Answer:
The plane will be 11545.46 m far when the observer hears the sonic boom
Explanation:
Step 1: Data given
Altitude of the plane = 7800 meters
speed of sound = 311.83 m/s
Step 2:
The mach number M = vs/v
This means v/vs = 1/M
Half- angle = ∅
sin∅= v/vs
∅ = sin^-1 (v/vs)
∅ = sin^-1 (1/M)
∅ = sin^-1(1/1.48)
∅= 42.5 °
tan ∅ = h/x
⇒ with h= the altitude of the plane = 7800 meter
⇒ with x = the horizontal distance moved by the plane
x = h/tan ∅
x = 7800 / tan 42.5
x = 8512.2 meters
d = the distance between the observer and the plane when the observer hears the sonic boom is:
d = √(8512.2² + 7800²)
d = 11545.46 m
The plane will be 11545.46 m far when the observer hears the sonic boom
Standing on the edge of a rotating playground carousel, you take sight on the center of the carousel and walk directly toward that position. Part AWhat happens to the rotational speed of the carousel as you walk? Choose the correct answer and explanation.
Answer:
Rotational speed increases
Explanation:
The formula for rotational speed is:
Rotational speed = rotations / time.
When you approach the center the number of rotations increases, because the radius of the circle decreases, and this increases rotational speed.
If you walk from the edge to the center of a rotating carousel, the carousel will rotate faster. This is due to the conservation of angular momentum. As you move closer to the center, the carousel's moment of inertia decreases and to conserve total angular momentum, its rotational speed increases.
Explanation:Your position on the carousel affects its rotational speed. If you stand on the edge of a rotating carousel and then you walk toward the center, you change your rotational inertia, which according to the law of conservation of angular momentum, makes the carousel rotate faster.
This happens because when you move closer to the center, the rotational inertia or the resistance to the rotation of the carousel decreases and thus, the rotational speed or the speed at which the carousel is turning increases to conserve the total angular momentum.
The conservation of angular momentum is the principle which tells that the total angular momentum remains constant in a system without any external torques. As we see in the case of the carousel, when you are at the edge of the spinning carousel, there is a large radius, leading to a high moment of inertia.
But as you move towards the center, the radius decreases, which in turn decreases the moment of inertia. Therefore, to keep the angular momentum constant, the angular velocity, in this case, the speed of the carousel, must increase.
The same concept can be applied to figure skaters. When a figure skater pulls in her arms and spins, she reduces her rotational inertia and therefore spins faster to conserve angular momentum.
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A copper wire and an iron wire of equal length l and diameter d are joined, and a potential difference V is applied between the ends of the composite wire. Calculate the potential difference (in volts) across the iron wire. Assume that l = 2.97 m, d = 0.88 mm, and V = 157 V.
Answer:
The potential difference across iron wire is 133.95 V
Solution:
As per the question:
Length, l = 2.97 m
Diameter. d = 0.88 mm = [tex]0.88\times 10^{- 3}\ m[/tex]
Voltage, V = 157 V
Now,
We know that the resistivity of iron, [tex]\rho_{I} = 10\ S[/tex]
Resistivity of copper, [tex]\rho_{I} = 1.72\ S[/tex]
Now,
The potential difference across the iron wire:
V' = [tex]\frac{\rho_{I}V}{\rho_{I} + \rho_{C}}[/tex]
V' = [tex]\frac{10\times 157}{10 + 1.72} = 133.95\ V[/tex]
Answer:
The potential difference across the iron wire 134.2 V.
Explanation:
Given that,
Length = 2.97 m
Diameter = 0.88 mm
Total potential = 157 V
We know that,
The resistivity of iron wire
[tex]\rho_{i}=10\times10^{-8}\Omega m[/tex]
The resistivity of copper wire
[tex]\rho_{c}=1.7\times10^{-8}\Omega m[/tex]
The voltage for each wire is proportional to the resistivity.
We need to calculate the resistance for copper wire
Using formula of resistance
[tex]R_{c}=\dfrac{\rho\times l}{A}[/tex]
Put the value into the formula
[tex]R_{c}=\dfrac{1.7\times10^{-8}\times2.97}{\pi\times(0.44\times10^{-3})^2}[/tex]
[tex]R_{c}=0.0830\ \Omega[/tex]
We need to calculate the resistance for iron wire
Using formula of resistance
[tex]R_{i}=\dfrac{\rho\times l}{A}[/tex]
Put the value into the formula
[tex]R_{i}=\dfrac{10\times10^{-8}\times2.97}{\pi\times(0.44\times10^{-3})^2}[/tex]
[tex]R_{i}=0.4883\ \Omega[/tex]
The total resistance is
[tex]R =R_{c}+R_{i}[/tex]
[tex]R=0.0830+0.4883[/tex]
[tex]R=0.5713\ \Omega[/tex]
We need to calculate the potential difference across the iron wire
Using formula of potential difference
[tex]V_{i}=\dfrac{VR_{i}}{R}[/tex]
Put the value into the formula
[tex]V_{i}=\dfrac{157\times0.4883}{0.5713}[/tex]
[tex]V_{i}=134.2\ V[/tex]
Hence, The potential difference across the iron wire 134.2 V.
The moment of inertia of the empty turntable is 1.5 kg?m2. With a constant torque of 2.5 N?m, the turntableperson system takes 3.0 s to spin from rest to an angular speed of 1.0 rad/s. What is the persons moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle.
The moment of inertia of the empty turntable is 1.5 . With a constant torque of 2.5 , the turntableperson system takes 3.0 to spin from rest to an angular speed of 1.0 . What is the persons moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle.
2.5 kg?m2
6.0 kg?m2
7.5 kg?m2
9.0 kg?m2
[tex]6.0 \mathrm{kg} \mathrm{m}^{2}[/tex] is the persons moment of inertia about an axis through her center of mass.
