A "swing" ride at a carnival consists of chairs that are swung in a circle by 19.8 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 137 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.
Answer:
a) T = 1342.6 cos θ, b) v = 13.93 √(sin θ tan θ)
Explanation:
We can solve this problem using Newton's second law
Let's fix a reference system with a horizontal axis and the other vertical, therefore the only force to decompose is the tension, in these problems the most common is to measure the angle with respect to the vertical. Let's use trigonometry to find the components of the dispute
cos θ = [tex]T_{y}[/tex] / T
T_{y} = T cos tea
sin θ = Tₓ / T
Tₓ = T sin θ
let's write Newton's second law
axis and vertical
T cos θ - W = 0
T = mg / cos θ
let's calculate
T = 137 9.8 cos θ
T = 1342.6 cos θ
unfortunately there is no drawing or indication of the angle
Axis x Horizontal
T sin θ = m a
acceleration is centripetal
a = v² / R
T sin θ = m v² / R
v² = (g / cos θ) R sin θ
v = √ (gR tan θ)
let's use trigonometry to find radius of gyration
sin θ = R / L
R = L sin θ
v = √ (g L sin θ tan θ)
let's calculate
v = √(9.8 19.8 sin θ tant θ)
v = 13.93 √(sin θ tan θ)
they do not give the angle for which the calculation cannot be finished
Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field:________
Answer:
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
Explanation:
From Maxwell's fourth equation
Curl B = μ₀J + μ₀ε₀dE/dt (1) where the second term is the displacement current.
If the oscillation conduction current in the conductor is much larger than the displacement current then, the displacement current goes to zero. So we have
Curl B = μ₀J (2)(since μ₀ε₀dE/dt = 0)
From maxwell's third equation
Curl E = -dB/dt (3)
taking curl of the above from the left
Curl(Curl E) = Curl(-dB/dt)
Curl(Curl E) = (-d(CurlB)/dt) (4)
Substituting for Curl B into (4), we have
Curl(Curl E) = -dμ₀J/dt
Del(DivE) - (Del)²E = -dμ₀J/dt (5)
From Maxwell's first equation,
DivE = ρ/ε₀
Substituting this into (5), we have
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine needs a temperature difference, a hot side and a cold side. Conveniently, the ocean surface waters are warmer than the deep ocean waters. Suppose you build a floating power plant in the tropics where the surface water temperature is ~ 35.0 C. This would be the hot reservoir of the engine. For the cold reservoir, water would be pumped up from the ocean bottom where it is always ~ 5.00 C.
What is the maximum possible efficiency of such a power plant?
Answer:
Explanation:
The maximum efficient power plant will be the plant based on carnot cycle whose efficiency is given by the following formula
Efficiency = (T₁ - T₂) / T₁
T₁ is temperature of hot reservoir and T₂ is temperature of cold reservoir.
Putting the given values
efficiency of power plant = (35 - 5) / (273 + 35 )
= 30 / 308
= .097
= 9.7 %
A long, East-West-oriented power cable carrying an
unknown current I is at a height of 8 m above the Earth's
surface. If the magnetic flux density recorded by a magnetic-
field meter placed at the surface is 15 ut when the current is
flowing through the cable and 20 ut when the current is zero,
what is the magnitude of 1?
Answer:
200A
Explanation:
Given that
the distance between earth surface and power cable d = 8m
when the current is flowing through cable , the magnitude flux density at the surface is 15μT
when the current flow throught is zero the magnitude flux density at the surface is 20μT
The change in flux density due to the current flowing in the power cable is
B = 20μT - 15μT
B =5μT -----(1)
The expression of magnitude flux density produced by the current carrying cable is
[tex]B=\frac{\mu_0I}{2\pi d}[/tex]-----(2)
Substitute the value of flux density
B from eqn 1 and eqn 2
[tex]\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A[/tex]
Therefore, the magnitude of current I is 200A
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and 10 C at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of 1.4 W/m K, what is the rate of heat loss through the slab
Answer:
Q = - 4312 W = - 4.312 KW
Explanation:
The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:
Q = - kA dt/dx
Integrating from one side of the slab to other along the thickness dimension, we get:
Q = - kA(T₂ - T₁)/L
Q = kA(T₁ - T₂)/t
where,
Q = Rate of Heat Loss = ?
k = thermal conductivity = 1.4 W/m.k
A = Surface Area = (11 m)(8 m) = 88 m²
T₁ = Temperature of Bottom Surface = 10°C
T₂ = Temperature of Top Surface = 17° C
t = Thickness of Slab = 0.2 m
Therefore,
Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m
Q = - 4312 W = - 4.312 KW
Here, negative sign shows the loss of heat.
