Answer:
Explanation:
Let the mass of the thin block be M = .072 kg .
mass of bullet m = 4.67 x 10⁻³ kg.
speed of bullet v = 556 m/s.
speed of block after billet exits the block = 18 m/s.
We shall apply law of conservation of momentum
momentum of bullet block system
.072 x 0 + 4.67 x 10⁻³ x 556 = .072 x 18 + 4.67 x 10⁻³ x V here V is velocity of bullet after exit
= 2.59652 = 1.296 + 4.67 x 10⁻³ x V
4.67 x 10⁻³ x V = 1.30052
V = .278 x 10³ m /s
= 278 m /s
A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The battery has been connected for a long time. Each plate has an area A=58.3 m2 . The left plate is at ground potential and the right plate is at positive potential. The separation between the capacitor plates is D= 0.3m and all the points in the picture are far from the edges of the plates. The capacitor is placed on board of a space station where there is no gravity present.
In the coordinate system shown in the picture (x is a "horizontal" axis, y is "vertical" although names are purely conventional since there's no gravity), where all lengths are in meters, a positive charge q=3e-05C and mass m=0.4g has been placed directly in point C(0.25,12) and let go. What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
You decided to repeat the experiment but this time you place the same charge in point C first, and then connect the same power supply to the capacitor via the same resistor to charge it. Do not overcomplicated things: assume that the effect of any magnetic fields is negligible. Assuming that the axes scale on all graphs are the same what is the graph that best represents the horizontal speed of your charge as function of time in this experiment?
To verify your qualitative reasoning, calculate the speed of your charge in your second experiment after the same interval of time it took your charge to move from point C to point A in the first experiment. This is quite a challenging question so don't get disappointed if your group can't get the correct answer. You will not loose any points if you miss this question\
Answer:
See explanation
Explanation:
Given:-
- The DC power supply, Vo = 600 V
- The resistor, R = 1845 MΩ
- The plate area, A = 58.3 cm^2
- Left plate , ground, V = 0
- The right plate, positive potential.
- The distance between the two plates, D = 0.3 m
- The mass of the charge, m = 0.4 g
- The charge, q = 3*10^-5 C
- The point C = ( 0.25 , 12 )
- The point A = ( 0.05 , 12 )
Find:-
What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
Solution:-
- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.
E = Vo / D
E = 600 / 0.3
E = 2,000 V / m
- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:
Fe = E*q
Fe = (2,000 V / m) * ( 3*10^-5 C)
Fe = 0.06 N
- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:
Fnet = m*a
Where, a : The acceleration of the object/particle.
- The only unbalanced force acting on the particle is (Fe):
Fe = m*a
a = Fe / m
a = 0.06 / 0.0004
a = 150 m/s^2
- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:
s = C - A
s = ( 0.25 , 12 ) - ( 0.05 , 12 )
s = 0.2 m
- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:
vf^2 = vi^2 + 2*a*s
vf^2 = 0 + 2*0.2*150
vf = √60
vf = 7.746 m/s
- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:
vf - vi = a*t
t = ( 7.746 - 0 ) / 150
t = 0.0516 s
- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:
[tex]Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)][/tex]
- Where, The constant c : The capacitance of the capacitor.
- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:
[tex]E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\[/tex]
- Again, apply the Newton's second law of motion and determine the acceleration (a):
Fe = m*a
a = Fe / m
[tex]a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}[/tex]
- Where the acceleration is rate of change of velocity "dv/dt":
[tex]\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\[/tex]
- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:
[tex]c = \frac{A*eo}{d}[/tex]
Where, eo = 8.854 * 10^-12 .... permittivity of free space.
[tex]K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\[/tex]
- The differential equation turns out ot be:
[tex]\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\[/tex]
- Separate the variables the integrate over the interval :
t : ( 0 , t )
v : ( 0 , vf )
Therefore,
[tex]\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )[/tex]
- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:
[tex]vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s[/tex]
- The final velocity of particle while charging the capacitor would be:
vf = 7.264 m/s ... slightly less for the fully charged capacitor
A car is moving down a flat, horizontal highway at a constant speed of 21 m/s when suddenly a rock dropped from rest straight down from a bridge crashes through the windshield. The rock strikes the windshield at a right angle, and the windshield makes a 60 degree angle with the horizontal.
a. Find the speed of the rock relative to the car when it hits the windshield.
b. Find the distance above the point of contact with the windshield from which the rock was dropped.
c. If the driver had seen what was about to happen just as the rock was being released, how much time would she have had to react before the rock strikes the windshield?
Answer:
a)9.8
Explanation:
because any object falling from any height is 9.8
(sorry that's the only one i know)
1. The speed of the rock relative to the car when it hits the windshield is [tex]\( {42.00 \, \text{m/s}} \).[/tex]
2. The distance above the point of contact with the windshield from which the rock was dropped is [tex]\( {67.5 \, \text{m}} \).[/tex]
3. The driver would have had [tex]\( {3.71 \, \text{s}} \)[/tex] to react before the rock strikes the windshield.
Let's solve each part of the problem step-by-step.
