed toward the electron and has a magnitude of ke/r. The electric field is directed away from the electron and has a magnitude of ke/2. The electric field is directed toward the electron and has a magnitude of ke/? The electric field is directed toward the electron and has a magnitude of ke?/r. The electric field is directed away from the electron and has a magnitude of ke/r.
Explanation:
The charge on the electron is, [tex]q=-1.6\times 10^{-19}\ C[/tex]
The electric field at a distance r from the electron is :
[tex]E=k\dfrac{q}{r^2}[/tex]
Where
k is the electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}[/tex]
We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.
So, the correct option is " the electric field is directed toward the electron and has a magnitude of [tex]k\dfrac{q}{r^2}[/tex]. Hence, this is the required solution.
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vector at the point P, a distance d above the midpoint of the wire. The field E-vector points along one of the primary axes, yWhat is the magnitude E of the electric field at point P? Throughout this part, express your answers in terms of the constant k, defined by k=1/(4πε)
Answer:
[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]
Explanation:
Given that
Length= 2L
Linear charge density=λ
Distance= d
K=1/(4πε)
The electric field at point P
[tex]E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta[/tex]
[tex]sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}[/tex]
[tex]r^2=d^2+x^2[/tex]
So
[tex]E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}[/tex]
Now by integrating above equation
[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]
The fastest measured pitched baseball left the pitcher's hand at a speed of 42.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?
Answer:
The acceleration and time are 588 m/s² and 0.071 sec respectively.
Explanation:
Given that,
Speed = 42.0 m/s
Distance = 1.50 m
(a). We need to calculate the acceleration
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
Put the value in the equation
[tex]a=\dfrac{42.0^2-0}{2\times1.50}[/tex]
[tex]a=588\ m/s^2[/tex]
(b). We need to calculate the time
Using equation of motion
[tex]v = u+at[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{42.0-0}{588}[/tex]
[tex]t=0.071\ sec[/tex]
Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.
In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it starts to careen down the steeply sloped track. In one event, the sled reaches a top speed of 9.2 m/s before starting down the initial part of the track, which is sloped downward at an angle of 4.0°. What is the sled's speed after it has traveled the first 140 m?
Answer:
[tex]v_f = 16.6 m/s[/tex]
Explanation:
As we know by force equation that force along the inclined planed due to gravity is given as
[tex]F_g = mg sin\theta[/tex]
so the acceleration due to gravity along the plane is given as
[tex]a = \frac{F_g}{m}[/tex]
now we have
[tex]a = g sin\theta[/tex]
[tex]a = (9.81 sin4.0)[/tex]
[tex]a = 0.68 m/s^2[/tex]
now we know that
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - 9.2^2 = 2(0.68)(140)[/tex]
[tex]v_f = 16.6 m/s[/tex]
Velocity of a object is the ratio of distance traveled by the object with the time taken
The sled's speed after it has traveled the first 140 m is 16.6 m/s.
What is velocity of a object?Velocity of a object is the ratio of distance traveled by the object with the time taken.
Given information-
The top speed reached by the sled is 9.2 m/s.
The angle of the slope is 4 degrees downward.
Distance traveled by the sled is 140 meters.
The force acting on a body is the product of mass and its acceleration.
The acceleration of the inclined plane with this definition can be given as,
[tex]a=\dfrac{F_g}{m} =\dfrac{mg\sin\theta}{m} \\a=g\sin\theta[/tex]
Here, [tex]F_g[/tex] is the force due to gravity and [tex]a[/tex] is the acceleration of the body.
Put the values in the above equation as,
[tex]a=9.81\times \sin(4^o)\\a=0.68 \rm m/s^2[/tex]
Thus the acceleration of the body is 0.68 m/s squared.
Now the sled's speed can be find out using the equation of motion as,
[tex]2ad=v_f^2-v_i^2[/tex]
As the initial speed is 9.2 seconds. Thus,
[tex]2\times0.68\times140=v_f^2-9.2^2\\v_f=16.6 \rm m/s[/tex]
Hence the sled's speed after it has traveled the first 140 m is 16.6 m/s.