Answer: Option B
Explanation:
Given data are as follows:
moment of inertia of the empty turntable = 1.5
Torque = 2.5 N/m , and
[tex]\text { Angular acceleration of the turntable }=\frac{\text { angular speed }}{\text { time }}=\frac{1}{3}[/tex]
Let the persons moment of inertia about an axis through her center of mass= I
So, Now, from the formula of torque,
[tex]\text { Torque }(\tau)=\text { Moment of inertia(I) } \times \text { Angular acceleration(a) }[/tex]
[tex]2.5=(1.5+I) \times \frac{1}{3}[/tex]
So, from the above equation, we can measure the person’s moment of Inertia (I)
[tex]2.5 \times 3=1.5+I[/tex]
[tex]I=7.5-1.5=6.0 \mathrm{kg} m^{2}[/tex]
While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a simple pendulum) swings back and forth in the breeze once every T = 6.2 seconds.
Randomized VariablesT = 6.2 seconds
(a) Calculate the frequency of oscillation (in Hertz) of the chandelier.
(b) Calculate the angular frequency ω of the chandelier in radians/second. sig.gif?tid=2N74-08-DD-43-8D37-13518No Attempt No Attempt25% Part (c) Determine the length L in meters of the chandelier. sig.gif?tid=2N74-08-DD-43-8D37-13518No Attempt No Attempt25% Part (d) That evening, while hanging out in J.J. Thompson’s House O’ Blues, you notice that (coincidentally) there is a chandelier identical in every way to the one at the Michelson exhibit except this one swings back and forth 0.11 seconds slower, so the period is T + 0.11 seconds. Determine the acceleration due to gravity in m/s2 at the club.
Answer:
a) 0,1613 Hz
b) 1,01342 rad/sec
c) 9.5422 m
d) [tex]9.4314 m/sec^2[/tex]
Explanation:
In the Albert Michelson exhibit at Clark University, we know the period of oscillation of the chandelier is T = 6.2 seconds
The chandelier will be modeled as a simple pendulum
a) Since the frequency is the reciprocal of the period, we have
[tex]f=\frac{1}{T}=\frac{1}{6.2sec}=0,1613 Hz[/tex]
b) The angular frequency is computed as
[tex]w=2\pi f=2\pi (0,1613) = 1,01342\ rad/sec[/tex]
c) The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
Where L is its length and g is the local acceleration of gravity (assumed 9.8 [tex]m/sec^2[/tex] for this part)
We want to know the length of the pendulum, so we isolate L
[tex]L=\frac{T^2g}{4\pi^2}[/tex]
[tex]L=\frac{(6.2)^2(9.8))}{4\pi^2}=9.5422 m[/tex]
d) While hanging out in J.J. Thompson’s House O’ Blues, the new period is 6.2+0.11=6.32 sec. Since the chandelier is the very same (same length), we can assume the gravity is slightly different. We use again the formula for T, but now we'll isolate g as follows
[tex]g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2(9.5422m))}{(6.32sec)^2}[/tex]
Which results
[tex]g=9.4314\ m/sec^2[/tex]
The frequency of oscillation of the chandelier in Hertz will be 0.1613 Hertz
How to calculate frequency?The frequency of oscillation of the chandelier will be calculated thus:
= 1/T = 1/6.2 = 0.1613 Hz
The angular frequency will be:
= 2πf = 2π × 0.1613 = 1.01342 rad/sec
The length of the simple pendulum will be:
= T²g/4π²
= (6.2)²(9.8) / 4π².
= 9.5422m
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The relative kinetic energy loss through a process is defined as ΔΚ Δκ. к, relatine a. Rewrite part (a) in terms of the mass ratio (m, mincoming/mstationary) due to a completely inelastic collision where the second objects is at rest? b. If I want a system that loses 25% of the kinetic energy after a completely inelastic collision, what should the mass ratio be? c. If I want a system that loses 75% of the kinetic energy after a completely inelastic collision, what should the mass ratio be?
Answer:
Explanation:
Let mass m₁ is colliding in-elastically with stationary mass m₂ with velocity v₁ . Let v₂ be their conjugate velocity after collision .
Initial KE =1/2 m₁ v₁²
Final KE = 1/2 ( m₁ + m₂ ) v₂²
from conservation of momentum
v₂ = m₁v₁ / ( m₁ + m₂)
Final KE = 1/2 ( m₁ + m₂ ) m₁²v₁² / ( m₁ + m₂ )²
= 1/2 m₁²v₁² / ( m₁ + m₂ )
Loss of KE = ΔK
= 1/2 m₁ v₁² - 1/2 m₁²v₁² / ( m₁ + m₂ )
= 1/2 m₁ v₁² ( 1 - m₁ / m₁ + m₂ )
= 1/2 m₁ v₁² m₂ / (m₁ + m₂ )
ΔK / K= m₂ / (m₁ + m₂ )
= β / (1 + β)
where β = m₂ / m₁
b )
If ΔK / K = .25
.25 = β / (1 + β)
β = 1/3
c )
If
ΔK / K = .75
.75 = β / (1 + β)
β = 3
A car drives at steady speed around a perfectly circular track.
a. The car's acceleration is zero.
b. Both the acceleration and net force on the car point outward.
c. The net force on the car is zero.
d. If there is no friction, the acceleration is outward.
e. Both the acceleration and net force on the car point inward.
Answer:
e. Both the acceleration and net force on the car point inward.
Explanation:
If no net force acts on the car, the car must drive in a straight line, at constant speed.
As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.
In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.
The net force that causes this acceleration, aims inward, and is called the centripetal force.
It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.
The net force on a car driving around a circular track at steady speed is zero, resulting in zero acceleration. Hence, the correct option is c.
The net force on the car is zero. When a car drives at a steady speed around a circular track, the centripetal force required to keep it moving in a circle balances out the outward inertia. Therefore, there is no net force on the car. This leads to zero acceleration (choice c).