Part (a): Find the speed of the rock relative to the car when it hits the windshield.
Given:
- Car speed, [tex]\( v_{\text{car}} = 21 \, \text{m/s} \)[/tex]
- Windshield angle with the horizontal,[tex]\( \theta = 60^\circ \)[/tex]
Since the rock is dropped from rest, its initial vertical velocity is [tex]\( 0 \, \text{m/s} \).[/tex]
The rock falls under the influence of gravity, so its vertical velocity [tex]\( v_{\text{vertical}} \)[/tex] at any time t is given by:
[tex]\[ v_{\text{vertical}} = g t \][/tex]
where [tex]\( g = 9.8 \, \text{m/s}^2 \).[/tex]
The horizontal velocity of the rock is [tex]\( 0 \, \text{m/s} \)[/tex] with respect to the ground.
To find the speed of the rock relative to the car when it hits the windshield, we need to consider the components of velocity with respect to the car:
1. The car's horizontal velocity component.
2. The rock's vertical velocity component.
The velocity components with respect to the car:
- Horizontal component relative to the car, [tex]\( v_{\text{horizontal, relative}} = v_{\text{car}} \)[/tex]
- Vertical component, [tex]\( v_{\text{vertical}} = g t \)[/tex]
The speed of the rock relative to the car can be found using the Pythagorean theorem:
[tex]\[ v_{\text{relative}} = \sqrt{(v_{\text{horizontal, relative}})^2 + (v_{\text{vertical}})^2} \][/tex]
We need to find t when the rock strikes the windshield.
Since the rock strikes the windshield at a right angle, and the windshield makes a 60° angle with the horizontal, the relative speed of the rock [tex]\( v_{\text{relative}} \)[/tex] must have components that align with the 60° angle.
Thus:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{relative}} \sin(60^\circ) \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{relative}} \cos(60^\circ) \][/tex]
Given:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{vertical}} = g t \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{car}} = 21 \, \text{m/s} \][/tex]
From the component relationship:
[tex]\[ \frac{v_{\text{relative, vertical}}}{v_{\text{relative, horizontal}}} = \tan(60^\circ) = \sqrt{3} \][/tex]
[tex]\[ \frac{g t}{21} = \sqrt{3} \][/tex]
[tex]\[ g t = 21 \sqrt{3} \][/tex]
[tex]\[ t = \frac{21 \sqrt{3}}{9.8} \][/tex]
Calculate t:
[tex]\[ t \approx \frac{21 \cdot 1.732}{9.8} \approx \frac{36.372}{9.8} \approx 3.71 \, \text{s} \][/tex]
Now, we can find the vertical velocity [tex]\( v_{\text{vertical}} \)[/tex]:
[tex]\[ v_{\text{vertical}} = g t = 9.8 \times 3.71 \approx 36.36 \, \text{m/s} \][/tex]
Now, the speed of the rock relative to the car:
[tex]\[ v_{\text{relative}} = \sqrt{(21)^2 + (36.36)^2} \approx \sqrt{441 + 1322.49} \approx \sqrt{1763.49} \approx 42.00 \, \text{m/s} \][/tex]
Part (b): Find the distance above the point of contact with the windshield from which the rock was dropped.
The distance h from which the rock was dropped can be found using the kinematic equation:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
Using [tex]\( t \approx 3.71 \, \text{s} \)[/tex]:
[tex]\[ h = \frac{1}{2} \times 9.8 \times (3.71)^2 \approx \frac{1}{2} \times 9.8 \times 13.76 \approx 67.5 \, \text{m} \][/tex]
Part (c): The time available for the driver to react is the time \( t \) it takes for the rock to fall, which we calculated in part (a):
[tex]\[ t \approx 3.71 \, \text{s} \][/tex]
A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. The central bright fringe becomes narrower. The effect cannot be determined unless the distance between the slit and the screen is known. The central bright fringe becomes wider. The central bright fringe remains the same size.
Answer:
Explanation:
General guidance
Concepts and reason
The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.
Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.
Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.
Fundamentals
The condition for diffraction maximum in the case of single slit diffraction is as follows:
sin Θ=λ/α
Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .
The incorrect options are as follows:
• The central bright fringe remains in same size.
The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.
• The effect cannot be determined unless the distance between the slit and screen is known.
Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.
• The central bright fringe becomes narrower.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.
If the width of the slit increases, then the central bright fringe width also increases.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.
The condition for diffraction maximum is as follows:
sin Θ=λ/α
The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
Therefore,
• The central bright fringe becomes wider is correct.
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.
Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.
If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.
Answer
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.
A charge of 9.0 x 10-5C is placed in an electric field with a strength of 4.0 x 104 N. What is the electrical force acting
on the charge?
Answer:
3.6 N
Explanation:
E=F/q
so
F=qxE
F= 9x10^-5x4x10^4
F= 3.6 N
The top surface of the passenger car of a train moving at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top surface is absorbing solar radiation at a rate of 200 W/m2, and the temperature of the ambient air is 308C. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface of the car.