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A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 7.5 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.
Answer:
450 pm
Explanation:
The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:
a = v^2 / r
The equation for the electric force is:
F = q1 * q2 / (4 * π * e0 * r^2)
Where
q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C
e0: electric constant (8.85*10^-12 F/m)
If we divide this force by the mass of the electron we get the acceleration
me = 9.1*10^-31 kg
a = q1 * q2 / (4 * π * e0 * me * r^2)
v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)
We can simplify r
v^2 = q1 * q2 / (4 * π * e0 * me * r)
Rearranging:
r = q1 * q2 / (4 * π * e0 * me * v^2)
r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a 36-N force is applied to the left side of the left cube causing all three cubes to accelerate to the right. If the cubes are each subject to a frictional force of 6.0 N, what is the magnitude of the force exerted on the right cube by the middle cube in this case
Answer:
Force exerted on the right cube by the middle cube:
F= 12.02N : in the positive direction of the x axis( +x)
Explanation:
We apply Newton's second law for forces in the direction of the x-axis.
∑Fx= m*a
∑Fx: algebraic sum of forces ( + to the right, - to the left)
m: mass
a: acceleration
Forces (x) in total mass : Newton's second lawWe apply Newton's first law for forces in the direction of the y-axis.
∑Fx= mt*a , mt: total mass = 6*3= 18 kg
36-6= 18*a
[tex]a=\frac{36-6}{18} = 1.67 \frac{m}{s}[/tex]
Forces (y) in total mass : Newton's first law∑Fy= 0
Nt-Wt=0 , Nt=Wt
Nt: total normal , Wt= total Weight: mt*g , g: acceleration due to gravity
Wt=18*9,8=176.4 N
Nt=176.4 N
Calculation of the coefficient of kinetic frictionμk=Ff/Nt Ff: friction force = 6 N
μk=6/176.4 = 0.034
Forces (x) on the first block (on the right): Newton's second law∑Fx= m₁*a
F-Ff₁= 6*1.67 , Equation (1)
F:Force exerted on the right cube by the middle cube
Ff₁= μk*N₁ , N₁=W₁ = 9.8*6= 58.8 N
Ff₁= 0.034*58.8 = 2 N
In the Equation (1):
F-2= 10.02
F= 2+10.02= 12.02N
F= 12.02N
Consider a physical pendulum with length of 99.5 cm and mass of 185 g. If the pendulum was released from an angle less than 10°, then calculate the angular speed of the pendulum. (g = 9.80 m/s^2)
Answer:
The angular speed of the pendulum is 3.138 rad/s.
Explanation:
Given that,
Length = 99.5 cm
Mass = 185 g
We need to calculate the angular speed
Using formula of angular speed
[tex]\omega=\sqrt{\dfrac{g}{l}}[/tex]
Where, l = length
g = acceleration due to gravity
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{9.80}{99.5\times10^{-2}}}[/tex]
[tex]\omega=3.138\ rad/s[/tex]
Hence, The angular speed of the pendulum is 3.138 rad/s.
A cyclist reaches the bottom of a hill with a speed of 18 m/s. Neglecting air resistance and other friction, to what maximum height can they coast up the hill without pedalling? a) 19 m b) 20 m c) 17 m d) 18 m
Answer:
c) 17 m
Explanation:
From the conservation of energy.
Change in kinetic energy= change in potential energy
Now,
[tex]\Delta KE=\Delta PE[/tex]
Now according to the situation.
[tex]\frac{1}{2}mv^{2}=mgh\\h=\frac{v^{2} }{2g}[/tex]
Here, v is the velocity, m is the mass of an object, h is the height, g is the acceleration due to gravity.
Given that the speed of the bicycle is
[tex]v=18m/s[/tex]
And the acceleration due to gravity is
[tex]g=9.8 m/s^{2}[/tex]
Substitute these values.