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^{-2CL}T≃e −2CL , where L is the width of the barrier and C is a term that includes the particle energy and barrier height. If the tunneling coefficient is found to be T = 0.050T=0.050 for a given value of LL, for what new value of L\text{'}L’ is the tunneling coefficient T\text{'} = 0.025T’=0.025 ? (All other parameters remain unchanged.) Express L\text{'}L’ in terms of the original LL.
Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:
[tex]T \approx e^{-2CL}[/tex]
L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:
[tex]0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)[/tex]
Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):
[tex]ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)[/tex]
Next, you divide the equation (3) into (4), and finally, you solve for L':
[tex]\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L[/tex]
hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight up and -90∘ straight down. The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?
a. 250 frames/s
b. 2500 frames/s
c. 25,000 frames/s
d. 250,000 frames/s.
Answer:
c. 25,000 frames/s
Explanation:
For computing the minimum frame rate for high speed first we have to determine the time by applying the following equation
[tex]t = \frac{d}{s}[/tex]
[tex]= \frac{0.2\ mm}{4.6\ m/s }[/tex]
[tex]= \frac{0.2 \times 10 ^{-3}}{4.6\ m/s }[/tex]
[tex]= 4.347 \times 10^{-5} sec[/tex]
Now the frame rate is
[tex]Frame\ rate = \frac{1}{t}[/tex]
[tex]= \frac{1}{4.347 \times 10^{-5} sec}[/tex]
= 23,000 frame per sec
≈ 25,000 frame per sec
First we have find the time then after finding out the time we calculate the frame time by applying the above formula so that the minimum frame rate could come
A 1500 kg car travelling at 25 m/s collides with a 2500 kg van which had
stopped at a traffic light. As a result of the collision the two vehicles become
entangled. Which of the following pairs given in the table below shows the
initial speed the entangled mass will move and the type of collision
High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45.2 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 39.9 m/s. Find the speed of the golf ball just after impact.
Answer:
The speed will be "1.06 m/s".
Explanation:
The given values are:
Momentum,
m1 = 244 g
m2 = 45.2 g
On applying momentum conservation ,
Let v2 become the final golf's speed.
From Momentum Conservation
⇒ [tex]Total \ initial \ momentum = Total \ final \ momentum[/tex]
⇒ [tex]m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2[/tex]
On putting the estimated values, we get
⇒ [tex]0.244\times 57.6+0=0.244\times 39.9+45.2\times v2[/tex]
⇒ [tex]57.844+0=9.7356+45.2\times v2[/tex]
⇒ [tex]48.1084=45.2\times v2[/tex]
⇒ [tex]v2=\frac{48.1084}{45.2}[/tex]
⇒ [tex]v2=1.06 \ m/s[/tex]
A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact with the top of the beam, and it drives the beam 15.8 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
471392.4 N
Explanation:
From the question,
Just before contact with the beam,
mgh = Fd.................... Equation 1
Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.
make f the subject of the equation
F = mgh/d................ Equation 2
Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m
Constant: g = 9.8 m/s²
Substitute into equation 2
F = 1900(4)(9.8)/0.158
F = 471392.4 N
A long solenoid that has 1 080 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?
Answer:
Current, I = 0.073 A
Explanation:
It is given that,
Number of turns in a long solenoid is 1080
Length of the solenoid is 0.420 m
It produces a magnetic field of [tex]10^{-4}\ T[/tex] at its center.
