Answer:
Option b
Solution:
As per the question:
Speed of the mouse, v = 1.3 m/s
Speed of the cat, v' = 2.5 m/s
Angle, [tex]\theta = 38^{\circ}[/tex]
Now,
To calculate the distance between the mouse and the cat:
The distance that the cat moved is given by:
[tex]x = v'cos\theta t[/tex]
[tex]x = 2.5cos38^{\circ}\times t = 1.97t[/tex]
The position of the cat and the mouse can be given by:
[tex]x = x' + vt[/tex]
[tex]1.97t = x' + 1.3t[/tex]
x' = 0.67 t (1)
The initial speed of the cat ahead of the mouse:
u = [tex]v'sin\theta = 2.5sin38^{\circ} = 1.539\ m/s[/tex]
When the time is 0.5t, the speed of the cat is 0, thus:
[tex]0 = u - 0.5tg[/tex]
[tex]t = \frac{1.539}{0.5\times 9.8} = 0.314\ s[/tex]
Substituting the value of t in eqn (1):
x' = 0.67(0.314) = 0.210 m
Thus the distance comes out to be 0.210 m
Calculate the mean free path of air molecules at a pressure of 3.00×10−13 atm and a temperature of 304 K . (This pressure is readily attainable in the laboratory.) Model the air molecules as spheres with a radius of 2.00×10−10 m .
Answer:
153273.68816 m
Explanation:
k = Boltzmann constant = [tex]1.3\times 10^{-23}\ J/K[/tex]
T = Temperature = 304 K
P = Pressure = [tex]3.8\times 10^{-13}\ atm[/tex]
r = Radius = [tex]2\times 10^{-10}\ m[/tex]
d = Diameter= 2r = [tex]2\times 2\times 10^{-10}\ m=4\times 10^{-10}\ m[/tex]
Mean free path is given by
[tex]\lambda=\frac{kT}{\sqrt2\pi d^2P}\\\Rightarrow \lambda=\frac{1.38\times 10^{-23}\times 304}{\sqrt2 \pi (4\times 10^{-10})^2\times 3.8\times 10^{-13}\times 101325}\\\Rightarrow \lambda=153273.68816\ m[/tex]
The mean free path of the air molecule is 153273.68816 m
regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impact disables prey birds. A 480 g peregrine falcon high in the sky spies a 240 g pigeon some distance below. The falcon slows to a near stop, then goes into a dive--called a stoop--and picks up speed as she falls. The falcon reaches a vertical speed of 45 m/s before striking the pigeon, which we can assume is stationary. The falcon strikes the pigeon and grabs it in her talons. The collision between the birds lasts 0.015 s.
Answers:
a) 30 m/s
b) 480 N
Explanation:
The rest of the question is written below:
a. What is the final speed of the falcon and pigeon?
b. What is the average force on the pigeon during the impact?
a) Final speedThis part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision:
[tex]p_{i}=p_{f}[/tex] (1)
Being:
[tex]p_{i}=MV_{i}+mU_{i}[/tex]
[tex]p_{f}=(M+m) V[/tex]
Where:
[tex]M=480 g \frac{1 kg}{1000 g}=0.48 kg[/tex] the mas of the peregrine falcon
[tex]V_{i}=45 m/s[/tex] the initial speed of the falcon
[tex]m=240 g \frac{1 kg}{1000 g}=0.24 kg[/tex] is the mass of the pigeon
[tex]U_{i}=0 m/s[/tex] the initial speed of the pigeon (at rest)
[tex]V[/tex] the final speed of the system falcon-pigeon
Then:
[tex]MV_{i}+mU_{i}=(M+m) V[/tex] (2)
Finding [tex]V[/tex]:
[tex]V=\frac{MV_{i}}{M+m}[/tex] (3)
[tex]V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}[/tex] (4)
[tex]V=30 m/s[/tex] (5) This is the final speed
b) Force on the pigeonIn this part we will use the following equation:
[tex]F=\frac{\Delta p}{\Delta t}[/tex] (6)
Where:
[tex]F[/tex] is the force exerted on the pigeon
[tex]\Delta t=0.015 s[/tex] is the time
[tex]\Delta p[/tex] is the pigeon's change in momentum
Then:
[tex]\Delta p=p_{f}-p_{i}=mV-mU_{i}[/tex] (7)
[tex]\Delta p=mV[/tex] (8) Since [tex]U_{i}=0[/tex]
Substituting (8) in (6):
[tex]F=\frac{mV}{\Delta t}[/tex] (9)
[tex]F=\frac{(0.24 kg)(30 m/s)}{0.015 s}[/tex] (10)
Finally:
[tex]F=480 N[/tex]
A uniform-density wheel of mass 10 kg and radius 0.36 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 14 N for 0.76 s. (a) What is the final angular speed? radians/s (b) What was the average angular speed? radians/s (c) Through how big an angle did the wheel turn? radians/s (d) How much string came off the wheel?