[tex]h=\frac{18^{2} }{2\times 9.8}\\h=16.53m[/tex]
Which means it is equivalent to 17 m
Therefore, the maximum height is 17 m
A traveler covers a distance of 217 miles in a time of 8 hours 32 minutes. (a) What is the average speed for this trip?
Answer:
Average speed, v = 11.37 m/s
Explanation:
Given that,
The distance covered by the traveler, d = 217 miles = 349228 meters
Time taken, [tex]t = 8\ hours \ 32\ minutes =8\ h+\dfrac{32}{60}\ h=8.53\ h[/tex]
or t = 30708 s
We need to find the average speed for this trip. The average speed is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{349228\ m}{30708\ s}[/tex]
v = 11.37 m/s
So, the average speed for this trip is 11.37 m/s. Hence, this is the required solution.
A golf ball is hit into the air, but NOT Staight Up, and encounters no significant air resistance. Which statement accurately describes its motion while it is in air? On the way up, both its horizontal and vertical velocity components are decreasing, on the way down they are both increasing. Its acceleration is zero at the highest point. Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change. On the way up its acceleration is 9.8m/s2 upward, on the way down its acceleration is 9.8m/s2 downward, Its velocity is zero at the highest point.
Answer:
Answered
Explanation:
A) False , horizontal velocity component is constant since there's no horizontal acceleration component
B) False, acceleration remains constant through out and only verticle acceleration is there horizontal acceleration is zero.
C) True , Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change ( because of acceleration in verticle direction)
D)False, the acceleration is constant through the motion.
E)False, Vertical velocity is Zero but there will be Horizontal velocity.
A charge 4q is at the origin, and a charge of -3q is on the positive x-axis at x = a. Where would you place a third charge so it would experience zero net electric force? (Note: Either enter your answer as an unevaluated expression, e.g., (3/2 + sqrt(3))a, or if evaluated, use 2 decimal places.) Your expression should be in terms of the given variable.
Answer:
6.45 a
Explanation:
Charge on O, q1 = 4q
Charge on A, q2 = - 3 q
OA = a
Let the net force is zero at point P, where AP = x , let a charge Q is placed at P.
The force on point P due to the charge q1 = The force on point P due to the
charge q2
By using Coulomb's law
[tex]\frac{Kq_{1}Q}{OP^{2}}=\frac{Kq_{2}Q}{AP^{2}}[/tex]
[tex]\frac {K4qQ}{(a+x)^{2}}= \frac {K3qQ}{x^{2}}[/tex]
[tex]\frac{a+x}{x}=\sqrt{\frac{4}{3}}[/tex]
[tex]\frac{a+x}{x}=1.155[/tex]
a + x = 1.155 x
0.155 x = a
x = 6.45 a
Thus, the force is zero at x = 6.45 a.
A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g of ethyl alcohol at 20 degrees. What is the final temperature?
The final temperature can be calculated using the principle of conservation of energy. Using the formulas Q = mcΔT and Q = mcΔT, we can find the heat lost by the iron and the heat gained by the ethyl alcohol. The final temperature of all 3 substances are, T_f = 12.08 °C.
Explanation:To calculate the final temperature of the substances, we can use the principle of conservation of energy. The heat gained by one substance is equal to the heat lost by another substance. First, we calculate the heat lost by the iron using the formula Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Then, we calculate the heat gained by the ethyl alcohol using the same formula. Finally, we use the principle of conservation of energy to find the final temperature.
We have:
Iron: m = 200 g, c = 452 J/g °C, Ti = 120 °CCopper: m = 150 g, c = 387 J/g °C, Ti = -50 °CEthyl alcohol: m = 300 g, c = 2.30 J/g °C, Ti = 20 °CUsing the formula Q = mcΔT, we can calculate:
Heat lost by iron: Qiron = (200 g)(452 J/g °C)(120 °C - T_f)Heat gained by ethyl alcohol: Qethyl alcohol = (300 g)(2.30 J/g °C)(T_f - 20 °C)Since the heat lost by the iron is equal to the heat gained by the ethyl alcohol, we can set up the equation:
(200 g)(452 J/g °C)(120 °C - T_f) = (300 g)(2.30 J/g °C)(T_f - 20 °C)
To solve for T_f, we can simplify and rearrange the equation:
(90400 - 452T_f) J = (T_f- 13800) J
Combining like terms:
13800 J = 1142T_f J
Dividing both sides by 1142:
T_f = 12.08 °C
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A glass vessel that can be repeatedly filled with precisely the same volume of liquid is called a pycnometer. A certain pycnometer, when empty and dry, weighed 25.296 g. When filled with water at 25 oC the pycnometer and water weighed 34.914 g. When filled with a liquid of unknown composition the pycnometer and its contents weighed 33.485 g. At 25 oC the density of water is 0.9970 g/ml. What is the density of the unknown liquid?