Answer:
33.1 degrees celcius
Explanation:
Look up attached file
The equilibrium temperature of the train’s top cannot be calculated without the specific heat transfer due to convection. The Stefan-Boltzmann law allows us to calculate heat transfer from radiation but not convection.
Explanation:The key to solving this problem is the Stefan-Boltzmann law of radiation, which describes the rate of heat transfer by emitted radiation. In brief, P = σeAT⁴, where σ = 5.67 × 10⁻⁸ J/s.m².K⁴ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, and T is its temperature in Kelvins.
Given that the top surface of the car absorbs solar radiation at a rate of 200 W/m² and its dimensions, we can calculate the area of the surface. The surface, if deemed perfectly insulated, only loses heat through convection, not radiation. However, the amount of heat transfer due to convection is not provided in the question, making it impossible to calculate the exact equilibrium temperature of the top surface of the car.
The net rate of heat transfer takes into account both the temperature of the object and the temperature of its surroundings. If T₁ represents the temperature of the surface to be investigated, and T₂ the ambient temperature, without the value of heat lost through convection, we cannot calculate T₁. Therefore, more information would be required to answer your question fully.
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You are asked to construct a mobile with four equal m = 165 kg masses, and three light rods of negligible mass and equal lengths. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached? L1 = (b) At what location on the level 2 rod should the free end of rod 3 be attached? L2 = (c) At what location on the level 3 rod should the whole assembly be suspended from so that the mobile is in equilibrium? L3 =
Answer:
Explanation:
the picture attached shows the whole explanation
4-True-False (experiment 6) : a) 1/(f^2) vs output current is a straight line ( f is frequency) b) Magnetic field of Helmholtz coil and earth are constant during the experiment. c) The sign of B_axial in experiment 6.2 can change the results for the value of N. d) In an ideal selenoid, the sign of B_radial is independent of sign of B_axial e) In experiment 6.1, If we use a magnetic dipole with parameters 4m and 2I, the intercept of 1/(T^2) vs current stays the same since the magnetic field of earth is constant.
Answer:
A.True
B. True
C False
D. False
E. True
Explanation:
A leaf spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 6.2 cm (0.062 m) from its original unstretched position. What is the increase in potential energy of the spring (in kJ)?
The increase in potential energy of the spring when compressed a distance of 0.062m from its unstretched position, with a spring constant of 87600N/m, is approximately 0.168 kJ.
Explanation:The increase in potential energy of the spring can be calculated using the formula U = ½kx². In this problem, k is the spring constant which is 87,600 N/m and x is the displacement of the leaf spring when it is compressed, which is 0.062 m. Plugging in these values, we get U = ½ * (87,600 N/m) * (0.062 m)² = 167.5424 Joules. This value is in Joules, so to convert it to kilojoules (kJ), we divide by 1000. So, the increase in potential energy of the spring is approximately 0.168 kJ.
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An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 46.8 H inductor, determine the inductive reactance and the rms current through the inductor.(a) the inductive reactance (in Ω)
(b) the rms current through the inductor (in A)
Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A
A uniform electric field, with a magnitude of 650 N/C, is directed parallel to the positive x-axis. If the potential at x = 3.0 m is 1 700 V, what is the potential at x = 1.0 m?
The potential at x = 1.0 m in the given uniform electric field, with the given parameters, would be 400 V.
Explanation:The subject question pertains to the concept of electric potential in a uniform electric field. The potential difference, V, in an uniform electric field is given by the equation V = Ed, where E is the field strength and d is the distance. Given an electric field strength, E, of 650 N/C and a potential, V, at x = 3.0 m of 1 700 V, we seek the potential at x = 1.0m. Since the electric field is uniform, we know that the change in potential is linear with distance. Because the distance decreases by 2 meters (from x = 3.0 m to x = 1.0 m), the potential will decrease by E*2, which is 650 N/C * 2 m = 1300 V. So, the potential at x = 1.0 m is 1 700 V - 1 300 V = 400 V.
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To find the electric potential at x = 1.0 m in a uniform electric field directed along the x-axis, subtract the potential difference due to the field over the distance moved from the initial potential; the answer is 400 V.
The question is asking to calculate the electric potential at a certain point given a uniform electric field and an initial electric potential at a different point. To find the potential at x = 1.0 m, we can use the relationship between electric field (E), electric potential (V), and distance (d), with the following formula: V = V_0 - E * d, where V is the potential at the point of interest, V_0 is the initial potential, E is the electric field strength, and d is the distance between the points.
Gor the given electric field of 650 N/C directed parallel to the positive x-axis, and with the potential at x = 3.0 m being 1700 V, the potential difference \\Delta V caused by moving from x = 3.0 m to x = 1.0 m can be calculated by multiplying the electric field strength by the distance over which the field is applied, which is 2.0 m (3.0 m - 1.0 m). Hence, \\Delta V = E * d = 650 N/C * 2.0 m = 1300 V.