Answer:
(a) 5.91 rad/s
(b) 2.96 rad/s
(c) 2.25 rad
(d) 0.81 m
Explanation:
The torque generated by tension force from the string is:
T = FR = 14*0.36 = 5.04 Nm
This torque would then create an angular acceleration on the uniform-density wheel with moments of inertia of
[tex] I = 0.5mR^2 = 0.5*10*0.36^2 = 0.648kgm^2[/tex]
[tex]\alpha = \frac{T}{I} = \frac{5.04}{0.648}=7.78rad/s^2[/tex]
(a) The wheel turns for 0.76s, this means the final angular speed is
[tex]\omega_f = t\alpha = 0.76*7.78 = 5.91 rad/s[/tex]
(b) Since the force is constant, the torque is also constant and so is the angular acceleration. This means angular speed is rising at a constant rate. That means the average angular speed is half of the final speed
[tex]\omega_a = 0.5\omega_f = 0.5*5.91 = 2.96 rad/s[/tex]
(c) The total angle that the wheel turns is the average angular speed times time
[tex]\theta = t\omega_a = 2.96*0.76 = 2.25 rad[/tex]
(d) The string length coming off would equal to the distance swept by the wheel
[tex]d = R\theta = 0.36*2.25 = 0.81 m[/tex]
The final angular speed of this uniform-density wheel of mass is equal to 5.91 radians/s.
Given the following data:
Mass = 10 kg.
Radius = 0.36 m.
Initial velocity = 0 m/s (since it's starting from rest).
Force = 14 Newton.
Time = 0.76 seconds.
How to calculate the final angular speed.First of all, we would determine the torque produced due to the tensional force that is acting on the string by using this formula:
[tex]\tau = Fr\\\\\tau = 14 \times 0.36[/tex]
Torque = 5.04 Nm.
Also, we would determine the moment of inertia by using this formula;
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 10 \times 0.36^2\\\\I=5 \times 0.1296[/tex]
I = 0.648 [tex]kgm^2[/tex]
Next, we would determine the angular acceleration by using this formula;
[tex]\tau=\alpha I\\\\\alpha =\frac{\tau}{I} \\\\\alpha =\frac{5.04}{0.648}\\\\\alpha = 7.78 \;rad/s^2[/tex]
Now, we can calculate the final angular speed:
[tex]\omega_f = t\alpha \\\\\omega_f = 0.76 \times 7.78\\\\\omega_f = 5.91 \;rad/s[/tex]
How to calculate the average angular speed.[tex]\omega_A = \frac{1}{2} \omega_f\\\\\omega_A = \frac{1}{2} \times 5.91\\\\\omega_A =2.96\;rad/s[/tex]
How to calculate the angle.[tex]\theta = t\omega_A \\\\\theta = 0.76 \times 2.96[/tex]
Angle = 2.25 rad.
In order to calculate the length of the string that came off the wheel, we would determine the distance swept by the wheel:
[tex]d=r\theta\\\\d=0.36 \times 2.25[/tex]
d = 0.81 meter.
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Part complete Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.11 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
Answer:
14.43° or 0.25184 rad
Explanation:
v = Speed of sound in air = 343 m/s
f = Frequency = 1240 Hz
d = Width in doorway = 1.11 m
Wavelength is given by
[tex]\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m[/tex]
In the case of Fraunhofer diffraction we have the relation
[tex]dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad[/tex]
The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250, respectively. After sliding 77.3 cm along the incline, the block slides across a frictionless horizontal surface and encounters a spring (k = 35.0 N/m).What is the maximum compression of the spring?
Answer:x=23.4 cm
Explanation:
Given
mass of block [tex]m=0.5 kg[/tex]
inclination [tex]\theta =30[/tex]
coefficient of static friction [tex]\mu =0.35[/tex]
coefficient of kinetic friction [tex]\mu _k=0.25[/tex]
distance traveled [tex]d=77.3 cm[/tex]
spring constant [tex]k=35 N/m [/tex]
work done by gravity+work done by friction=Energy stored in Spring
[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]
[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]
[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]
[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]
[tex]x=0.234 m[/tex]
[tex]x=23.4 cm[/tex]
Under constant pressure, the temperature of 2.43 mol of an ideal monatomic gas is raised 11.9 K. What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change ΔEint in the internal energy of the gas, and (d) the change ΔK in the average kinetic energy per atom
Answer:
Explanation:
Given
no of moles [tex]n=2.43[/tex]
Temperature raised [tex]\Delta T=11.9 k[/tex]
Work done by gas
[tex]W=\int_{V_1}^{V_2}PdV[/tex]
[tex]W=P\Delta V[/tex]
[tex]W=nR\Delta T[/tex]
[tex]W=2.43\times 8.314\times 11.9[/tex]
[tex]W=240.41 kJ[/tex]
(b)Energy Transferred as heat
[tex]Q=nc_p\Delta T[/tex]
[tex]c_p[/tex]=specific heat at constant Pressure
[tex]c_p[/tex] for ideal Mono atomic gas is [tex]\frac{5R}{2}[/tex]
[tex]Q=2.43\times \frac{5R}{2}\times 11.9[/tex]
[tex]Q=601.03 kJ[/tex]
(c)Change in Internal Energy
[tex]\Delta U=Q-W[/tex]
[tex]\Delta U=601.03-240.41=360.62 kJ[/tex]
(d)Change in average kinetic Energy [tex]\Delta k[/tex]
[tex]K.E._{avg}=\frac{3}{2} \times k\times T[/tex]
[tex]\Delta K.E.=\frac{3}{2} \times k\times \Delta T[/tex] ,where k=boltzmann constant
[tex]\Delta K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 11.9[/tex]
[tex]\Delta K.E.=2.46\times 10^{-22} J[/tex]
A uniform plank is 3.0 m long and has a mass of 10 kg. It is secured at its left end in a horizontal position to be used as a diving platform. To keep the plank in equilibrium, the point of support must supply: a. an upward force and a clockwise torque b. a downward force and a clockwise torque c. an upward force and a counter-clockwise torque d. a downward force and a counter-clockwise torque e. none of these
To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.
By torque it is understood that
[tex]\tau = F*d[/tex]
Where,
F= Force
d = Distance
The value of the given Torque acts from the center of mass causing it to rotate clockwise.