Answer:
density of liquid 0.848 g/ml
Explanation:
from the information given in the question
mass of water = 34.914 - 25.296 = 9.618 g
volume of pycnometer = volume of water
which will be equal to [tex]= \frac{ mass}{density}[/tex]
[tex]= \frac{9.618}{0.9970} = 9.646 ml[/tex]
mass of liquid =33.485-25.296 = 8.189 ml
density of liquid[tex]= \frac{mass}{volum\ of\ liquid}[/tex]
= [tex]\frac{8.189}{9.646} =0.848 g/ml[/tex]
two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume of the left and is initially evacuated. Detrmine the pressure of water after the paertiion has been removed nd enough heat has been transfered so that the temperature of the water is 40F.
Answer:
pressure of water will be 49.7 atm
Explanation:
given data
pressure = 30 psi = 2.04 atm
water = 2 pound = 907.18
mole of water vapor = 907.19 /2 = 50.4 mole
volume = 4 ft³ = 113.2 L
temperature = 40 F = 277.59 K
to find out
pressure of water
solution
we will apply here ideal gas condition
that is
PV = nRT .......................1
put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole
and we get here temperature
PV = nRT
2.04 × 113.2 = 50.4×0.0821×T
solve it and we get
T = 55.8 K
so we have given right chamber has twice the volume of the left chamber i.e
volume = twice of volume + volume
volume = 2(113.2) + 113.2
volume = 339.6 L
so from equation 1 pressure will be
PV = nRT
P(339.6) = 50.4 × ( 0.0821) × (277.59)
P = 3.3822 atm = 49.7 atm
so pressure of water will be 49.7 atm
A banked circular highway is designed for traffic moving
at60km/h. The radius of the curve is 200m. Traffic ismoving along
the highway at 40km/h on a rainy day. What isthe minimum
coefficient of friction between tires and road thatwill allow cars
to negotiate the turn without sliding off theroad?
Answer:
0.063
Explanation:
velocity of the car, v = 40 km/h = 11.11 m/s
radius, r = 200 m
Let the coefficient of friction is μ.
The coefficient of friction relates to the velocity on banked road is given by
[tex]\mu =\frac{v^{2}}{rg}[/tex]
where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.
By substituting the values, we get
[tex]\mu =\frac{11.11^{2}}{200\times 9.8}[/tex]
μ = 0.063
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
Answer:
31.75 m/s
Explanation:
h = 41.7 m
Let the initial velocity of the second stone is u
Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.
For first stone:
Use second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7
So, 41.7= 0 + 0.5 x 9.8 x t^2
41.7 = 4.9 t^2
t = 2.92 s ..... (1)
For second stone:
Use second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity
[tex]h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2[/tex] .... (2)
By equation the equation (1) and (2), we get
[tex]41.7=1.12 u +4.9 \times 1.12^{2}[/tex]
u = 31.75 m/s
Magnetic field is measured in SI units of a tesla (T), and a current through a wire generates a field around the wire. The largest fields that we can make with most normal conducting materials carrying current are limited by the resistance of the wire and are around 1 tesla. For comparison, Earth's magnetic field is roughly 50 micro tesla You may have heard of a field unit called a gauss. It takes 10,000 G to make 1 T. What would be the maximum force on a wire 0.1 m long carrying a current of 80 A in a uniform magnetic field of 1 T? 8 N when the wire is perpendicular to the field
80 N when the wire is perpendicular to the field
8 N when the wire is parallel to the field
80 N when the wire is parallel to the field
Answer:
The answer is 8 N
Explanation:
The Lorentz force for a current carrying wire is
f = I * L x B
So, for magnetic forces to manifest the current must not be parallel to the magnetic field. So the cases where the wire is parallel to the field would result in a force of zero applied on the wires by the magnetic field because the cross product becomes zero.