Therefore, the potential at x = 1.0 m is V = 1700 V - 1300 V = 400 V.
An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 4.29 cm in a uniform magnetic field with B = 1.53 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
Answer:
A) speed = 3144287.425 m/s
B) period = 8.57x10^-8 s
C) KE = 3.302x10^-14 J
D) potential difference = 103187.5 V
Explanation:
Detailed explanation and calculation is shown in the image below.
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Answer:
The order is 2>4>3>1 (TE)
Explanation:
Look up attached file
Global Precipitation Measurement (GPM) is a tool scientist use to forecast weather. Which statements describe GPM
Answer:
Correct answers:
Explanation:
• It is a satellite that collects data about rain and snow.
GPM is part of NASA's Earth Systematic Missions program and its function is to track precipitation from space with satellites and to provide accurate information of the time, place an amount of rain or snow anywhere in the world.
• Its orbit covers 90 percent of Earth's surface.
GPM has unique perspective on measuring precipitation from space because of their collected satellite retrievals.
• The sensors measure microwaves.
GPM Microwave Imager (GMI) is a sensor that observes the microwave energy emitted by the Earth and atmosphere, so it improves the accuracy of rain and snowfall estimates.
Light of wavelength 199 nm shines on a metal surface. 4.08 eV is required to eject an electron. What is the kinetic energy of
(a) the fastest and
(b) the slowest ejected electrons?
(c) What is the stopping potential for this situation?
(d) What is the cutoff wavelength for this metal?
Answer:
a) = 6.23eV
b) = 0eV
c) = 6.23V
d) = 304.12nm
Explanation:
λ = 199nm
E = 4.08eV = Φ
hf = K(max) + Φ
Φ = work function of the target material
hf = photon energy
K(max) = k.e of the metal
K(max) = [(6.626*10⁻³⁴) * (3.0*10⁸)] / 199*10⁻⁹
K(max) = 9.99*10⁻¹⁹J
1.602*10⁻¹⁹J = 1eV
9.99*10⁻¹⁹ = 6.23eV
K(max) = 6.23eV
b). The slowest moving electron has a kinetic energy of zero
c). The stopping potential is the potential difference required to stop the fastest electron.
eV = K(max)
V = K(max) / e
V = 6.23eV / e
V = 6.23V
d)
The cut- off wavelength occurs when the maximum K.E is zero.
Φ = hc / λ
λ₀ = hc / Φ
λ₀ = (6.626*10⁻³⁴ * 3.0*10⁸) / (4.08 * 1.602*10⁻¹⁹)
λ₀ = 1.9878*10⁻²⁵ / 6.5361*10⁻¹⁹
λ₀ = 3.04*10⁻⁷m
λ₀ = 304.12nm
) Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 1.50 × 105 kg and a velocity of 0.30 m/s to the right, and the second having a mass of 1.10 × 105 kg and a velocity of 0.12 m/s to the left. What is their final velocity?
Answer:
Their final velocity is 0.122m/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision. The bodies move with a common velocity after collision.
Momentum = mass × velocity
BEFORE COLLISION
momentum of the first train
= 1.5×10^5 × 0.3
= 0.45×10^5kgm/s
Momentum of second train
= 1.1×10^5 × (-0.12) (velocity is negative since the body is moving towards the left)
= -0.132 × 10^5kgm/s
AFTER COLLISION
Momentum of both bodies is given as:
(1.5×10^5+1.1×10^5)V
= 2.6×10^5V
V is their common velocity
According to the law:
0.45×10^5+(-0.132×10^5) =2.6×10^5V
0.45×10^5 - 0.132×10^5 = 2.6×10^5V
0.318×10^5 = 2.6×10^5V
0.318 = 2.6V
V = 0.318/2.6
V = 0.122m/s
7. A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of 30.0 ???? applied over 0.30 seconds. What is the length of the pendulum if it has a period of 5.0 seconds? What is the maximum angle of displacement of the swinging pendulum?
Answer:
Explanation:
For pendulum , time period is
T = 2π √ l / g
T² = 4π² l/ g
l = / g T² / 4π²
= 9.8 x 5² / 4π²
= 6.21 m
Angular impulse = Torque x time
= 30 x 6.21 x .3 = increase in angular momentum
30 x 6.21 x .3 = I ω ; I is moment of inertia and ω is angular velocity .
I ω = 55.89
I = 1/3 m l²
= 1/3 x 5 x 6.21²
= 64.27
Angular kinetic energy = 1/2 I ω²
= 1/2(I ω)²/ I
= .5x 55.89²/ 64.27
= 24.30 J
Due to it , the centre of mass of pendulum increases by height h so that
h = l/2 ( 1 - cos θ )
1/2 I ω² = mg l/2 ( 1 - cos θ)
24.3 = 5 x 9.8 x 3.1 ( 1 - cos θ)
( 1 - cos θ) = .16
cos θ = .84
θ = 33 degree.