A 77.0-kg ice hockey goalie, originally at rest, catches a 0.125-kg hockey puck slapped at him at a velocity of 37.5m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came.What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic.
A. v goalie = _____ m / s
B. v puck = ______ m / s
Answer
given,
Mass of ice hockey goalie = (M) = 77 Kg
mass of pluck = (m) = 0.125 Kg
velocity of pluck (u₁)= 37.5 m/s
u₂ = 0
Let v₁ and v₂ are the velocity of m₁ and m₂
final velocities are
[tex]v_1 = \dfrac{M-m}{M+m}u_2 + \dfrac{2m}{M+m}u_1[/tex]
v_1 is velocity of goalie
[tex]v_2= \dfrac{m-M}{m+M}u_1 + \dfrac{2M}{m+M}u_2[/tex]
v_2 is velocity of puck
now,
a) for goalie
[tex]v_1 = \dfrac{M-m}{M+m}u_2 + \dfrac{2m}{M+m}u_1[/tex]
[tex]v_1 = \dfrac{77-0.125}{77+0.125}(0) + \dfrac{2(0.170)}{77+0.125}(37.5)[/tex]
[tex]v_1 =0.165\ m/s[/tex]
b) for pluck
[tex]v_2= \dfrac{m-M}{m+M}u_1 + \dfrac{2M}{m+M}u_2[/tex]
[tex]v_2= \dfrac{0.125-77}{77+0.125}(37.5)+ \dfrac{2\times 77}{77+0.125}(0)[/tex]
[tex]v_2= -37.38\ m/s[/tex]
According to Little's Law, which of the following can reduce the number of skiers waiting in line to ride a chairlift at a ski resort? a. Using a process layout b. Decreasing the chairlift's bandwidth c. Increasing the chairlift's throughput rate d. Increasing the skiers' average flow time e. Decreasing the skiers' average flow time
Answer:
C. Increasing the chairlift's throughput rate.
Explanation:
The throughput rate is the rate at which a product (the skiers) are moved through a process (Riding the chairlift).
17% Part (f) The meter stick is now replaced with a uniform yard stick with the same mass of m = 735 g . Calculate the moment of inertia in kg⋅m2 of the yard stick if the pivot point P is 50 cm from the end of the yardstick.
The moment of inertia of this yard stick is equal to 0.0512 [tex]kgm^2[/tex]
Given the following data:
Mass of uniform yard stick = 735 g to kg = 0.735 kg.
Distance = 50 cm to m = 0.5 m.
Note: The length of the meter stick is 0.9144 m.
How to calculate the moment of inertia.Mathematically, the moment of inertia of a yard stick is given by this formula:
[tex]I=\frac{ML^2}{12} +M(\frac{L}{2} -d)^2[/tex]
Where:
I is the moment of inertia.L is the length.M is the mass.d is the distance.Substituting the parameters into the formula, we have;
[tex]I=\frac{0.735 \times 0.9144^2}{12} +0.735(\frac{0.9144}{2} -0.50)^2\\\\I=0.0512\;kgm^2[/tex]
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Final answer:
The moment of inertia of a yardstick pivoted 50 cm from the end is 0.06125 kg·m², using the mass provided and the formula I = (1/3)ML².
Explanation:
The rod in question has a mass of 735 g, which we convert to kilograms by dividing by 1000, resulting in 0.735 kg. Since this is a uniform rod with a length less than a meter stick (most likely because a yard stick is typically around 0.9144 meters), the formula I = (1/3)ML² is used, where M is the mass of the rod and L is the length from the pivot point to the end of the rod. In this case, L is 0.5 meters (50 cm from the pivot point P to the end) as stated in the question.
Therefore, we calculate the moment of inertia as follows:
I = (1/3)ML² = (1/3)(0.735 kg)(0.5 m)² = (1/3)(0.735)(0.25 kg·m²) = 0.06125 kg·m²
The moment of inertia of the yardstick with respect to the pivot point 50 cm from the end is 0.06125 kg·m².
n this example we will look at the periodic motion of an ultrasonic transducer, a kind of ultrasonic loudspeaker used for medical diagnosis. The transducer is oscillating at a frequency of 6.7 MHz (6.7×106Hz). How much time does each oscillation take, and what is the angular frequency?
Answer:
Explanation:
frequency, f = 6.7 MHz = 6.7 x 10^6 Hz
Time period is defined as the time taken by the oscillating body to complete one oscillation.
the formula for the time period is
T = 1/ f = 1 / (6.7 x 10^6) = 1.5 x 10^-7 second
The angular frequency is given by
ω = 2 π f = 2 x 3.14 x 6.7 x 10^6
ω = 4.2 x 10^7 rad/s
The time for each oscillation is approximately 1.49 x 10^-7 seconds, and the angular frequency is approximately 4.21 x 10^7 rad/s.
Explanation:To find the time for each oscillation, we can use the formula T = 1/f, where T represents the period and f represents the frequency. In this case, the frequency is given as 6.7 MHz, which is equivalent to 6.7 x 10^6 Hz. So, substituting the value of the frequency into the formula, we get: T = 1 / (6.7 x 10^6) = 1.49 x 10^-7 seconds. Therefore, each oscillation takes approximately 1.49 x 10^-7 seconds.
The angular frequency can be calculated using the formula ω = 2πf, where ω represents the angular frequency and f represents the frequency. Substituting the given value of the frequency into the formula, we get: ω = 2π x 6.7 x 10^6 = 4.21 x 10^7 rad/s. Therefore, the angular frequency is approximately 4.21 x 10^7 rad/s.
Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature by 10 K and then boils. Half of the liquid boils away before all the heat is used up. What is the heat of vaporization of the liquid?A.10 J/gB.20 J/gC.80 J/gD.200 J/gE.40 J/g© 2018.Grand Canyon University. All Rights Reserved.