For the perpendicular cases:
f = I * L * B
f = 80 * 0.1 * 1 = 8 N
The Mariana Trench in the western Pacific Ocean includes the greatest known ocean depth at approximately 6.8 miles. The atmosphere exerts a pressure of 14.7 lbf/in.2 at the ocean surface. Modeling the ocean seawater as static and assuming constant local acceleration of gravity of 32.1 ft/s2 and constant seawater density of 64.2 lb/ft3 , determine the absolute pressure, in lbf/in.2 , at this depth
Final answer:
The absolute pressure at the bottom of the Mariana Trench is calculated by adding the atmospheric pressure to the hydrostatic pressure due to the water column, resulting in approximately 509,367.85 lbf/in.².
Explanation:
To calculate the absolute pressure at the bottom of the Mariana Trench, we start by understanding that pressure in a static fluid increases linearly with depth. The increase in pressure, ΔP, due to the water column can be calculated using the formula ΔP = ρgh, where ρ is the density of seawater, g is the acceleration due to gravity, and h is the depth. Given the constants, ρ = 64.2 lb/ft³, g = 32.1 ft/s², and h = 6.8 miles (35,856 ft), we first convert the depth into feet as pressure calculations require consistent units. The calculation is as follows: ΔP = 64.2 lb/ft³ * 32.1 ft/s² * 35,856 ft = 73,346,473.6 lb/ft². Converting this to lbf/in.², we divide by 144 (since 1 ft² = 144 in.²), resulting in ΔP approximately 509,353.15 lbf/in.². Adding the atmospheric pressure of 14.7 lbf/in.² at the surface, the absolute pressure at the bottom of the Mariana Trench in lbf/in.² is approximately 509,367.85 lbf/in.².
The absolute pressure at the depth of the Mariana Trench is approximately [tex]\(7.42 \times 10^9 \, \text{lbf/in}^2\),[/tex]calculated using the hydrostatic pressure formula.
To find the absolute pressure at the depth of the Mariana Trench, we can use the hydrostatic pressure formula:
[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]
Where:
- P is the absolute pressure at the depth,
- P0 is the atmospheric pressure at the surface (given as 14.7 lbf/in²),
- rho is the density of seawater (given as 64.2 lb/ft³),
- g is the acceleration due to gravity (given as 32.1 ft/s²), and
- h is the depth of the trench (given as 6.8 miles).
First, let's convert the depth from miles to feet:
[tex]\[ 6.8 \text{ miles} \times 5280 \text{ ft/mile} = 35904 \text{ ft} \][/tex]
Now, we can plug in the values into the formula:
[tex]\[ P = 14.7 \text{ lbf/in}^2 + (64.2 \text{ lb/ft}^3) \times (32.1 \text{ ft/s}^2) \times (35904 \text{ ft}) \][/tex]
Let's calculate this value.
To find the absolute pressure at the depth of the Mariana Trench, we'll first calculate the pressure due to the water column using the hydrostatic pressure formula:
[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P_0 = 14.7 \, \text{lbf/in}^2 \)[/tex] is the atmospheric pressure at the surface,
-[tex]\( \rho = 64.2 \, \text{lb/ft}^3 \)[/tex] is the density of seawater,
- [tex]\( g = 32.1 \, \text{ft/s}^2 \)[/tex] is the acceleration due to gravity, and
- [tex]\( h = 6.8 \, \text{miles} \times 5280 \, \text{ft/mile} = 35904 \, \text{ft} \)[/tex]is the depth of the trench in feet.
Plugging in the values:
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
Let's calculate this.