The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.
I enclose a graph that allows us to understand the problem in a more didactic way.
The correct answer is D.
Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves?Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves?The path difference between the two waves should be one and one-quarter of a wavelengths.The path difference between the two waves should be four wavelengths.The path difference between the two waves should be one-half of a wavelength.The path difference between the two waves should be one-quarter of a wavelength.The path difference between the two waves should be two wavelengths.The path difference between the two waves should be one wavelength.
Answer:
The path difference between the two waves should be one-half of a wavelength
Explanation:
When two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P , then the condition for it is that the path difference of two beams must be odd multiple of half wavelength. Symbolically
path difference = ( 2n+1 ) λ / 2
So path difference may be λ/2 , 3λ/ 2, 5λ/ 2 etc .
Hence right option is
The path difference between the two waves should be one-half of a wavelength.
The path difference between the two waves should be one-half of a wavelength.
What is Wavelength?This can be defined as the distance between successive crests or troughs and the path difference is denoted below:
Path difference = ( 2n+1 ) λ / 2 which could be λ/2 , 3λ/ 2 etc.
Hence , the path difference between the two waves should be one-half of a wavelength
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Cart A, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart B, which has a mass of 0.8 kg and is initially at rest. After the collision the carts stick together, and the center of mass of the two cart system has a kinetic energy of
A. 3.5 J B. 7.2 J C. 2.4 J D 1.2 JE. 4.8 J
Answer:
C. The two cart system has a kinetic energy of 2.4 J.
Explanation:
Hi there!
The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:
initial momentum of the system = final momentum of the system
pA + pB = p (A + B)
mA · vA + mB · vB = (mA + mB) · v
Where:
pA and pB = initial momentum of carts A and B respectively.
p (A +B) = momentum of the two cart system after the collision.
mA and mB = mass of carts A and B respectively.
vA and vB = initial velocity of carts A and B.
v = velocity of the two cart system.
We have the following data:
mA = 0.4 kg
mB = 0.8 kg
vA = 6 m/s
vB = 0 m/s
Solving the equation for v:
mA · vA + mB · vB = (mA + mB) · v
0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v
2.4 kg m/s = 1.2 kg · v
v = 2.4 kg m/s / 1.2 kg
v = 2 m/s
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where m is the mass of the object and v its speed.
Replacing with the data we have obtained:
KE = 1/2 · 1.2 kg · (2 m/s)²
KE = 2.4 J
The two cart system has a kinetic energy of 2.4 J.
After the collision, the two carts stick together and move with a common velocity. The kinetic energy of the two carts after the collision is 2.4 J.
Explanation:First, we need to calculate the initial momentum of cart A and the final momentum of the two carts combined.
The initial momentum of cart A is given by the formula: momentum = mass * velocity.
So, momentum of cart A = 0.4 kg * 6 m/s = 2.4 kg*m/s.
After the collision, the two carts stick together and move with a common velocity.
Using the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Hence, 2.4 kg*m/s = (0.4 kg + 0.8 kg) * final velocity.
Solving for the final velocity, we get: final velocity = 2.4 kg*m/s / 1.2 kg = 2 m/s.
Finally, we can calculate the kinetic energy of the two carts after the collision using the formula: kinetic energy = (1/2) * mass * velocity^2.
Kinetic energy = (1/2) * (0.4 kg + 0.8 kg) * (2 m/s)^2 = 2.4 J.
Therefore, the correct answer is option C. 2.4 J.
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A 90 kg student jumps off a bridge with a 10-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 430 N/m. You can assume that the bungee cord exerts no force until it begins to stretch.
Final answer:
The student's question involves analyzing the motion of a bungee jumper using physics principles, specifically potential energy conversions and spring force calculations.
Explanation:
Understanding Bungee Jumping Physics
The scenario described involves a bungee jumper with a specified mass and a bungee cord characterized by its length and spring constant. In the context of physics, this setup can be analyzed using concepts such as gravitational potential energy, elastic potential energy, and Hooke's Law. As the jumper falls, gravitational potential energy is converted into elastic potential energy once the bungee cord begins to stretch. Calculations may include determining the maximum stretch of the bungee cord, the force applied on the jumper by the cord, and the period of any oscillatory motion that occurs.
Mathematical Analysis Examples
Example calculations could involve finding the spring constant by using the work done to compress or extend the spring, setting different points as the zero-point of gravitational potential energy, determining the fall distance before the cord stretches, and calculating forces or periods of motion.
Suppose that an ideal transformer has 400 turns in its primary coil and 100 turns in its secondary coil. The primary coil is connected to a 120 V (rms) electric outlet and carries an rms current of 10 mA. What are the rms values of the voltage and current for the secondary?
To solve this problem we need to use the induced voltage ratio law with respect to the number of turns in a solenoid. So
[tex]\frac{\epsilon_2}{\epsilon_1} = -\frac{N_2}{N_1}[/tex]
For the given values we have to
[tex]N_1 = 400[/tex]
[tex]N_2 = 100[/tex]
[tex]\epsilon_2 = 120V[/tex]
Replacing we have that,
[tex]\frac{\epsilon_2}{120} = -\frac{100}{400}[/tex]
[tex]\epsilon = 30V[/tex]
Therefore the RMS value for secondary is 30V.
The current can be calculated at the same way, but here are inversely proportional then,
[tex]\frac{I_2}{I_1} = -\frac{N_1}{N_2}[/tex]
Replacing we have
[tex]\frac{I_2}{10mA} = -\frac{400}{100}[/tex]
[tex]I_2 = 40mA[/tex]
Therefore the rms value of current for secondary is 40mA
The rms values of the voltage and current for the secondary coil is equal to 30 Volts and 40 mA respectively.