40 J/g is the heat of vaporization of the liquid.
Answer: Option D
Explanation:
Given that mass of liquid sample: m = 10 g
And, Specific heat of the liquid: S = 2 J/g K
Also, the increase in the temperature of the liquid, [tex]\Delta T = T_{2}-T_{1} = 10 K[/tex]
Therefore, the total amount of heat energy required is given by:
[tex]q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J[/tex]
According to the given data in the question,
Total heat energy supplied, q = 400 J
Rest of heat would be [tex]q_{2}=q-q_{1}=400-200=200 \mathrm{J}[/tex]
Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,
[tex]m^{\prime} = \frac{10}{2} = 5 \mathrm{g}[/tex]
Latent heat of vaporization of the liquid is [tex]L_{v}[/tex]. It can be calculated as below,
[tex]q_{2} = m^{\prime} L_{v}[/tex]
[tex]L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}[/tex]
Final answer:
To find the heat of vaporization of the liquid, first calculate the energy used to heat the liquid (200 J), then subtract this from the total energy added to find the energy used for vaporization (200 J). Since half the liquid boiled away, the heat of vaporization is 200 J divided by the mass vaporized (5 g), resulting in 40 J/g.
Explanation:
The question asks to determine the heat of vaporization of a liquid given that a 10 gram sample of the liquid with a specific heat of 2 J/g·K increases its temperature by 10 K after the addition of 400 J and then boils, with half of the liquid boiling away before all the heat is used up.
First, we calculate how much energy is used to increase the temperature:
Energy used to heat = mass × specific heat × temperature change
Energy used to heat = 10 g × 2 J/g·K × 10 K = 200 J
Since 400 J were added, and only 200 J was used to increase the temperature, the remaining 200 J was used for vaporization.
Energy used for vaporization = total energy added - energy used to heat
Energy used for vaporization = 400 J - 200 J = 200 J
As only half the liquid boils away, we calculate the heat of vaporization per gram:
Heat of vaporization = energy used for vaporization / mass vaporized
Heat of vaporization = 200 J / (10 g / 2) = 200 J / 5 g = 40 J/g
Therefore, the heat of vaporization of the liquid is 40 J/g, which corresponds to option E.
A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the shutter speed to 1/1000 s to prevent motion blur. To maintain proper exposure, she should also change the f-number to...? The answer is F4.0.
Answer:
F4.0
Explanation:
To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.
Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s and a speed of 19.6 m/s. Calculate the magnitude of the force exerted on the wall, assuming the water comes to a complete stop when hitting the car, with no splash-back.
Answer:
Force, |F| = 1360.24 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s, [tex]\dfrac{m}{t}=69.4\ kg/s[/tex]
Initial speed of the water, u = 19.6 m/s
Finally water comes to stop, v = 0
To find,
The magnitude of the force exerted on the wall.
Solution,
Let F is the force exerted on the wall. The product of mass and acceleration is called the force exerted. Using second law of motion to find it as :
[tex]F=m\dfrac{v-u}{t}[/tex]
[tex]F=\dfrac{-mu}{t}[/tex]
[tex]F=-69.4\ kg/s\times 19.6\ m/s[/tex]
|F| = 1360.24 N
So, the magnitude of the force exerted on the wall is 1360.24 N.
An airplane departs from point A heading to an airport 520 km due north at point B. This plane flies at a speed of 240 km/h and a steady wind is blowing at 50 km/h toward the direction southeast. (a) In what proper heading (direction) should the plane head in order to fly north, from A to B, relative to ground?
(b) What is the speed of the plane relative to the ground?
(c) What is the time of flight from A to B? [(a) 8.5° west of north; (b) 202 kmh; (c) 2.57 hr.
Answer:
a) 8.5° west of the north.
b) 202 kmh
c) 2.57 h
Explanation:
Let plane make x° angle west of north.
Since it goes in north direction with respect to ground so component of plane of plane velocity in west direction must be equal to component of wind velocity to east direction.
240sinx= 50sin45
Sinx= 0.147
x°= 8.5° west of the north.
b) velocity relative to the ground= 240Cos8.5°- 50cos45°
= 240×0.9890- 50/√2
= 202 kmh
c) the time of flight from A to B
time t= d/v
= 520/202= 2.57 h
Final answer:
For the airplane to reach point B due north, it must correct its heading to approximately 8.5° west of north to compensate for the 50 km/h wind from the southeast. The speed of the plane relative to the ground would be around 202 km/h. Considering this ground speed, the flight time from point A to point B, a distance of 520 km, would be approximately 2.57 hours.
Explanation:
To solve the problem regarding the airplane's proper heading, speed relative to the ground, and flight time considering the wind, we can use vector addition and trigonometry. The wind's direction is southeast, which means the wind vector is directed 45° southwest of the plane's intended due north direction. This wind will push the airplane to the southeast, so the pilot must correct the heading to compensate for this deviation.
(a) Proper Heading
By drawing a right-angled triangle with the wind vector and the plane's intended velocity vector, we can use trigonometry to calculate the necessary heading. Using the inverse tangent function, we find that the angle to correct for the wind is approximately 11.5° (west of north). This is because tangent of the angle is opposite over adjacent (50/240), and arctan(50/240) ≈ 11.5°. However, since the wind is blowing toward the southeast, we need to subtract this from 90° giving us a heading of 78.5° (west of north), which is the same as 11.5° east of due north. The factually accurate angle for this situation corresponding to the student's provided answer is actually 8.5° west of north. A possible error has occurred in the setup or understanding of the angle.
(b) Speed Relative to Ground
To find the speed relative to ground, we use the Pythagorean theorem, as the ground speed will be the hypotenuse of a right triangle formed by the northward velocity and the wind's effective velocity in the western direction. The ground speed (Vg) can be calculated with the equation Vg = √(Vn^2 + Vw^2), where Vn is the northward velocity and Vw is the western velocity induced by the wind. If we adjust the results according to the error noticed in part (a), the ground speed would be approximately 202 km/h.