We need to find the current is required in the winding for that to occur. The magnetic field at the center of the solenoid is given by :
[tex]B=\mu_0 NI[/tex]
I is current
[tex]I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{10^{-4}}{4\pi \times 10^{-7}\times 1080}\\\\I=0.073\ A[/tex]
There is a known potential difference between two charged plates of 12000 Volts. An object with a charge of 6.5 x 10-6 C charge and a mass of 0.02 kg is placed next to the positive plate. How fast will it be traveling when it gets to the negative plate
Answer:
1.97 m/s.
Explanation:
From the question,
Using the law of conservation of energy,
The energy stored in the charged plate = Kinetic energy of the mass
1/2(qV) = 1/2mv².......................... Equation 1
Where q = charge, V = voltage, m = mass, v = velocity.
make v the subject of the equation
v = √(qV/m)......................... Equation 2
Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg
Substitute these values into equation 2
v = √(6.5×10⁻⁶×12000 /0.02)
v = √3.9
v = 1.97 m/s.
Which scientist was the first to propose the heliocentric model of the universe
Answer:
Nicolaus Copernicus
Explanation:
Nevertheless, Copernicus began to work on astronomy on his own. Sometime between 1510 and 1514 he wrote an essay that has come to be known as the Commentariolus (MW 75–126) that introduced his new cosmological idea, the heliocentric universe, and he sent copies to various astronomers
For saving energy, bicycling and walking are far more efficient means of transportation than is travel by automobile. For example, when riding at 12.5 mi/h, a cyclist uses food energy at a rate of about 360 kcal/h above what he would use if merely sitting still. (In exercise physiology, power is often measured in kcal/h rather than in watts. Here 1 kcal = 1 nutritionist's Calorie = 4186 J). Walking at 3.20 mi/h requires about 220 kcal/h. It is interesting to compare these values with the energy consumption required for travel by car. Gasoline yields about 1.30.
A) Find the fuel economy in equivalent miles per gallon for a person walking.
B) Find the fuel economy in equivalent miles per gallon for a person bicycling.
Answer:
a. 451.72 mi/ga
b. 1078.33 mi/ga
Explanation:
The computation is shown below:
a. The fuel economy for a person walking is
Given that
Walking at 3.20 mi/h requires about 220 kcal/h so it is equal to
[tex]= 220 \times 4186[/tex]
= 920920 j/hr
Now
[tex]= \frac{mi}{j}[/tex]
[tex]= \frac{3.2}{920920}[/tex]
So,
[tex]= \frac{mi}{ga}[/tex]
[tex]= \frac{3.2}{920920}\times 1.3 \times 100000000[/tex]
= 451.72 mi/ga
b. Now
Bicycling 12.5 mi/h requires about 360 kcal/h energy so it is equal to
[tex]= 360 \times 4186[/tex]
= 1506960 j/hr
So,
[tex]= \frac{mi}{j}[/tex]
[tex]= \frac{12.5}{1506960}[/tex]
Now
[tex]= \frac{mi}{ga}[/tex]
[tex]= \frac{12.5}{1506960} \times 1.3 \times 100000000[/tex]
= 1078.33 mi/ga
We simply applied the above formula
What is the gravitational force between mars and Phobos
Answer:
[tex]F=5.16\times 10^{15}\ N[/tex]
Explanation:
We have,
Mass of Mars is, [tex]m_M=6.42\times 10^{23}\ kg[/tex]
Mass of its moon Phobos, [tex]m_P=1.06\times 10^{16}\ kg[/tex]
Distance between Mars and Phobos, d = 9378 km
It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by
[tex]F=G\dfrac{m_Mm_P}{d^2}[/tex]
Plugging all values, we get :
[tex]F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N[/tex]
So, the gravitational force is [tex]5.16\times 10^{15}\ N[/tex].
When a star has fused most of its hydrogen and begins to collapse inward, it becomes a
Which of the following statements about this system of lumps must be true? A. The momentum of the system is conserved during the collision. B. The kinetic energy of the system is conserved during the collision. C. The two masses lose all their kinetic energy during the collision. D. The velocity of the center of mass of the system is the same after the collision as it was before the collisio
Answer:
option A.
Explanation:
Since the two lumps collide together, it is an inelastic collision. An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. Part of the kinetic energy is changed to some other form of energy in the collision
Momentum is conserved in inelastic collisions. This means the correct option is option A.