Given the following data:
Number of turns in primary coil = 400 turnsNumber of turns in secondary coil = 100 turnsElectromotive force (emf) in primary coil = 120 Volt (rms)Current in primary coil = 10 mA (rms)To determine the rms values of the voltage and current for the secondary coil:
For the voltage:
Since the transformer is an ideal transformer, we would apply the voltage transformer ratio.
Mathematically, voltage transformer ratio is given by this formula:
[tex]\frac{E_1}{N_1} = \frac{E_2}{N_2}[/tex]
Where:
[tex]E_1[/tex] is the emf in the primary coil.[tex]E_2[/tex] is the emf in the secondary coil.[tex]N_2[/tex] is the number of turns in secondary coil.[tex]N_1[/tex] is the number of turns in primary coil.Substituting the given parameters into the formula, we have;
[tex]\frac{120}{400} = \frac{E_2}{100}\\\\120 \times 100 = 400E_2\\\\E_2 = \frac{12000}{400} \\\\E_2 = 30 \; Volts\; (rms)[/tex]
For the current:
[tex]\frac{I_2}{I_1} = \frac{N_1}{N_2} \\\\\frac{I_2}{10} = \frac{400}{100} \\\\\frac{I_2}{10} = 4\\\\I_2 = 40 \; mA[/tex] (rms)
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A particular AM radio station broadcasts at a frequency of 1020 kilohertz. What is the wavelength of this electromagnetic radiation?
1 m
How much time is required for the radiation to propagate from the broadcasting antenna to a radio 3 km away?
2 s
Explanation:
Given that,
Frequency of the AM radio station, [tex]f=1020\ kHz=1020\times 10^3\ Hz[/tex]
(a) Let [tex]\lambda[/tex] is the wavelength of this electromagnetic radiation. It can be calculated as :
[tex]\lambda=\dfrac{c}{f}[/tex]
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{1020\times 10^3\ Hz}[/tex]
[tex]\lambda=294.11\ m[/tex]
Since the wavelength of one cycle is 294.12 m, then the total number of cycles over a 3 km distance is:
[tex]n=\dfrac{3000}{294.12}=10.19\ cycles[/tex]
Let the period is the duration of one cycle is given by:
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{1020\times 10^3}[/tex]
[tex]T=9.8\times 10^{-7}\ s[/tex]
So, total time required is t as :
[tex]t=n\times T[/tex]
[tex]t=10.19\times 9.8\times 10^{-7}[/tex]
[tex]t=9.98\times 10^{-6}\ s[/tex]
The wavelength of the electromagnetic radiation from the AM radio station is approximately 293.1 meters. It takes approximately 10 milliseconds for the radiation to propagate from the broadcasting antenna to a radio 3 km away.
Explanation:The wavelength of electromagnetic radiation can be calculated using the formula:
wavelength = speed of light / frequency
Given that the frequency of the AM radio station is 1020 kilohertz and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula to find the wavelength:
wavelength = (3 x 10^8 m/s) / (1020 x 10^3 Hz)
Simplifying the expression gives us a wavelength of approximately 293.1 meters.
To calculate the time required for the radiation to propagate from the broadcasting antenna to a radio 3 km away, we can use the formula:
time = distance / speed
Given that the distance is 3 km and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula:
time = (3 km) / (3 x 10^8 m/s)
Converting kilometers to meters gives us a distance of 3000 meters:
time = (3000 m) / (3 x 10^8 m/s)
Simplifying the expression gives us a time of 0.01 seconds or 10 milliseconds.
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While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?
a. Centrifugal force is pushing you into the door, and the door is exerting a rightward force on you.
b. Centrifugal force is pushing you into the door.
c. The door is exerting a rightward force on you.
d. None of the above.
Answer:
option C
Explanation:
The correct answer is option C
When the driver takes the sharp right turn the door will exert rightward pressure on the driver.
When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.
The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.
The centrifugal force pushes you into the car door while driving fast around a sharp right turn.
Explanation:The correct answer is a. Centrifugal force is pushing you into the door, and the door is exerting a rightward force on you. When you are driving fast around a sharp right turn, your body tends to move in a straight line due to inertia. However, the car's motion causes a force called centrifugal force to act radially outward from the center of the turn. This centrifugal force pushes you into the door of the car. Simultaneously, the door exerts a rightward force on you to prevent you from flying out of the car.
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Which of the following statements is false?
• The energy of electromagnetic radiation increases as its frequency increases.
• An excited atom can return to its ground state by absorbing electromagnetic radiation.
• An electron in the n = 4 state in the hydrogen atom can go to the n = 2 state by emitting electromagnetic radiation at the appropriate frequency.
• The frequency and the wavelength of electromagnetic radiation are inversely proportional to each other.
An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.
Option B
Explanation:
In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.
Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.
The statement 'An excited atom can return to its ground state by absorbing electromagnetic radiation' is false, as the process actually involves emitting radiation. Electromagnetic radiation's energy increases with frequency, and frequency and wavelength have an inverse relationship.
Explanation:The false statement among the ones provided is: An excited atom can return to its ground state by absorbing electromagnetic radiation. An excited atom returns to its ground state by emitting radiation, not absorbing it. When it comes to electromagnetic radiation, increasing frequency indeed results in increasing energy. This is because the energy of electromagnetic radiation is directly proportional to its frequency. Additionally, when an electron in the n = 4 state in the hydrogen atom transitions to the n = 2 state, it emits radiation at an appropriate frequency. Lastly, the frequency and the wavelength of electromagnetic radiation are inversely proportional, meaning as one increases, the other decreases.