The length of the pendulum is about 6.25m and the maximum angle of displacement would occur when the pendulum is kicked with the maximum force of 30N.
Explanation:The length of a pendulum is determined by the formula T=2π√(L/g). Given that the period T is 5.0 seconds and g (acceleration due to gravity) is 9.8 m/s², we can rearrange the equation to solve for L (length), resulting in L = (T² * g) / (4π²) = (5.0)² * 9.8 / (4π²) ≈ 6.25m. This regard to the maximum angle of displacement, when a pendulum is in simple harmonic motion, which is typically for displacements less than 15 degrees, the displacement is directly proportional to its force. Therefore, the maximum angle would occur when the force is the greatest, in this case, when the student kicks the pendulum with a force of 30 N.
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An earthquake on the ocean floor in the Gulf of Alaska induces a tsunami (sometimes called a "tidal wave") that reaches Hilo, Hawaii, 4780 km away, after a time of 9.99 h. Tsunami’s have enormous wavelengths (100- 200 km), and for such waves the propagation speed is v ≈ qgd , where d is the average depth of the water, d is in meters, v is in m/s, and g = 9.8 m/s^2 . From the information given, find the average ocean depth between Alaska and Hawaii.
Answer:
23785m
Explanation:
V= 4780/9.9= 482.8km/h
d= v²/g = 482.8²/9.8= 23785m
The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the coil and ΦB is in webers. At one instant, the coil is observed to be rotating at a rate of 8.70 ✕ 102 rev/min. (Assume that t is in seconds.) (a) What is the induced emf in the coil as a function of time for this angular speed? (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that e m f is in volts. Do not include units
Answer:
Explanation:
Given that a coil has a turns of
N = 110 turns
And the flux is given as function of t
ΦB = 9.75 ✕ 10^-3 sin(ωt),
Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min
ω = 8.70 ✕ 10² rev/min
Converting this to rad/sec
1 rev = 2πrad
Then,
ω = 8.7 × 10² × 2π / 60
ω = 91.11 rad/s
Now, we want to find the induced EMF as a function of time
EMF is given as
ε = —NdΦB/dt
ΦB = 9.75 ✕ 10^-3 sin(ωt),
dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)
So,
ε = —NdΦB/dt
ε = —110 × 9.75 × 10^-3•ω Cos(ωt)
Since ω = 91.11 rad/s
ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)
ε = —97.71 Cos(91.11t)
The EMF as a function of time is
ε = —97.71 Cos(91.11t)
Extra
The maximum EMF will be when Cos(91.11t) = -1
Then, maximum emf = 97.71V
A 1.005 m chain consists of small spherical beads, each with a mass of 1.00 g and a diameter of 5.00 mm, threaded on an elastic strand with negligible mass such that adjacent beads are separated by a center-to-center distance of 10.0 mm. There are beads at each end of the chain. The strand has a spring constant of 28.8 N/m. The chain is stretched horizontally on a frictionless tabletop to a length of 1.50 m, and the beads at both ends are fixed in place.
1. What is the linear mass density of the chain? What is the tension in the chain?
2. With what speed would a pulse travel down the chain?
3. The chain is set vibrating and exhibits a standing-wave pattern with four antinodes. What is the frequency of this motion?
4. If the beads are numbered sequentially from 1 to 101, what are the numbers of the five beads that remain motionless?
5. The 13th bead has a maximum speed of 7.54 m/s. What is the amplitude of that bead's motion?
6. If x0=0 corresponds to the center of the 1st bead and x101=1.50m corresponds to the center of the 101st bead, what is the position xn of the nth bead? Calculate the first four coordinates.