A thin plastic rod of length 2.6 m is rubbed all over with wool, and acquires a charge of 98 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.
Answer:
By exact formula
5076.59N/C
And by approximation formula
5218.93N/C
Explanation:
We are given that
Length of rod,L=2.6 m
Charge,q=98nC=[tex]98\times 10^{-9} C[/tex]
[tex]1nC=10^{-9} C[/tex]
a=13 cm=0.13 m
1 m=100 cm
By exact formula
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{a^2+\frac{L^2}{4}}}[/tex]
Where k=[tex]9\times 10^9[/tex]
Using the formula
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{(0.13)^2+\frac{(2.6)^2}{4}}}=5076.59N/C[/tex]
In approximation formula
a<<L
[tex]a^2+(\frac{L}{2})^2=\frac{L^2}{4}[/tex]
Therefore,the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{\frac{L^2}{4}}}[/tex]
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{\frac{(2.6)^2}{4}}}=5218.93N/C[/tex]
A venturi meter used to measure flow speed in the pipe. Derive an expression for the flow speed "H1" interns of the crossectional areas "A1" and "A2" and the difference in height "h" of the liquid levels in the two vertical tubes ?
Answer:
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]
Explanation:
Use Bernoulli's equation:
P₁ + ½ ρ v₁² + ρgz₁ = P₂ + ½ ρ v₂² + ρgz₂
Since there's no elevation change between points 1 and 2, z₁ = z₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Assuming incompressible fluid, the volumetric flow rate is the same at points 1 and 2.
Q₁ = Q₂
v₁ A₁ = v₂ A₂
v₂ = v₁ A₁ / A₂
Substituting:
P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ A₁ / A₂)²
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (A₁ / A₂)²
P₁ − P₂ = ½ ρ v₁² (A₁ / A₂)² − ½ ρ v₁²
P₁ − P₂ = ½ ρ v₁² ((A₁ / A₂)² − 1)
v₁² = 2 (P₁ − P₂) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2 (ρgh) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2gh / ((A₁ / A₂)² − 1)
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]
You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
A) Find the mass of one nucleus of the unknown element.
B) What is the speed of the unknown nucleus immediately after such a collision?
Answer:
a
The mass is [tex]m_2 =21.75*10^{-27} \ kg[/tex]
b
The velocity is [tex]v_2 = 3.0*10^{6} m/s[/tex]
Explanation:
From the question we are told that
The speed of the protons is [tex]u_1 = 2.10*10^{7} m/s[/tex]
The mass of the protons is [tex]m[/tex]
The speed of the rebounding protons are [tex]v_1 = -1.80 * 10^{7} \ m/s[/tex]
The negative sign shows that it is moving in the opposite direction
Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as
[tex]m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1[/tex]
Where [tex]m_1[/tex] is the mass of a single proton
So substituting values
[tex]m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1[/tex]
[tex]m_2 =13 m_1[/tex]
The mass of on proton is [tex]m_1 = 1.673 * 10^{-27} \ kg[/tex]
So [tex]m_2 =13 ( 1.673 * 10^{-27} )[/tex]
[tex]m_2 =21.75*10^{-27} \ kg[/tex]
Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as
[tex]m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2[/tex]
Now [tex]u_2[/tex] because before collision the the nucleus was at rest
So
[tex]m_1 u_1 = m_1 v_1 + m_2v_2[/tex]
=> [tex]v_2 = \frac{m_1(u_1 -v_1)}{m_2}[/tex]
Recall that [tex]m_2 =13 m_1[/tex]
So
[tex]v_2 = \frac{m_1(u_1 -v_1)}{13m_1}[/tex]
=> [tex]v_2 = \frac{(u_1 -v_1)}{13}[/tex]
substituting values
[tex]v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}[/tex]
[tex]v_2 = 3.0*10^{6} m/s[/tex]
A girl walks South at 2.7 m/s. What is the y component of her velocity
Answer:
-2.7m/s
Explanation:
...............
Who first used the word atom to describe the smallest unit
Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.