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
[tex]\[ P = 14.7 + (206368.8 \times 35904) \][/tex]
[tex]\[ P = 14.7 + 7417798272 \][/tex]
[tex]\[ P ≈ 7417798286.7 \, \text{lbf/in}^2 \][/tex]
Therefore, the absolute pressure at the depth of the Mariana Trench is approximately [tex]\( 7.42 \times 10^9 \, \text{lbf/in}^2 \).[/tex]
What is the significance of the direction of an electric field line at some point on the line?
Answer with Explanation:
The direction of the electric field line at any point gives us the direction of the electric force that will act on a positive charge if placed at the point. We know that if we place a charge in an electric field it will experience a force, as we know that force is a vector quantity hence it requires both magnitude and direction for it's complete description. The direction of this electric force that acts on a charge is given by the direction of the electric field in the space. In case the charge is negatively charged electric force will act on it in the direction opposite to the direction of electric field at the point.
A jetliner rolls down the runway with constant acceleration from rest, it reaches its take off speed of 250 km/h in 1 min. What is its acceleration? Express in km/h^2, does this result make sense?
Answer:
Acceleration, [tex]a=14970.05\ km/h^2[/tex]
Explanation:
Given that,
Initially, the jetliner is at rest, u = 0
Final speed of the jetliner, v = 250 km/h
Time taken, t = 1 min = 0.0167 h
We need to find the acceleration of the jetliner. The mathematical expression for the acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{250}{0.0167}[/tex]
[tex]a=14970.05\ km/h^2[/tex]
So, the acceleration of the jetliner is [tex]14970.05\ km/h^2[/tex]. Hence, this is the required solution.
Is a sonic boom heard by the passengers traveling in a supersonic aircraft? Why or why not? a.) they do not experience any shift in frequency because they are traveling at the same speed as that of the sound source.
b.) They do not experience any shift because they are traveling at a lower speed than that of the sound source.
c.) They will experience a small shift in frequency because they are traveling at a lower speed than that of the sound source.
d.) They will experience a large shift in frequency because they are traveling at a greater speed than that of the sound source.
Answer:
No the sonic boom is not heard by the passenger travelling in a supersonic aircraft. They do not experience any shift in the frequency as they are travelling at the speed of the source itself.Thus option 'a' is correct answer.
Explanation:
Sonic boom is a shock wave of sound associated when a source travels faster than the speed of sound in the ambient conditions. In a shock wave associated with the sonic boom there exists high gradients of pressure, density and temperature of air across the shock wave. The shock wave travels at the speed of the sound but the source (in our case the passenger) travels at a faster speed than the sound in ambient conditions.
A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of the electrostatic force between those two parts?
Answer:
1.72 x 10³ N.
Explanation:
When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.
So the charges will be each equal to
31/2 = 15.5 x 10⁻⁶ C
F = K Q q / r²
= [tex]\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}[/tex]
= 1.72 x 10³ N.
A container is filled to the brim with 1.5 L of mercury at 20°C. As the temperature of the container and mercury is increased to 55°C, a total of 7.8 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container.
Answer:
The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]
Explanation:
Given that,
Volume = 1.5 l
Temperature = 20°C
Increased temperature = 55°C
Volume = 7.8 mL
We need to calculate the linear expansion coefficient of the material
Using formula of linear expansion
[tex]\Delta V=V_{0}\beta \Delta T[/tex]
[tex]\Delta V=V_{0}(3\alpha \Delta T)[/tex]
[tex]\alpha =\dfrac{\Delta V}{3V_{0}\Delta T}[/tex]
Put the value into the formula
[tex]\alpha=\dfrac{7.8\times10^{-3}}{3\times1.5\times(55-20)}[/tex]
[tex]\alpha=0.00004952[/tex]
[tex]\alpha=4.952\times10^{-5}\ K^{-1}[/tex]
Hence, The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]
A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067 g d) 0.015 g
Answer:
The mass of the bullet is 15 g.
(b) is correct option.