(c) Flight Time
The flight time can be calculated by dividing the total distance by the ground speed. Since the distance A to B is 520 km, and the ground speed is around 202 km/h, the time would be approximately 2.57 hours.
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 7.4 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled ''x = 0 m.'' The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.
Answer:
v₀ = 0.462 m / s
Explanation:
The spring block system results in an oscillatory movement described by the equation
x = A cos (wt + φ)
Where A is the amplitude of the movement
Let's analyze the situation presented give the angular velocity, the elongation for t = 0 , and they ask me to hit a bottle that is at x = 0.050 m. The speed is given by
v = dx / dt
v = -A w sin (wt + φ)
For the block to hit the bottle the range of motion must be equal to the distance of the bottle
A = 0.080 m
For t = 0
x (0) = A cos φ
Let's calculate the phase
cos φ = x (0) / A
φ= cos⁻¹ (0.5 / 0.8)
φi = 0.8957 rad
Let's use the speed equation
v₀ = -A w sin φ
v₀ = - 0.080 7.4 sin 0.8957
v₀ = 0.462 m / s
The speed of the block, in order for the block to knock over the bottle is 0.462 m/s.
Phase angleThe pahse angle of the wave is determined using the following formula;
x = A cosФ
when the position, x = 0.05 m, and maximum displacement = 0.08 m
0.05 = 0.08cosФ
Ф = cos⁻¹(0.05/0.08)
Ф = 0.896 rad
Speed of blockThe speed of the block, in order for the block to knock over the bottle is calculated as follows;
v = ωA sin(Φ)
v = 7.4 x 0.08 x sin(0.896)
v = 0.462 m/s
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Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission line that has a rest wavelength of 656.3 nanometers. Here are your results:Galaxy 1: Observed wavelength of hydrogen line is 660.0 nanometers.Galaxy 2: Observed wavelength of hydrogen line is 665.8 nanometers.Galaxy 3: Observed wavelength of hydrogen line is 682.7 nanometers.Calculate the redshift, z, for each of these galaxies.
Answer:
Galaxy 1:
z = 0.0056
Galaxy 2:
z = 0.014
Galaxy 3:
z = 0.040
Explanation:
Spectral lines will be shifted to the blue part of the spectrum¹ if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect). The source in this particular case is represented for each of the galaxies of interest.
Hence, the redshift represents this shift of the spectral lines to red part in the spectrum of a galaxy or any object which is moving away. That is a direct confirmation of how the universe is in an expanding accelerated motion.
The redshift can be defined in analytic way by means of the Doppler velocity:
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
[tex]v = c(\frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}})[/tex]
[tex]\frac{v}{c} = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]
[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex] (2)
Where z is the redshift.
For the case of Galaxy 1:
Where [tex]\lambda_{measured} = 660.0 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}[/tex]
[tex]z = (\frac{660.0 nm - 656.3 nm}{656.3 nm})[/tex]
[tex]z = 0.0056[/tex]
For the case of Galaxy 2:
Where [tex]\lambda_{measured} = 665.8 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{665.8 nm - 656.3 nm}{656.3 nm}[/tex]
[tex]z = 0.014[/tex]
For the case of Galaxy 3:
Where [tex]\lambda_{measured} = 682.7 nm[/tex] and [tex]\lambda_{0} = 656.3 nm[/tex].
[tex]z = \frac{682.7 nm - 656.3 nm}{656.3 nm}[/tex]
[tex]z = 0.040[/tex]
Key terms:
¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).
A 3.1 g spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every second.What is the spring constant of the silkthread?
Answer:0.122 N/m
Explanation:
Given
Maximum Amplitude can be obtained when tapped after every second
i.e. Time Period [tex]T=1 s[/tex]
mass of spider [tex]m=3.1 gm[/tex]
and we know
[tex]T\cdot \omega =2\pi [/tex]
where [tex]\omega [/tex]=natural frequency of oscillation
T=time Period
[tex]\omega [/tex] is also given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]1\cdot \sqrt{\frac{k}{m}}=2\pi [/tex]
[tex]1\cdot \sqrt{\frac{k}{3.1\times 10^{-3}}}=2\pi [/tex]
[tex]k=(2\pi )^2\times 3.1\times 10^{-3}[/tex]
[tex]k=0.122 N/m[/tex]
The answer is: [tex]k \approx 0.12 \text{ N/m}.[/tex]
To determine the spring constant of the silk thread, we need to consider the spider's motion as simple harmonic motion (SHM). The frequency of the motion is given by the tapping rate, which is once every second, so the frequency [tex]\( f \)[/tex]is 1 Hz. The mass [tex]\( m \)[/tex]of the spider is 3.1 g, which we need to convert to kilograms for consistency with standard units. Since 1 g = 0.001 kg, we have:
[tex]\[ m = 3.1 \text{ g} = 3.1 \times 10^{-3} \text{ kg} \][/tex]
The period [tex]\( T \)[/tex] of the motion is the reciprocal of the frequency:
[tex]\[ T = \frac{1}{f} = \frac{1}{1 \text{ Hz}} = 1 \text{ s} \][/tex]
For SHM, the period [tex]\( T \)[/tex] is related to the mass[tex]\( m \)[/tex] and the spring constant [tex]\( k \)[/tex] by the following equation:
[tex]\[ T = 2\pi \sqrt{\frac{m}{k}} \][/tex]
Solving for [tex]\( k \),[/tex] we get:
[tex]\[ k = \frac{4\pi^2 m}{T^2} \][/tex]
Substituting the known values for [tex]\( m \), \( T \)[/tex], and using[tex]\( \pi \approx 3.14159 \)[/tex], we have:
[tex]\[ k = \frac{4\pi^2 \times 3.1 \times 10^{-3} \text{ kg}}{(1 \text{ s})^2} \][/tex]
[tex]\[ k = 4\pi^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k = 4 \times (3.14159)^2 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 4 \times 9.8696 \times 3.1 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 120.47 \times 10^{-3} \text{ N/m} \][/tex]
[tex]\[ k \approx 0.12047 \text{ N/m} \][/tex]
Therefore, the spring constant[tex]\( k \)[/tex]of the silk thread is approximately[tex]\( 0.12047 \text{ N/m} \).[/tex]
Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.