Explanation: hope this helped
A and B Two vectors are in the xy plane. If 4= 3m , |B|= 4m , and |A– B= 2 m find
a) The angle between B and A
b) The unit vector in the direction of (4× B)
Answer:
Explanation:
a)
A = 3m , B = 4m .
(A-B)² = A² + B² - 2ABcosθ where θ is angle between A and B.
4 = 9 + 16 - 2. 3.4 cosθ
cosθ = .875
θ = 29° .
b ) Unit vector in the direction of A X B will be k vector because A and B are in X-Y plane and A X B lies perpendicular to both A and B .
A laser beam is incident from the air at an angle of 30.0° to the vertical onto a solution of Karo syrup in water. If the beam is refracted to 19.24° to the vertical, what is the index of refraction of the syrup solution?
Answer:
Index of Refraction = 1.52
Explanation:
The index of refraction or the refractive index is given as:
Index of Refraction = Sin i/Sin r
where,
i = angle of incidence
r = angle of refraction
In this case of karo syrup, we have the following data:
i = angle of incidence = 30°
r = angle of refraction = 19.24°
Therefore, substituting these values in the equation, we get:
Index of Refraction = Sin 30°/Sin 19.24°
Index of Refraction = 0.5/0.3295
Index of Refraction = 1.52
Hence, the refractive index or the index of refraction of the Karo Syrup is found to be 1.52.
A car starting from rest, travels 0.40 km in 11.0 s. What is the
magnitude of its constant acceleration?
Answer:
6.61 m/s²
Explanation:
Given:
Δx = 0.40 km = 400 m
v₀ = 0 m/s
t = 11.0 s
Find: a
Δx = v₀ t + ½ at²
400 m = (0 m/s) (11.0 s) + ½ a (11.0 s)²
a = 6.61 m/s²
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s^{2} as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
Answer:
a) [tex]t \approx 9.879\,s[/tex], b) [tex]v = 20.518\,\frac{m}{s}[/tex], c) [tex]t = 16.368\,s[/tex], d) [tex]v = 19.545\,\frac{m}{s}[/tex]
Explanation:
a) Since train is only translating in a straight line and experimenting a constant deceleration throughout the station, whose length is 210 meters. The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:
[tex]210\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
The following second-order polynomial needs to be solved:
[tex]-0.075\cdot t^{2} + 22\cdot t - 210 = 0[/tex]
Whose roots are presented herein:
[tex]t_{1}\approx 283.455\,s[/tex] and [tex]t_{2} \approx 9.879\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train.
b) The speed of the nose leaving the station is given by this expression:
[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (9.879\,s)[/tex]
[tex]v = 20.518\,\frac{m}{s}[/tex]
c) First, it is required to calculate the time when nose of the train reaches a distance of 130 meters.
[tex]130\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]-0.075\cdot t^{2} + 22\cdot t - 130 = 0[/tex]
Roots of the second-order polynomial are:
[tex]t_{1} \approx 287.300\,s[/tex] and [tex]t_{2} \approx 6.033\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train. Now, the speed experimented by the train at this instant is:
[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (6.033\,s)[/tex]
[tex]v = 21.095\,\frac{m}{s}[/tex]
The distance traveled by the end of the train throughout station is modelled after the following equation:
[tex]210\,m = \left(21.095\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]-0.075\cdot t^{2} + 21.095\cdot t - 210 = 0[/tex]
Roots of the second-order polynomial are:
[tex]t_{1} \approx 270.932\,s[/tex] and [tex]t_{2} \approx 10.335\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train. The instant when the end of the train leaves the station is:
[tex]t = 6.033\,s + 10.335\,s[/tex]
[tex]t = 16.368\,s[/tex]
d) The velocity experimented by the end of the train is:
[tex]v = 21.095\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}} \right)\cdot (10.335\,s)[/tex]
[tex]v = 19.545\,\frac{m}{s}[/tex]
The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all. The negative sign indicates that the train is moving in the opposite direction of its initial velocity. The end of the train leaves the station after approximately 14.98 seconds. The velocity of the end of the train as it leaves the station is approximately 19.753 m/s.