7. What is the maximum speed of the 30th bead?
Answer:
1) μ = 1.33 10⁻³ kg / m , F = - 14,256 , 2) v= 103.53 m/s, 3) f = 138.04 Hz , 4) 1, 25, 50, 76, 101 , 5) A = 0.00869 m , 6) # _position = (# _account-1) (1.5m / 100 accounts)
Explanation:
1) Linear density is the mass per unit length
μ = m / L
μ = 2 1 10⁻³ / 1,5
μ = 1.33 10⁻³ kg / m
this is the density when the chain is stretched, which is when the pulse occurs
we can find the tension with
F = - k (x₁-x₀)
where k is the spring constant
F = - 28.8 (1.5 -1.005)
F = - 14.256 N
the negative sign indicates that the force is restorative
2) the pulse speed is
v = √ T /μ
v = √ 14,256 / 1,33 10⁻³
v = 103.53 m / s
3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is
2 λ = L
λ = L / 2
wave speed is related to frequency and wavelength
v = λ f
f = v / λ
f = v 2 / L
f = 103.53 2 / 1.5
f = 138.04 Hz
4) The marbles are numbered, the marbles that remain motionless are
the first (1) and the last (101)
Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in
#_node = m λ / 2 = m L / 4
#_node position (m)
1 1.5 / 4 = 0.375
2 2 1.5 / 4 = 0.75
3 3 1.5 / 4 = 1,125
Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes
#_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33
# _canica = 25
#_canica = 0.75 67.33
#_canica = 50
# _canica = 1,125 67.33
#canica = 75.7 = 76
in short the number of the fixed marbles is
1, 25, 50, 76, 101 canic
5) The movement of the account is oscillatory at this point, which is why it is described by
y = A cos wt
[tex]v_{y}[/tex]= -A w sin wt
the speed is maximum for when the breast is worth ±1
v_{y} = Aw
A = v_{y} / w
angular velocity related to frequency
w = 2π f
A = v_{y} / 2πf
A = 7.54 / (2π 138.04)
A = 0.00869 m
6) for the position of each account we can use a direct proportion rule
in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount
# _position = (# _account-1) (1.5m / 100 accounts)
#_canic position(m)
1 0
2 0.015
3 0.045
4 0.06
7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression
y = Acos wt
the maximum speed is
[tex]v_{y}[/tex] = -Aw sin wt
speed is maximum when the sine is ±1
v_{y} = A w
to calculate the amplitude of the count we use that for a standing wave
y = 2Asin kx
y / A = 2 sin (2π /λ x)
the wavelength is
λ = 0.75 m
the position is
x (30) = 29 1.5 / 100 = 0.435 m
y (30) A = 2 sin (2pi 0.435 / 0.75)
y (30) / A = 0.96 m
A T-shirt cannon is made of pipe within which a light spring, with k=100 N/m, can be compressed to launch the bundled shirt of mass 0.3 kg. Imagine the spring compressed 0.15 m and then released. Ignoring friction, calculate the launch velocity (speed at which the T-shirt leaves the spring) in two cases: A horizontal launch A vertical launch HTML EditorKeyboard Shortcuts
Answer:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Explanation:
The launch speed is calculated by means of the Principle of Energy Conservation:
[tex]U_{k} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = x \cdot \sqrt{\frac{k}{m} }[/tex]
[tex]v = (0.15\,m)\cdot \sqrt{\frac{100\,\frac{N}{m} }{0.3\,kg} }[/tex]
[tex]v \approx 2.739\,\frac{m}{s}[/tex]
The velocities for each scenario are presented herein:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Final answer:
The launch velocity of the T-shirt, when the spring with a spring constant of 100 N/m is compressed by 0.15 m, is calculated using conservation of energy. It turns out to be approximately 2.74 m/s for both the horizontal and vertical launches.
Explanation:
To calculate the launch velocity of the T-shirt from the cannon in both horizontal and vertical scenarios, we'll use the conservation of energy principle. This principle states that the potential energy stored in the compressed spring is converted into kinetic energy of the moving T-shirt.
Calculation of Launch Velocity (Horizontal and Vertical)
The energy stored in the compressed spring can be given by the formula:
[tex]E = 1/2 k x^2[/tex]
where E is the energy in joules, k is the spring constant, and x is the compression distance.
For the given spring with k = 100 N/m compressed by 0.15 m, the energy stored is:
[tex]E = 1/2 (100) (0.15)^2 = 1.125 joules[/tex]
Now we set this equal to the kinetic energy of the T-shirt upon release:
[tex]KE = 1/2 m v^2[/tex]
where KE is the kinetic energy, m is the mass of the T-shirt, and v is the velocity.
To find the launch velocity, we solve for v given the mass m = 0.3 kg and energy E = KE:
[tex]v = sqrt((2 * E) / m) = sqrt((2 * 1.125) / 0.3) = sqrt(7.5) = approximately 2.74 m/s[/tex]
This velocity will be the same for both the horizontal and the vertical launch as friction is ignored and the energy is purely transferred into the kinetic energy of the T-shirt.
A coil of 43 turns in the shape of a rectangle with width 5 cm and length 10 cm is dropped from a position where magnetic field is 0 T to a position where the field is 0.5 T and is directed perpendicular to the plane of the coil. The displacement occurs in a time of 0.431 s. Calculate the resulting average emf induced in the coil
Answer:
Explanation:
Given that,
Number of turns
N = 43 turns
Dimensions of rectangle
Length L = 10cm = 0.1m
Width W = 5cm = 0.05
Then, area of rectangle
A = L×W = 0.1 × 0.05 = 0.005m²
Initial magnetic field
Bi = 0T
Final magnetic field
Bf = 0.5T
The displacement time
∆t = 0.431s
Induce emf ε?
EMF is given as
ε = —NdΦ/dt
Where magnetic flux
Φ = BA
Note A is constant
Then,
ε = —NdΦ/dt = —Nd(BA)/dt
ε = —N•A•dB/dt
ε = —N•A•(∆B/∆t)
ε = —N•A•(Bf - Bi) / (∆t)
ε = —43 × 0.005 × (0.5 - 0) /0.431
ε = —43 × 0.005 × 0.5 /0.431
ε = —0.249 V
The average induced EMF is 0.249V
4
An 80 kg skier stands at the top of a 40 meter slope. What is her gravitational potential energy
before she skies down the slope?