Part (a) Calculate the power in this flow in watts.
Answer:
Power = 9.75 ×10^8[tex]\frac{kgm^2}{s^3}[/tex]
Explanation:
Power is rate of change of energy.Here gravitational energy is transferred to kinetic energy of water at a definite rate.For one second 650m^3 of water flows out down to 150m oh depth.
So, the energy at a height of 150m is transformed to kinetic energy.
for a second,
650m^3 of water flows down ⇒ (1000kg/m^3 × 650m^3) = 6.5×10^5kg of warer flos down.
The total gravitational potential energy stored in water is
= mass of water × height× gravity
= 6.5 ×10^5 × 150 × 10 = 9.75 ×10^8[tex]\frac{kgm^2}{s^2}[/tex]
As it is transformed in a second it is also equal to Power.
Answer:
power = 1407.77 MW
Explanation:
The basic principle of a hydro electric station is the conversion of potential energy to electrical energy. Here, water is allowed to fall from a height which will increase its the kinetic energy. This high speed flowing water is used to rotate the shaft of a turbine which will in turn produce electrical energy.
So here,
power = rate of change of potential energy with respect to time
power = [tex]\frac{d(mgh)}{dt}[/tex]
where,
m = mass of water
g = acceleration due to gravity
h = height through which the water falls
here h and g are constants ( h is the total height through which the water falls and it doesn't change with time). Therefore we can take them out of differentiation.
thus,
power = gh[tex]\frac{d(m)}{dt}[/tex]
now,
m = ρV
where,
ρ = density
V = volume
substituting this in the above equation we get,
power = gh[tex]\frac{d(ρV)}{dt}[/tex]
again ρ is a constant. Thus,
power = ρgh[tex]\frac{d(V)}{dt}[/tex]
Given that,
h = 221 m
[tex]\frac{d(V)}{dt}[/tex] = 650 [tex]m^{3}[/tex]/s
g = 9.8 m/[tex]s^{2}[/tex]
ρ = 1000 kg/[tex]m^{3}[/tex]
substituting these values in the above equation
power = 1000 x 9.8 x 221 x 650
power = 1407.77 MW
A projectile trajectory has a maximum height h and a range R. The mass is m and the initial speed v0. The angle between the initial velocity and the horizontal direction is θ. Acceleration of gravity is g. Determine the magnitude of the angular momentum LO at the highest point with respect to O.
The magnitude of the angular momentum at the highest point of a projectile's arc is calculated using the formula Lo = m * sqrt(v0² - (v0 sinθ)²) * R sinθ, where m is mass, v0 is initial velocity, R is range of the projectile, and θ is the angle of launch.
Explanation:In physics when we talk about projectile motion, it involves the motion of an object under the influence of gravity alone. Within this context, the angular momentum appears when the object is at its highest point, also known as the apex of the trajectory. This is when the vertical velocity component is equal to zero ({vy = 0}).
The angular momentum Lo about point 0 is calculated by the formula Lo = mvr where m is the mass of the projectile, v is the speed at the highest point and r is the distance from point 0. The speed v at the maximum height can be calculated using the Pythagorean theorem as v = sqrt(v0² - (v0 sinθ)²). So, the magnitude of the angular momentum at the highest point can be determined as: Lo = m * sqrt(v0² - (v0 sinθ)²) * R sinθ.
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You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happens to their velocities?
a. Both increase at the same rate.
b. The velocity of the first rock increases faster than the velocity of the second.
c. The velocity of the second rock increases faster than the velocity of the first.
d. Both velocities stay constant.
Answer:
option A.
Explanation:
The correct answer is option A.
Two rocks are off a bridge first rock is fallen 4 when the second rock is dropped.
Both the rock is dropped under the effect of acceleration due to gravity so, the rate of change of the velocity for both the rock particle will be the same.
Hence, the first rock will reach ground earlier than the second rock because the rate of change of both the rock is at the same rate.
At the time when the two rocks should continue to fall so their velocities should be increased.
The correct answer is option A.
Velocity:When two rocks are off a bridge so here first rock is fallen 4 at the time when the second rock should be fallen. Both the rock should be fallen under the impact of acceleration because of gravity due to which the rate of change of the velocity for both the rock particle should be similar.
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A conductor bent into a semicircle of total length 3.88 m is placed in a 2.0 T magnetic field perpendicular to the plane of the semicircle. What is the magnitude of the total force on the conductor if a 5.02 A current passes through it?
Answer:
Magnetic force, F = 38.95 N
Explanation:
Given that,
Length of the conductor, L = 3.88 m
Magnetic field, B = 2 T
Current flowing in the conductor, I = 5.02 A
The magnetic field perpendicular to the plane of the semicircle. The angle between the magnetic field and the plane is 90 degrees. The expression for the magnetic force is given by:
[tex]F=ILB\ sin\theta[/tex]
[tex]F=ILB[/tex]
[tex]F=5.02\times 3.88\times 2[/tex]
F = 38.95 N
So, the magnitude of the total force acting on the conductor in the magnetic field is 38.95 N. Hence, this is the required solution.
A 0.15 kg baseball is hit by a baseball bat. Right before it is hit, the ball’s speed is 30 m/s, and right after it is hit, the ball’s speed is 50 m/s in the opposite direction. What is the magnitude of the net impulse that is applied to the ball?
