(a) When m₂ is released, its acceleration will be approximately -3.27 m/s².
(b) The tension in the string is approximately -13.08 N.
To determine the acceleration of m₂ when it is released and the tension in the string, we need to consider the forces acting on the system.
(a) Acceleration of m₂:
Since the pulley is assumed to be frictionless, the tension in the string is the same on both sides of the pulley. We can consider the system consisting of m₁ and m₂ as one body. The net force acting on this system is the difference between the weight of m₁ and the weight of m₂:
Net force = m₁g - m₂g
Applying Newton's second law, F = ma, where F is the net force and a is the acceleration, we have:
m₁g - m₂g = (m₁ + m₂)a
Rearranging the equation to solve for the acceleration, we get:
a = (m₁g - m₂g) / (m₁ + m₂)
Substituting the given values, m₁ = 2.00 kg and m₂ = 4.00 kg, and the acceleration due to gravity, g = 9.8 m/s², we can calculate the acceleration:
a = ((2.00 kg)(9.8 m/s²) - (4.00 kg)(9.8 m/s²)) / (2.00 kg + 4.00 kg)
a = (19.6 N - 39.2 N) / 6.00 kg
a = -19.6 N / 6.00 kg
a = -3.27 m/s²
Therefore, when m₂ is released, its acceleration will be approximately -3.27 m/s². The negative sign indicates that the acceleration is in the opposite direction of the gravitational force.
(b) Tension in the string:
The tension in the string can be determined by considering the forces acting on m₂. The net force on m₂ is equal to its mass multiplied by its acceleration:
Net force = m₂a
Substituting the given values, m₂ = 4.00 kg and a = -3.27 m/s², we can calculate the tension:
Tension = (4.00 kg)(-3.27 m/s²)
Tension = -13.08 N
Therefore, the tension in the string is approximately -13.08 N. The negative sign indicates that the tension acts in the opposite direction of the weight of m₂.
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The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colors) to the shortest wavelengths (the deepest violet colors) our eyes can detect. Wavelengths near 655 nm are perceived as red. Those near 515 nm are green and those near 475 nm are blue. Calculate the frequency of light (in Hz) with a wavelength of 655 nm, 515 nm, and 475 nm.
HINT
(a)
655 nm
Hz
(b)
515 nm
Hz
(c)
475 nm
Hz
Answer:
The frequency of light can be calculated using the formula:
`c = λv`
Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.
The speed of light in a vacuum is `3.00 × 10^8 m/s`.
To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.
Thus, the frequency of light with a wavelength of 655 nm is:
`v = c/λ`
`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`
`v = 4.58 × 10^14 Hz`
Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.
Similarly, the frequency of light with a wavelength of 515 nm is:
`v = c/λ`
`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`
`v = 5.83 × 10^14 Hz`
Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.
Finally, the frequency of light with a wavelength of 475 nm is:
`v = c/λ`
`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`
`v = 6.32 × 10^14 Hz`
Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.
So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.
A block lies on a horizontal frictionless surfaces.A horizontal force of 100N is applied to the block giving rise to an acceleration of 3.0 .(a) determine the mass of the block . (b) calculate the distance of the block will travel in the force applied for 10s .(c) calculate the speed of the block after the force has been applied for 10s
Answer:
Explanation:
A) The mass of the block is 33.3 Kg.
We know that F=ma
m=F/a
Where,
F⇒Force
m⇒mass
a⇒acceleration
Given,
F=100N
a=3m/s^2
m=100/3
m= 33.3Kg.
Therefore the mass of the block is 33.3 kg.
B) Distance traveled in 10s is 150m.
According to Newton's laws of motion,
s=ut+1/2at^2
s⇒displacement
u⇒initial velocity
t⇒time
Given,
u=0
t=10s
a=3m/s^2
s=0+1/2*3*100
s=150m.
Therefore distance traveled by the block after 10s is 150m.
C)The speed of the block after 10s will be 30m/s.
We know that
v^2=u^2+2as
v⇒Final velocity
u⇒initial velocity
v^2=0+2*3*150
v^2=900
v=30m/s.