Explanation:
Given that,
Momentum = 2.8 kg m/s
Speed = 187 m/s
We need to calculate the mass of the bullet
Using formula of momentum
[tex]P = mv[/tex]
[tex]m = \dfrac{P}{v}[/tex]
Where, P = momentum
v = speed
Put the value into the formula
[tex]m = \dfrac{2.8}{187}[/tex]
[tex]m = 0.015\ kg[/tex]
[tex]m = 15\ g[/tex]
Hence, The mass of the bullet is 15 g.
What additional information is obtained from measurements on an ordinal scale compared to measurements on a nominal scale? Comparison of the magnitudes of two measurements (such as one being twice as large as the other) Direction of difference (greater or less) between two measurements Magnitude of difference between two measurements
The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuck in the sandy bottom of the pond. If the stick is 19 cm tall, how long is the shadow?
Answer:
shadow length 7.67 cm
Explanation:
given data:
refractive index of water is 1.33
by snell's law we have
[tex]n_{air} sin30 =n_{water} sin\theta[/tex]
[tex]1*0.5 = 1.33*sin\theta[/tex]
solving for[tex] \theta[/tex]
[tex]sin\theta = \frac{3}{8}[/tex]
[tex]\theta = sin^{-1}\frac{3}{8}[/tex]
[tex]\theta = 22 degree[/tex]
from shadow- stick traingle
[tex]tan(90-\theta) = cot\theta = \frac{h}{s}[/tex]
[tex]s = \frac{h}{cot\theta} = h tan\theta[/tex]
s = 19tan22 = 7.67 cm
s = shadow length
The calculated shadow length is approximately 10.97 cm.
To determine the length of the shadow cast by the stick, we can use basic trigonometry.
Specifically, we'll use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.
Given:
The height of the stick (opposite side) is 19 cm.The angle of elevation of the sun is 60°.The formula for tangent is:
tan(θ) = opposite / adjacent
Here, the opposite side is the height of the stick, and the adjacent side is the length of the shadow. Thus, we can rearrange this formula to solve for the length of the shadow (adjacent side):
adjacent = opposite / tan(θ)
Substituting the given values:
adjacent = 19 cm / tan(60°)
We know that tan(60°) is √3 or approximately 1.732.
So:
adjacent = 19 cm / 1.732 ≈ 10.97 cm
Therefore, the length of the shadow cast by the stick is approximately 10.97 cm.
If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, your speed is decreased by the water's speed. The current in a river flows at 0.52 m/s. In still water you can swim at 1.73 m/s. If you swim downstream a certain distance, then back again upstream, how much longer, in percent, does it take compared to the same trip in still water?
Swimming downstream and upstream in a river with a current of 0.52 m/s results in approximately an 86% increase in time compared to swimming the same distance in still water at a speed of 1.73 m/s.
1. Understanding the problem:
You can swim 1.73 m/s in still water.
The river current flows at 0.52 m/s.
We need to find the percentage increase in time to swim downstream and back upstream compared to swimming the same distance in still water.
2. Effective speeds:
Downstream: Your speed increases with the current.
Effective speed = swimming speed + current speed = 1.73 m/s + 0.52 m/s = 2.25 m/s
Upstream: Your speed decreases against the current.
Effective speed = swimming speed - current speed = 1.73 m/s - 0.52 m/s = 1.21 m/s
3. Calculating time for each leg:
Let D be the distance for the entire round trip.
Time downstream: D / 2.25 m/s
Time upstream: D / 1.21 m/s
4. Finding the time difference:
Time difference = Time upstream - Time downstream
Time difference = (D / 1.21 m/s) - (D / 2.25 m/s)
5. Calculating percentage increase:
To find the percentage increase in time compared to still water, divide the time difference by the time taken to swim downstream and multiply by 100.
Percentage increase = [(D / 1.21 m/s) - (D / 2.25 m/s)] / (D / 2.25 m/s) * 100
6. Simplifying the equation:
After simplifying, you'll find the percentage increase to be approximately 86%.
Therefore, it takes approximately 86% longer to complete the round trip compared to swimming the same distance in still water.