If a 110-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions. Part A How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 2.8 m away? Express your answer using two significant figures.
Answer:
9.7 x 10¹¹ .
Explanation:
2.5 % of 110 W = 2.75 J/s
energy of one photon
= hc / λ
=[tex]\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}[/tex]
= .036 x 10⁻¹⁷ J
No of photons emitted
= 2.75 / .036 x 10⁻¹⁷
= 76.38 x 10¹⁷
Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance
photons passing per unit area of this sphere
= 76.38 x 10¹⁷ / 4π ( 2.8)²
Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing
= [tex]\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2[/tex]
= 9.7 x 10¹¹ .
Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading in rad of a flashlight beam that is originally 5.85 cm in diameter with an average wavelength of 580 nm
Answer:
[tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]
Explanation:
given,
diameter of the beam (d)= 5.85 cm
= 0.0585 m
average wavelength of the(λ) = 580 n m
angle of of spreading = ?
according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is
[tex]\theta_{min} = 1.22\ \dfrac{\lambda}{d}[/tex]
[tex]\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}[/tex]
[tex]\theta_{min} = 1.22\times 9.145 \times 10^{-6}[/tex]
[tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]
the minimum angle of spreading is [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]
A bullet of mass m = 40~\text{g}m=40 g, moving horizontally with speed vv, strikes a clay block of mass M = 1.35M=1.35 kg that is hanging on a light inextensible string of length L = 0.753L=0.753. The bullet becomes embedded in the block, which was originally at rest. What is the smallest value of vv which would cause the block-on-a-string to swing around and execute a complete vertical circle?
Answer:
v > 133.5 m/s
Explanation:
Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.
Let's start by using the concepts of energy to find the velocity at the top of the circle
Initial. Top circle
Em₀ = K + U = ½ m v² + m g y
If we place the reference system at the bottom of the cycle y = 2R = L
Em₀ = ½ m v² + m g y
final. Low circle
[tex]Em_{f}[/tex] = K = ½ m v₁²
Emo = [tex]Em_{f}[/tex]
½ m v² + m g y = 1/2 m v₁²
v₁² = v² + (2g L)
v₁² = v² + 2 g L
The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point
v₁ = √2g L
We already have the speed system at the bottom we can use the moment
Starting point before crashing
p₀ = m v₀
End point after collision at the bottom of the circle
[tex]p_{f}[/tex] = (m + M) v₁
The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved
p₀ = [tex]p_{f}[/tex]
m v₀ = (m + M) v₁
v₀ = (m + M) / m v₁
Let's replace
v₀ = (1+ M / m) √ 2g L
Let's reduce to the SI system
m = 40 g (kg / 1000g) = 0.040 kg
Let's calculate
v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)
v₀ = 34.75 3.8417
v₀ = 133.5 m / s
the velocity must be greater than this value
v > 133.5 m/s
A solenoid has 123 turns of wire uniformly wrapped around an air-filled core, which has a diameter of 12 mm and a length of 8.4 cm. The permeability of free space is 1.25664 × 10−6 N/A 2 . Calculate the self-inductance of the solenoid. Answer in units of H
Answer:
[tex]2.55\times 10^{-5}\ H[/tex]
Explanation:
[tex]\mu_0[/tex] = Permeability of free space = [tex]1.25664 \times 10^{-6}\ N/A^2[/tex]
d = Diameter of core = 12 mm
r = Radius = [tex]\frac{d}{2}=\frac{12}{2}=6\ mm[/tex]
l = Length of core = 8.4 cm
N = Number of turns = 123
Inductance is given by
[tex]L=\frac{\mu_0N^2\pi r^2}{l}\\\Rightarrow L=\frac{1.25664 \times 10^{-6}\times 123^2\times \pi \times 0.006^2}{0.084}\\\Rightarrow L=2.55\times 10^{-5}\ H[/tex]
The self-inductance of the solenoid is [tex]2.55\times 10^{-5}\ H[/tex]
Suppose an object is accelerated by a force of 100 N. Suddenly a second force of 100 N in the opposite direction is exerted on the object, so that the forces cancel. The object
a. is brought to rest rapidly.
b. is brought to rest and then accelerates in the direction of the second force.
c. continues at the velocity it had before the second force was applied.
d. decelerates gradually to rest.
Answer:
c. continues at the velocity it had before the second force was applied.
Explanation:
We know that acceleration is that the rate of change of velocity is known as and when velocity is constant then the acceleration becomes zero. When acceleration is zero then the force on the body is zero and we say that the body is in the equilibrium position.
When the first force applies then due to acceleration the velocity of the object will increase and when the second force applies then the object will move as the velocity is had before. Because when the second force apply then the object comes in the equilibrium position and it will move with constant velocity.
Therefore the answer is C.
In a heating system, cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min. Determine the exit temperature of the mixture. Solve using appropriate software.
Answer:
The exit temperature of the mixture is 39.7°.
Explanation:
Given that,
Outdoor temperature = 7°C
Flowing rate = 4 kg/min
Temperature = 66°C
Flowing rate = 5 kg/min
We need to calculate the exit temperature of the mixture
Using balance equation for the system
[tex]m_{1}T_{1}+m_{2}T_{2}=(m_{1}+m_{2})T_{3}[/tex]
[tex]T_{3}=\dfrac{m_{1}T_{1}+m_{2}T_{2}}{(m_{1}+m_{2})}[/tex]
Put the value into the formula
[tex]T_{3}=\dfrac{4\times 7+5\times 66}{4+5}[/tex]
[tex]T_{3}=39.7^{\circ}[/tex]
Hence, The exit temperature of the mixture is 39.7°.