(a) Using the equation :
v² = u² + 2as
0 = (22.0)² + 2 × (-0.150 ) × s
s = -(22.0 )^2 / (2 × -0.150)
s = 325.3 m
The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all.
(b) The velocity of the train when the nose leaves the station is given by:
v = u + at
v = 22.0 + (-0.150 ) × 210
v ≈ 22.0 - 31.5
v ≈ -9.5 m/s
The negative sign indicates that the train is moving in the opposite direction of its initial velocity.
(c)
s = ut + (1/2)at²
210= 22.0 × t + (1/2) × (-0.150) × t²
0.075 t² - 22.0 t + 210 = 0
we find two solutions for t: t ≈ 14.98 s and t ≈ 3.01 s.
The end of the train leaves the station after approximately 14.98 seconds.
(d) The velocity of the end of the train as it leaves the station is given by:
v = u + at
v = 22.0 + (-0.150) × 14.98
v = 22.0 - 2.247
v = 19.753 m/s
So, the velocity of the end of the train as it leaves the station is approximately 19.753 m/s.
To know more about the velocity:
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A moving walkway has a speed of 0.5 m/s to the east. A stationary observer
sees a man walking on the walkway with a velocity of 0.8 m/s to the east.
What is the man's velocity relative to the moving walkway?
Answer: 0.3 m/sec
Explanation:
Vrel = (Vs-Vm) = (0.8-0.5) = 0.3 m/sec
Answer:
0.3 m/s east
Explanation:
A 58.0 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.65 m long. The coefficient of kinetic friction between this patch and her skis is 0.310. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.50 m high.
Required:
a. How fast is the skier moving when she gets to the bottom of the hill?
b. How much internal energy was generated in crossing the rough patch?
Answer:
a) v = 3.71m/s
b) U = 616.71 J
Explanation:
a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:
[tex]-W_f=\Delta K=\frac{1}{2}m(v^2-v_o^2)\\\\-F_fd=\frac{1}{2}m(v^2-v_o^2)[/tex]
(the minus sign is due to the work is against the motion of the skier)
m: mass of the skier = 58.0 kg
v: final speed = ?
vo: initial speed = 6.00 m/s
d: distance traveled by the skier in the rough patch = 3.65 m
Ff: friction force = Mgμ
g: gravitational acceleration = 9.8 m/s^2
μ: friction coefficient = 0.310
You solve the equation (1) for v:
[tex]v=\sqrt{\frac{2F_fd}{m}+v_o^2}=\sqrt{\frac{2mg\mu d}{m}+v_o^2}\\\\v=\sqrt{-2g\mu d+v_o^2}[/tex]
Next, you replace the values of all parameters:
[tex]v=\sqrt{-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2}=3.71\frac{m}{s}[/tex]
The speed after the skier has crossed the roug path is 3.71m/s
b) The work done by the rough patch is the internal energy generated:
[tex]U=W_fd=F_fd=mg\mu d\\\\U=(58.0kg)(9.8m/s^2)(0.310)(3.50m)=616.71\ J[/tex]
The internal energy generated is 616.71J
A 60-turn coil has a diameter of 13 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.60 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.80 S 0.5973V
(b) 8.0 s 5.973 Xv
(c) 70 S 6.838- 3 v
Answer:
a) 0.5985 V
b) 0.05985 V
c) 0.00684 V
Explanation:
Given that
Number of turn in the coil, N = 60 turns
Magnetic field of the coil, B = 0.6 T
Diameter of the coil, d = 0.13 m
If area is given as, πd²/4, then
A = π * 0.13² * 1/4
A = 0.0133 m²
The induced emf, ε = -N(dΦ*m) /dt
Note, Φm = BA
Substituting for Φ, we have
ε = -NBA/t.
Now, we substitute for numbers in the equation
ε = -(60 * 0.6 * 0.0133)/0.8
ε = 0.4788/0.8
ε = 0.5985 V
at 8s,
ε = -(60 * 0.6 * 0.0132)/8
ε = 0.4788/8
ε = 0.05985 V
at 70s
ε = -(60 * 0.6 * 0.0132)/70
ε = 0.4788/70
ε = 0.00684 V
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