O 3200
O 31,360
19.63
O 15,680
Answer:
31,360J
Explanation:
Gravitation potential energy (gpe) is calculated from the formula mgh.
That implies, gpe = mgh
Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J
Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.
The gravitational potential energy of the skier at the top of the slope, calculated using the formula PE = mgh, is 31,360 joules.
Explanation:To find the gravitational potential energy of the skier, we use the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration because of gravity, and h is the height. The acceleration because gravity is 9.8 m/s².
So, we substitute these values into the formula:
PE = (80 kg) × (9.8 m/s²) × (40 m) = 31,360 joules.
Therefore, the gravitational potential energy of the skier before she skips down the slope is 31,360 joules.
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HELP ASAP PLEASE!!!
The earth is:
A. a perfect sphere
B. thinner at the equator
C. slightly flatter at the poles
D. made of mostly carbon and iron
The Earth is slightly flatter at the poles, making the correct answer option C. It is an oblate spheroid with a denser core made primarily of iron and nickel.
The correct answer to the question, 'The earth is:', is C: slightly flatter at the poles. This is because the Earth's shape is an oblate spheroid, meaning it's somewhat squished at the poles and bulges a bit at the equator due to its rotation. The Earth's equatorial radius is greater than its polar radius, which results in this slight flattening at the poles. Also, the Earth's core is composed predominantly of iron and nickel, making it very dense.
While it is true that the Earth is not a perfect sphere, it is also not thinner at the equator as option B suggests. Option D mentions that Earth is made of mostly carbon and iron, which is incorrect; the Earth is composed primarily of iron, oxygen, silicon, and other elements, but not mostly carbon.
According to the theory, the average force Favg, applied to the mass multiplied by the duration the force is applied should equal the mass change in momentum. For each run calculate the change in momentum of the mass using the average force, then find the % difference between that and change in momentum using the change in velocity.
Answer:
The change in momentum for average force is: Δ[tex]P_{average force} = F_{avg}[/tex]Δt and it is calculated below.
For thick and thin spring, the %difference between change in momentum for average force and change in momentum for velocities is; 178.27% and 159.72% respectively
Explanation:
The average force applied to an object can be calculated by multiplying the average net external force by the time interval over which the force acts. The change in momentum can be calculated using the average force, as well as the change in velocity. The percentage difference between the two values can be found by dividing the difference by the change in momentum using the change in velocity and then multiplying by 100.
Explanation:The average force applied to an object can be calculated using the average net external force multiplied by the time interval over which the force acts. This relationship is known as the impulse-momentum theorem, which states that the change in momentum is equal to the impulse. Impulse is the product of the average force and the time interval.
To calculate the change in momentum of an object using the average force, you can multiply the average force by the duration the force is applied. This will give you the change in momentum caused by the force. You can then calculate the change in momentum using the change in velocity by multiplying the mass of the object by the change in velocity. Finally, you can find the percentage difference between the two values by dividing the difference between the two values by the change in momentum using the change in velocity and then multiplying by 100.
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Because the direction of Earth's motion around the Sun continually changes during the year, the apparent position of a star in the sky moves in a small loop, known as the aberration of starlight. In order to better understand this phenomenon, it is sometimes helpful to use visual analogies. In these visual analogies, the car is analogous to the Earth, and the rainfall is analogous to starlight. Determine which visual analogies correspond to the following scenarios:
A) The Earth moving around the Sun and interacting with light from a distant star
B) A person on the moving Earth observing the light from a distant star
C) A person on a motionless Earth observing the light from a distant star
Answer:
B) A person on the moving Earth observing the light from a distant star
Explanation:
The aberration of starlight is a phenomenon in Astronomy used to describe the 'seeming' or apparent movement of a star around its true position. This seeming' or apparent movement of a star is caused by and dependent on the velocity of the observer. The movement of the observer relative to the star creates the notion that the star is moving. However, in reality, the star is static. The star is not moving, what is really moving is the observer; this movement of the observer is what makes it seem that the star was moving. It is not an optical illusion, although, the effect seems close enough.
For example, in the analogy given:
A car moving in the rain or under the rainfall
The car is the one moving in the same way that the Earth is the one revolving. The rainfall drops vertically in the same way that the starlight is static. The movement of the car relative to the rainfall is what makes it seem that the rain is falling 'diagonally'. The same visual analogy is observe when a person on the moving Earth observing the light from a distant star
Hence, the correct option is B
Question 1
The orange line is measuring the
Answer:
Approximately 4.5 square miles
Explanation:
(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)
Final answer:
The 'orange line' referred to in the question likely represents data in a graph related to either the wavelength or frequency of orange light in the visible spectrum or a trend in statistical analysis. The exact answer depends on the specific context of the graph or data set from which the 'orange line' is derived.
Explanation:
The question 'The orange line is measuring the' seems to be related to analyzing data, most likely from a graph or chart that measures certain variables, potentially in the context of physics or mathematics dealing with light spectra or statistical analysis. Given the details provided, including mention of the visible light spectrum and specific wavelengths denoted by numbers, it appears that the 'orange line' may refer to a line on a graph that represents the wavelength or frequency of the color orange within the visible light spectrum.