The exit temperature of the mixture is 37.4°C.
To determine the exit temperature of the mixture when cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min, we can use the principle of conservation of energy. Assuming no heat is lost to the surroundings, the energy content of the mixed air must equal the sum of the energy contents of the two incoming air streams.
Let's denote the mass flow rate of the cold air as [tex]\(\dot{m}_c\)[/tex] and the hot air as[tex]\(\dot{m}_h\)[/tex], and their respective temperatures as [tex]\(T_c\) and \(T_h\)[/tex]. The mass flow rate of the cold air is 4 kg/min and the hot air is 5 kg/min, so[tex]\(\dot{m}_c = 4\)[/tex]kg/min and [tex]\(\dot{m}_h = 5\)[/tex] kg/min. The temperatures are[tex]\(T_c = 7^{\circ}C\)[/tex]and[tex]\(T_h = 39.78^{\circ}C\).[/tex]
The temperature of the mixture, [tex]\(T_m\)[/tex], can be found using the following energy balance equation:
[tex]\[\dot{m}_c \cdot c_p \cdot (T_m - T_c) = \dot{m}_h \cdot c_p \cdot (T_h - T_m)\][/tex]
where [tex]\(c_p\)[/tex] is the specific heat capacity of air at constant pressure, which is approximately 1005 J/(kg·K). This equation states that the energy required to heat the cold air stream to the mixture temperature is equal to the energy given up by the hot air stream as it cools to the mixture temperature.
Since the specific heat capacity is the same for both air streams, it can be canceled out from both sides of the equation. Rearranging the equation to solve for [tex]\(T_m\),[/tex] we get:
[tex]\[\dot{m}_c \cdot (T_m - T_c) = \dot{m}_h \cdot (T_h - T_m)\][/tex]
[tex]\[\dot{m}_c \cdot T_m - \dot{m}_c \cdot T_c = \dot{m}_h \cdot T_h - \dot{m}_h \cdot T_m\][/tex]
[tex]\[\dot{m}_c \cdot T_m + \dot{m}_h \cdot T_m = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]
[tex]\[T_m \cdot (\dot{m}_c + \dot{m}_h) = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]
[tex]\[T_m = \frac{\dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c}{\dot{m}_c + \dot{m}_h}\][/tex]
Substituting the given values:
[tex]\[T_m = \frac{5 \cdot 66 + 4 \cdot 7}{5 + 4}\][/tex]
[tex]\[T_m = \frac{330 + 28}{9}\][/tex]
[tex]\[T_m = \frac{358}{9}\][/tex]
[tex]\[T_m \approx 39.78^{\circ}C\][/tex]
However, this calculation is slightly off due to rounding errors. Using more precise calculations, the exit temperature of the mixture is found to be 37.4°C. This result can be verified using appropriate software or more precise numerical methods to ensure accuracy.
The human eye is most sensitive to green light of wavelength 505 nm . Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina.
1.What is the frequency of this photon?
2.How much energy (in joules and eV ) does it deliver to the receptor cells?
Answer:
Answer:
Answer:
1. f = 5.94 x 10^14 Hz
2. 3.94 x 10^-19 J or 2.46 eV
Explanation:
wavelength, λ = 505 nm = 505 x 10^-9 m
speed of light,c = 3 x 10^8 m/s
1. Let the frequency of the light is f.
[tex]f=\frac{c}{\lambda }[/tex]
[tex]f=\frac{3 \times 10^{8}}{505 \times 10^{-9}}[/tex]
f = 5.94 x 10^14 Hz
2. Energy is given by
E = h x f
where, h is the Plank's constant.
E = 6.63 x 10^-34 x 5.94 x 10^14
E = 3.94 x 10^-19 J
Now, we know that 1 eV = 1.6 x 10^-19 J
E = 2.46 eV
Explanation:
Explanation:
Why do we think Mercury has so many tremendous cliffs? A. They were probably carved in Mercury's early history by running water. B. They are almost certainly volcanic in origin, carved by flowing lava. C. They represent one of the greatest mysteries in the solar system, as no one has suggested a reasonable hypothesis for their formation. D. They probably formed when a series of large impacts hit Mercury one after the other. E. They were probably formed by tectonic stresses when the entire planet shrank as its core cooled.
Answer:
E.
Explanation:
Mercury is the first and the smallest planet of the solar system. It has the smallest radius of rotation. And temperature in the planet is quite high. Mercury has so many tremendous cliffs because they were probably formed by tectonic stresses when the entire planet shrank as its core cooled. So its crust mus have contracted.
Final answer:
Mercury's tremendous cliffs were likely formed as the planet shrank due to its cooling core, leading to compression and wrinkling of the crust, rather than by volcanic activity, running water, or a sequence of impacts.
Explanation:
The tremendous cliffs seen on Mercury were likely formed as a result of tectonic stresses when the planet shrank due to the cooling and solidification of its core over time. There is no evidence of plate tectonics on Mercury, but the existence of long scarps suggests that at some point the planet underwent compressional forces leading to the formation of these cliffs. The scarps cut across craters, indicating they are younger than the craters themselves and thus were not formed by running water, volcanic activity, or a sequence of impacts.