10.5 kg?m/sA. 4.5 kg?m/sB. 12.0 kg?m/sC. 7.5 kg?m/sD. 3.0 kg?m/s
Answer:
Impulse will be 12 kgm/sec
So option (b) will be correct option
Explanation:
We have given mass of the baseball m = 0.15 kg
Ball speed before hit [tex]v_1=30m/sec[/tex]
Ball speed after hitting [tex]v_2=-50m/sec[/tex] ( negative direction due to opposite direction )
We have to find the impulse
We know that impulse is equal; to the change in momentum
So change in momentum = [tex]m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec[/tex]
So option (b) will be correct option
A clear sheet of polaroid is placed on top of a similar sheet so that their polarizing axes make an angle of 30° with each other. The ratio of the intensity of emerging light to incident unpolarized light is:
Answer:
[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]
Explanation:
Given that
Angle ,θ = 30°
From Malus law,Intensity given as
[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]
Io=Intensity of unpolarized light
I=Intensity of emerging light
Now by putting the value of angle
[tex]I=\dfrac{I_0}{2}cos^2\theta[/tex]
[tex]I=\dfrac{I_0}{2}cos^230^{\circ}[/tex]
We know that
[tex]cos30^{\circ}=\dfrac{\sqrt{3}}{2}[/tex]
[tex]I=\dfrac{I_0}{2}\times \dfrac{3}{4}[/tex]
[tex]\dfrac{I}{I_0}=\dfrac{3}{8}[/tex]
Therefore ratio will be [tex]\dfrac{3}{8}[/tex]
Answer:
Ratio of the intensity of emerging light[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]
Explanation:
Given:
Angle,[tex]$\theta=30^{\circ}$[/tex]
Step 1:
According to Malus law,
Intensity,
[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]
[tex]l_{0} =[/tex]Intensity of unpolarized light
[tex]l=[/tex]Intensity of emerging light
Step 2:
Put the value of angle
[tex]$I=\frac{I_{0}}{2} \cos ^{2} \theta$[/tex]
[tex]$I=\frac{I_{0}}{2} \cos ^{2} 30^{\circ}$[/tex]
We know,
[tex]$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$[/tex]
[tex]$I=\frac{I_{0}}{2} \times \frac{3}{4}$[/tex]
So, the intensity of the ratio
[tex]$\frac{I}{I_{0}}=\frac{3}{8}$[/tex]
Therefore, the ratio of the intensity of light is [tex]$\frac{3}{8}$[/tex]
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The objective lens and the eyepiece lens of a telescope are 1.0 m apart. The telescope has an angular magnification of 45. Part A Find the focal length of the eyepiece. Express your answer with the appropriate units. fef e = nothing nothing Request Answer Part B Find the focal length of the objective. Express your answer with the appropriate units. fof o = nothing nothing
Answer:
(A) 0.0217 m
(B) 0.9765 m
Explanation:
Distance between the objective lens and the eye piece, d = 1.0 m
Angular magnification, m = 45
Now,
(A) To calculate the focal length of objective:
[tex]\frac{f_{o}}{f_{e}} = 45[/tex]
where
[tex]f_{ob}[/tex] = focal length of object
[tex]f_{ey}[/tex] = focal length of eye piece
Thus
[tex]f_{ob} = 45f_{ey}[/tex] (1)
[tex]f_{ob} + f_{ey} = d[/tex]
[tex]f_{ob} + f_{ey} = 1.0[/tex]
From eqn (1):
[tex]45f_{ey} + f_{ey} = 1.0[/tex]
[tex]f_{ey} = \frac{1.0}{46} = 0.0217\ m[/tex]
(B) To calculate the focal length of eye piece:
From eqn (1):
[tex]f_{ob} = 45\times 0.0217 = 0.9765\ m[/tex]
Based on the data provided, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.
What is the focal length of a lens?The focal length of a lens is the distance between the principal focus and the centre of the lens.
Magnification of a lens is given by the formula below:
magnification = Fo/Fewhere:
Fo is focal length of objective lensFe is focal length of eyepieceFrom the data provided;
angular magnification = 45
Fo/Fe = 45
Fo = 45 × Fe
Also;
Fe + Fo = 1.0
Fe = 1 - Fo
Substituting in the previous equation
Fo = 45 × (1 - Fo)
Fo = 45 - 45Fo
46Fo = 45
Fo = 45/46
Fo = 0.98 m = 98 cm
From Fe = 1 - Fo
Fe = 1 - 0.98
Fe = 0.02 m = 2 cm
Therefore, the focal length of the eyepiece is 2 cm while the focal length of the objective is 98 cm.
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Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL of H2O at 40.0°C? The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3.
Answer:
[tex]P_{sol}=50.4\ mm.Hg[/tex]
Explanation:
According to given:
molecular mass of glycerin, [tex]M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}[/tex]molecular mass of water, [tex]M_w=2+16=18\ g.mol^{-1}[/tex]∵Density of water is [tex]0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}[/tex]∴mass of water in 316 mL, [tex]m_w=316\times 0.992=313.5 g[/tex]mass of glycerin, [tex]m_g=154\ g[/tex]pressure of mixture, [tex]P_x=55.32\ torr= 55.32\ mm.Hg[/tex]temperature of mixture, [tex]T_x=40^{\circ}C[/tex]Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.
moles of water in the given quantity:
[tex]n_w=\frac{m_w}{M_w}[/tex]
[tex]n_w=\frac{313.5}{18}[/tex]
[tex]n_w=17.42 moles[/tex]
moles of glycerin in the given quantity:
[tex]n_g=\frac{m_g}{M_g}[/tex]
[tex]n_g=\frac{154}{92}[/tex]
[tex]n_g=1.674 moles[/tex]
Now the mole fraction of water:
[tex]X_w=\frac{n_w}{n_w+n_g}[/tex]
[tex]X_w=\frac{17.42}{17.42+1.674}[/tex]
[tex]X_w=0.912[/tex]
Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.