However, the context provided is ambiguous. For instance, the chapter headings and the Conclusion provided suggests statistical analysis, perhaps in the field of queueing theory and how waiting times are affected by different systems, which implies a more mathematical or business-oriented question. In the field of mathematics, 'orange line' could also be a reference to a trend line or data line in a statistical plot.
(hrw8c13p51) A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as 0.11. What is the semimajor axis of the comet's orbit
Answer:
Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m
Explanation:
From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month
Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds
Now, the square of the period of any planet is given by the formula;
T² = (4π²/GM)α³
Where;
G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²
M is mass of sun which has a value of 1.99 x 10^(30) kg
α is the semi major axis of the comets orbit.
Thus, making α the subject, we have;
α = ∛(GMT²/4π²)
So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)
α = 1.89 x 10^(13) m
Write the ordered pair that represents vec yz .Then find the magnitude of vec yz Y(- 4, 12), (1, 19)
Answer with Explanation:
We are given that
Y=(-4,12)
Z=(1,19)
We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.
Vector YZ=Z-Y=<1,19>-<-4,12>
Vector YZ=<5,7>
Magnitude of vector r
[tex]r=xi+yj[/tex]
[tex]\mid r\mid=\sqrt{x^2+y^2}[/tex]
Using the formula
[tex]\vec{YZ}=\sqrt{5^2+7^2}[/tex]
[tex]\vec{YZ}=\sqrt{25+49}=\sqrt{74} units[/tex]
Assume that a uniform magnetic field is directed into this page. If an electron is released with an initial velocity directed from the bottom edge to the top edge of the page, which of the following describes the direction of the resultant force acting on the electron?
a. out of the page
b. to the right
c. to the left
d. into the page.
Final answer:
The direction of the resultant force acting on an electron moving upward on the page through a magnetic field directed into the page is to the left, as per the left-hand rule.
Explanation:
The question deals with the Lorentz force experienced by a charged particle, in this case an electron, moving through a magnetic field. When an electron with initial velocity directed from the bottom edge to the top edge of the page moves through a uniform magnetic field directed into the page, the force acting on it can be determined using the left-hand rule, since electrons have a negative charge. Holding your left thumb up (direction of the electron's velocity) and the fingers into the page (direction of the magnetic field), the resultant force will be directed to the left of the page. Therefore, the correct answer to the direction of the resultant force acting on the electron is c. to the left.
ii. At the instant shown in Figure 3, the blocks are moving toward each other with the same speed of 0.35m/s. The blocks collide 0.50 seconds later. What is the speed of the two-block system’s center of mass just before the blocks collide?
Answer:
2.9 m/s
Explanation:
The speed of the two-block system center of mass will not be changed by the collision, since it is an internal force. The only force acting upon the two-block system, then, will be gravity. We can calculate the effects of gravity upon the speed of the blocks along the slope by taking the sine of the angle of the slope (37°) and multiplying it by the magnitude of Fg in the vertical direction. sin(37°)*(3.92 N) = 2.36 N. This force acts upon the system and accelerates it down the slope, which can be modeled with the equation 2.36 N = (0.40 kg)*a. Solving for a, we find that gravity accelerates the block at 5.9 m/s2 along the slope of the block. Using this figure, we can find the speed of the system with the equation v = v0 + at. Initial velocity of the system is zero, as both blocks are moving towards each other at equal speed, so v = (5.9 m/s2)*(0.50 s). Velocity of the system after 0.50 seconds = 2.9 m/s.
The center-of-mass of an isolated two-block system moving towards each other at the same speed remains the same before the collision; therefore, the center-of-mass velocity is also 0.35 m/s just before the blocks collide.
Explanation:In this scenario, we are dealing with the principle of conservation of momentum in an isolated system. In two-dimensional collisions involving identical masses moving toward each other at the same speed, the center-of-mass velocity does not change due to the collision.
To find the speed of the two-block system's center of mass, we can use the formula for center-of-mass velocity, that is the weighted average of the individual velocities. Since the blocks have identical mass, the center-of-mass velocity is just the average of their individual velocities. Given both blocks are moving toward each other at a speed of 0.35 m/s, the speed of the center-of-mass of the system before the blocks collide is also 0.35 m/s.
Given two identical blocks moving towards each other at a speed of 0.35 m/s, we can denote the velocity of each block as +0.35 m/s and -0.35 m/s. The positive and negative signs apply due to the blocks moving towards each other.The center-of-mass velocity (V_CM) is found by using the formula: V_CM = (m1*v1 + m2*v2) / (m1 + m2) where m1 and m2 are the masses of the blocks and v1 and v2 are their respective velocities.Since both blocks have the same mass and velocity, except for direction, the center-of-mass velocity is 0.Learn more about Center-of-Mass Velocity here:https://brainly.com/question/32099134
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