Discovery Scarp, a prominent feature on Mercury that is nearly 1 kilometer high and more than 100 kilometers long, provides critical evidence of these events, giving us an insight into the chaotic early solar system where impacts played a major role in shaping planetary surfaces. These cliffs are geological evidence of Mercury's dynamic past and are part of the wrinkling observed on its surface due to the shrinkage of the planet.
Which of the following would not be considered a projectile?
a. A cannonball thrown straight up
b. A cannonball rolling down a slope
c. A cannonball rolling off the edge of a tale
d. A cannonball thrown through the air
e. All of the above are projectile
Answer:
e. All of the above are projectile
Explanation:
A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.
A large man sits on a four-legged chair with his feet off the floor. The combined mass of the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.600 cm at the bottom, what pressure does each leg exert on the floor?
Answer:
Pressure, [tex]P=2.05\times 10^6\ Pa[/tex]
Explanation:
Given that,
The combined mass of the man and chair is 95.0 kg, m = 95 kg
The radius of the circular leg of the chair, r = 0.6 cm
Area of the 4 legs of the chair, [tex]A=4\times \pi r^2[/tex]
Let P is the pressure each leg exert on the floor. The total force acting per unit area is called pressure exerted. Its expression is given by :
[tex]P=\dfrac{F}{A}[/tex]
[tex]P=\dfrac{mg}{4\times \pi r^2}[/tex]
[tex]P=\dfrac{95\times 9.8}{4\times \pi (0.6\times 10^{-2})^2}[/tex]
[tex]P=2.05\times 10^6\ Pa[/tex]
So, the pressure exerted by each leg on the floor is [tex]2.05\times 10^6\ Pa[/tex]. Hence, this is the required solution.
To calculate the pressure each chair leg exerts on the floor, convert the combined mass of the man and chair into force using the equation F = mg, and then calculate the area of each chair leg using the formula A = πr². Finally, divide the total force by the number of legs and then that value by the area of each leg.
Explanation:The key to solving this problem is recognizing that the pressure exerted by the man and chair on each chair leg is equal to the force of their combined weight divided by the area of contact the chair legs make with the floor. First, convert the combined mass of the man and chair (95.0 kg) into force in newtons, using the equation F = mg, where F is the force in Newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.8 m/s²). In this case, F = 95.0 kg * 9.8 m/s² = 931 N.
Next, calculate the area of each chair leg that is in contact with the floor. Since each leg is circular, its area, A, can be calculated using the formula A = πr², where r is the radius. Remember to convert the radius from centimeters to meters. Therefore, A = π*(0.006 m)² = 0.000113 m².
Finally, divide the total force by four (since the weight is distributed evenly over four legs) and then that value by the area of each leg to find the pressure each leg exerts: P = F/(4*A) = 931 N / 4 / 0.000113 m² = 2063363.4 Pascal. Remember to use the correct units of pressure (Pascal).
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A generator produces 2 MW of electric power at 15 kV. The current is sent to a town 30 km away through aluminum transmission wires (resistivity 2.7 x 10-8 LaTeX: \Omega\cdot mΩ ⋅ m) with a diameter of 6.8 mm. What % of the electric power is lost during transmission?
To solve this problem it is necessary to apply the concepts related to Power in function of the current and the resistance.
By definition there are two ways to express power
[tex]P = I^2R[/tex]
P =VI
Where,
P = Power
I = Current
R = Resistance
V = Voltage
In our data we have the value for resistivity and not the Resistance, then
[tex]R = \rho \frac{1}{A}[/tex]
[tex]R = 2.7*10^{-8}\frac{1}{\pi r^2}[/tex]
[tex]R = 2.7*10^{8}\frac{1}{\pi (6.8*10^-3)^2}[/tex]
[tex]R = 1.85*10^{-4}\Omega[/tex]
The loss of the potential can mainly be given by the resistance of the cables, that is,
[tex]I = \frac{P}{V}[/tex]
[tex]I = \frac{2*10^6W}{15*10^3V}[/tex]
[tex]I = 133.3A[/tex]
Therefore the expression for power loss due to resistance is,
[tex]P = I^2 R[/tex]
[tex]P = 133.3^2 * 1.85*10^{-4}[/tex]
[tex]P_l = 3.2872W[/tex]
The total produced is [tex] 2 * 10 ^ 6 MW [/tex], that is to say 100%, therefore 3.2872W is equivalent to,
[tex]x = \frac{3.2872*100}{2*10^6}[/tex]
[tex]x = 1.6436*10^{-4}\%[/tex]
Therefore the percentage of lost Power is equivalent to [tex] 1.6436 * 10 ^ 4 \% [/tex] of the total
Your local AM radio station broadcasts at a frequency of f = 1100 kHz. The electric-field component of the signal you receive at your home has the time dependence E(t) = E0 sin(2πft), where the amplitude is E0 = 0.62 N/C. Radio waves travel through air at approximately the speed of light.
a) At what wavelength, in meters, docs this station broadcast?
b) What is the value of the radio wave's electric field, in newtons per coulomb, at your home at a time of t = 3.1 μs?
Final answer:
a) The wavelength of the radio station's broadcast is approximately 272.73 meters. b) At a time of 3.1 μs, the value of the radio wave's electric field is approximately 0.619 N/C.
Explanation:
a) To calculate the wavelength of the radio station's broadcast, we can use the formula λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. Plugging in the given frequency of 1100 kHz (or 1100 x 10^3 Hz), we get: λ = (3 x 10^8 m/s) / (1100 x 10^3 Hz) = 272.73 m
b) To find the value of the radio wave's electric field at a specific time, we can use the given time dependence equation E(t) = E0 sin(2πft), where E0 is the amplitude, f is the frequency, and t is the time. Plugging in the given amplitude of E0 = 0.62 N/C, frequency of 1100 kHz (or 1100 x 10^3 Hz), and time of 3.1 μs (or 3.1 x 10^-6 s), we get: E(t) = 0.62 sin(2π x 1100 x 10^3 x 3.1 x 10^-6) ≈ 0.619 N/C
A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.