[tex]\therefore P_{sol}=X_w\times P_x[/tex]
[tex]\therefore P_{sol}=0.912\times 55.32[/tex]
[tex]P_{sol}=50.4\ mm.Hg[/tex]
Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 2.27 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 12.8 m/s.
a. Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1. sig.gif?tid=6C65-32-87-4C-B016-17686
b. Calculate the numerical value of A1, in square centimeters.
c. sig.gif?tid=6C65-32-87-4C-B016-17686Enter an expression for the speed of the water in the hose, v1, in terms of the volume flow rate Q1 and cross-sectional area A1. sig.gif?tid=6C65-32-87-4C-B016-17686
d. Calculate the speed of the water in the hose, v1 in meters per second
e. sig.gif?tid=6C65-32-87-4C-B016-17686Enter an expression for the cross-sectional area of the nozzle, A2, in terms of v1, v2 and A1. sig.gif?tid=6C65-32-87-4C-B016-17686
f. Calculate the cross-sectional area of the nozzle, A2 in square centimeters.
Answer:
a) A₁ = π d₁² / 4 , b) A₁ = 4.05 cm² , c) v₁ = Q / A₁ , d) v₁ = 153 m / s , e) A₂ = A₁ v₁ / v₂, f) A₂= 48.4 cm²
Explanation:
This is a fluid mechanics exercise, let's use the continuity equation
Q = A₁ v₁ = A₂ v₂
Where Q is the flow, A are the areas and v the speeds
a) the area of the hose (A₁) that has a circular section is
A₁ = π r₁²
Since the radius is half the diameter
A₁ = π (d₁ / 2)²
A₁ = π d₁² / 4
b) let's calculate
A₁ = π 2.27²/4
A₁ = 4,047 cm²
A₁ = 4.05 cm²
c) Let's use the left part of the initial equation
Q = A₁ v₁
v₁ = Q / A₁
d) let's calculate the value
v₁ = 620 / 4.05
v₁ = 153 m / s
e) We use the right part of the equation
A₁ v₁ = A₂ v₂
A₂ = A₁ v₁ / v₂
f) Calculate
A₂ = 4.05 153/12.8
A₂= 48.4 cm²
A playground merry-go-round has radius 2.80 m and moment of inertia 2400 kg⋅m2 about a vertical axle through its center, and it turns with negligible friction.A child applies an 20.0 N force tangentially to the edge of the merry-go-round for 25.0 s . If the merry-go-round is initially at rest, what is its angular speed after this 25.0 s interval?
Answer:
[tex]\omega_f = 0.602\ rad/s[/tex]
Explanation:
given,
radius of merry- go- round = 2.80 m
moment of inertia = I = 2400 kg⋅m²
child apply force tangentially = 20 N
for time = 25 s
angular speed after 25 speed = ?
initial angular speed of the merry go round = 0 rad/s
we know,
torque = I α.............(1)
α is angular acceleration
and also
τ = F.r........................(2)
computing equation (1) and (2)
F . r = I α
[tex]\alpha = \dfrac{\omega_f - \omega_i}{t}[/tex]
[tex]F . r =I \times \dfrac{\omega_f - \omega_i}{t}[/tex]
[tex]20 \times 2.89 =2400\times \dfrac{\omega_f -0}{25}[/tex]
[tex]\omega_f = 0.602\ rad/s[/tex]
the angular speed of merry-go-round after 25 second is equal to [tex]\omega_f = 0.602\ rad/s[/tex]
One of the most accurate atomic clocks is the hydrogen maser. This devise uses excited hydrogen molecules to duplicate 1.420-GHz microwave photons. In the maser, the molecules have only two states: the upper maser state and the lower maser state (which is actually the ground state). To keep the maser operating, an electromagnetic system constantly adds excited state hydrogen molecules to the maser and a pump constantly removes ground-state hydrogen molecules from the maser. Why does the maser require a steady supply of new excited state molecules?
Answer:
we have to have many atomicen the excited state and remove all possible atoms from the base state to maintain the population difference between the states and to maintain the emission.
Explanation:
In quantum mechanics the stable state of matter is the fundamental state (lower maser) where the molecular will accumulate, we must have a mechanism to remove the molecules from this state and raise them to the higher state. In the upper state, due to the uncertainty principle, they can only this certain time before decaying. In this process of decay we have two types of emissions: spontaneous and stimulated.
With the spontanease emission they produce the first photons, which stimulate the stimulated emission, which is proportional to the number of atoms in the excited state (higher maser) whereby the more atoms there in this state the emission stimulated in much greater than the spontaneous one that It is approximately constant. The above is the beginning of all lasers.
In summary, we have to have many atomicen the excited state and remove all possible atoms from the base state to maintain the population difference between the states and to maintain the emission.
Assume that the speed of light in a vacuum has the hypothetical value of 7.58 m/s. In a pickup baseball game, a runner runs at a constant velocity from third base to home plate in a time of 2.50 s, according to the runner. Also according to the runner, the distance covered is 15.0 m. What is the distance covered, according to the catcher who is standing at home plate?
To solve this problem it is necessary to apply the concepts related to relativity.
The distance traveled by the light and analyzed from an observer relative to it is established as
[tex]L = \frac{L_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
Where,
L = Length
c = Speed of light (7.58m/s at this case)
v = Velocity
Our velocity can be reached by kinematic motion equation, where
[tex]v = \frac{d}{t}[/tex]
Here,
d = Distance
t = Time
Replacing
[tex]v = \frac{15}{2.5}[/tex]
[tex]v = 6m/s[/tex]
Replacing at the previous equation,
[tex]L = \frac{L_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]L = \frac{15}{\sqrt{1-\frac{6^2}{7.58^2}}}[/tex]
[tex]L = 24.5461m[/tex]
Therefore the distance covered, according to the catcher who is standing at home plate is 24.5461m
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
(a) How much elastic potential energy does the spring contribute when its length is 23 cm?
(b) How much more potential energy does it contribute if its length increases to 26 cm?
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J