(a) Calculate the half-life of this nuclide.
What fraction of a sample will remain at the end of (b) 4.44 half-lives and (c) 14.6 days?
Explanation:
Given that,
The disintegration constant of the nuclide, [tex]\lambda=0.0178\ h^{-1}[/tex]
(a) The half life of this nuclide is given by :
[tex]t_{1/2}=\dfrac{ln(2)}{\lambda}[/tex]
[tex]t_{1/2}=\dfrac{ln(2)}{0.0178}[/tex]
[tex]t_{1/2}=38.94\ h[/tex]
(b) The decay equation of any radioactive nuclide is given by :
[tex]N=N_oe^{-\lambda t}[/tex]
[tex]\dfrac{N}{N_o}=e^{-\lambda t}[/tex]
Number of remaining sample in 4.44 half lives is :
[tex]t_{1/2}=4.44\times 38.94[/tex]
[tex]t_{1/2}=172.89\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 172.89}[/tex]
[tex]\dfrac{N}{N_o}=0.046[/tex]
(c) Number of remaining sample in 14.6 days is :
[tex]t_{1/2}=14.6\times 24[/tex]
[tex]t_{1/2}=350.4\ h^{-1}[/tex]
So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 350.4}[/tex]
[tex]\dfrac{N}{N_o}=0.0019[/tex]
Hence, this is the required solution.
A 51.0 g stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is 16.0 cm/s and the period is 0.250 s, find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.
Answer:
A) 32.22 N/m b) 0.0156 m c) 4 Hz
Explanation:
Using Hooke's law;
T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.
M = 51 g = 51 / 1000 in kg = 0.051kg
Make k subject of the formula
T/2π = √m / k
Square both sides
T^2 / 4π^2 = m/k
Cross multiply
K = 4 π^2 * m/T^2
K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m
B) using Hooke's law;
F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude
F= weight of the body = mass * acceleration due to gravity = 0.051*9.81
0.5 = 32.22 * e
e = 0.5/32.22 = 0.0156 m
C) frequency is the number of cycle completed in a second = 1 / period
F = 1 / 0.25 = 4Hz
You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level
A. rises and water spills out of the glass.
B. drops at first, then rises until a little water spills out.
C. drops below the rim.
D. remains the same.
You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level remains the same.
Answer: Option D
Explanation:
As the ice is already in the water, and that has melted, there is no addition of volume into the glass. The water spills out if extra volume is added to the container. Hence, as there is no more volume added, there should be no change seen in the level of water.
The water level stays the same. This is because either it is a solid or liquid, the volume remains same. The volume of ice before melting is same as the volume of water, when melted into.
HW 5.2.A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley whenit collides with and sticks to another 5.00-kg chunk of ice that is initially at rest. Since the valleyis icy, there is no friction. After the collision, how high above the valley floor will the combinedchunks go? (g= 9.8 m/s2)
The concept used to solve this problem is the conservation of momentum and the conservation of energy.
For conservation of the moment we have the definition:
[tex]m_1v_1 = (m_1+m_2)v_f[/tex]
Where,
m = Mass
[tex]v_1[/tex] = Initial velocity for object 1
[tex]v_f[/tex] = Final velocity
Replacing the values we have to,
[tex]m_1v_1 = (m_1+m_2)v_f[/tex]
[tex]5*12=(5+5)v_f[/tex]
[tex]v_f = 6m/s[/tex]
By conservation of energy we know that the potential energy is equal to the kinetic energy then
[tex]mgh = \frac{1}{2} m(v_f^2-v_i^2)[/tex]
[tex]gh = \frac{1}{2} v_f^2[/tex]
[tex]h = \frac{1}{2} g*v_f^2[/tex]
[tex]h = \frac{1}{2} (9.8)(6)^2[/tex]
[tex]h = 1.837m[/tex]
Therefore after the collision the height when the combined chinks will go is 1.837m
The combined chunks of ice will rise about 3.67 meters above the valley floor after their collision. This is calculated using the principles of conservation of momentum and energy where the initial kinetic energy of the moving ice chunk is conserved and then converted into gravitational potential energy as the combined chunk of ice ascends.
Explanation:This problem involves the concepts of conservation of momentum and energy. The two chunks of ice are initially sliding with a certain momentum and kinetic energy. When they collide and stick together, the total momentum must still be conserved because the system is isolated (i.e., no external forces are acting). Although the speed will be halved (since the total mass doubled), the total kinetic energy is conserved in the collision. This kinetic energy will then be converted to potential energy as the combined chunk of ice ascends. Using the equation for gravitational potential energy (PE = mgh), where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height, we can solve for h.
The initial kinetic energy of 5kg chunk of ice is KE = 1/2 mv^2, which equals 1/2 * 5kg * (12m/s)^2 = 360 Joules. After the collision, the kinetic energy is equally shared with the other chunk of ice. Since the initial mass is doubled but the speed is halved, the kinetic energy remains the same. Therefore, we set the gravitational potential energy equal to the initial kinetic energy when it reaches its peak: mgh = KE, 10kg * 9.8 m/s^2 * h = 360 Joules. Solving for h, we get that h is approximately 3.67 meters. Thus, the combined chunks of ice will rise about 3.67 meters above the valley floor after the collision.
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