Therefore the velocity of the block will be 30m/s after 10s.
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A rope pulls a Tesla out of mud. The guy pulls a force F⊥ of 300N, and theta = 4.2°. The tension force T is ___ Newton.
A rope pulls a Tesla out of mud. The guy pulls a force F⊥ of 300N, and theta = 4.2°. The tension force T is 298.44__ Newton.
The problem describes a Tesla that is stuck in the mud and needs to be pulled out using a rope. the guy pulls a force F⊥ of 300N and that the angle between the rope and the horizontal plane is θ = 4.2°. The goal is to find the tension force T exerted by the rope.To solve for T, we'll need to use trigonometry. We can break the force vector into its horizontal and vertical components as follows:
Fx = F⊥ cosθ and Fy = F⊥ sinθ.
Since the rope is pulling the Tesla horizontally, the horizontal component of the force will be the tension force T. So we have:
T = Fx = F⊥ cosθ = (300 N) cos(4.2°) ≈ 298.44 N
Taking the cosine of the angle is necessary since it's the adjacent side that we're interested in, which is the horizontal component of the force. Therefore, the tension force exerted by the rope is approximately 298.44 N.
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Part 3 Waves on a string-with a loose end The reflected
wave interferes with the original wave and creates standing wave composed of
nodes and antinodes if the frequency is just right: Instead of a node an antinode
will always exist at the loose end: (This happens because the phase of the wave
is not inverted upon reflection from loose end and therefore always constructively
interfere at that position:) Draw and measure the frequency of the 1st harmonic
(node near driver end followed by an antinode on loose end) Settings: amplitude:
0.05 cm tension: high damping: none turn on: Loose End What fraction of a
wavelength is this? Hz Click Restart' to observe the standing wave. 2. Predict the
frequencies of several higher harmonics: Use the wave simulator to test each of
your calculated harmonics Draw and label the standing waves for each of the
harmonics you discovered: Divide each higher harmonic by the first harmonic:
Are the higher harmonics even-number or odd-number multiples of the first
harmonic?
The first harmonic of the standing wave on a string with a loose end represents half a wavelength.
The fraction of a wavelength represented by the first harmonic is 1/2.
The higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.
1. The first harmonic of a standing wave on a string with a loose end occurs when there is a node near the driver end and an antinode at the loose end. To measure the frequency of the first harmonic, we need to determine the fraction of a wavelength represented by this standing wave.
The first harmonic of the standing wave on a string with a loose end represents half a wavelength.
The first harmonic of a standing wave on a string with a loose end consists of a node near the driver end and an antinode at the loose end. This configuration creates the simplest standing wave pattern.
In a standing wave, a node is a point where the amplitude of the wave is always zero, representing a point of minimum displacement. An antinode, on the other hand, is a point of maximum displacement, where the amplitude is at its highest.
Since the loose end does not invert the phase of the wave upon reflection, the reflected wave and the original wave constructively interfere at the loose end, resulting in an antinode.
In the first harmonic, there is exactly half a wavelength between the node near the driver end and the antinode at the loose end.
Therefore, the fraction of a wavelength represented by the first harmonic is 1/2.
2. To predict the frequencies of higher harmonics, we can use the relationship that the frequency of each harmonic is a multiple of the frequency of the first harmonic. The higher harmonics can be calculated as follows:
Second Harmonic: The second harmonic consists of two nodes and one additional antinode compared to the first harmonic. The fraction of a wavelength for the second harmonic is 1/2 * 2 = 1. Thus, the second harmonic has a frequency that is twice that of the first harmonic.
Third Harmonic: The third harmonic consists of three nodes and two additional antinodes compared to the first harmonic. The fraction of a wavelength for the third harmonic is 1/2 * 3 = 1.5. Thus, the third harmonic has a frequency that is three times that of the first harmonic.
Fourth Harmonic: The fourth harmonic consists of four nodes and three additional antinodes compared to the first harmonic. The fraction of a wavelength for the fourth harmonic is 1/2 * 4 = 2. Thus, the fourth harmonic has a frequency that is four times that of the first harmonic.
In general, the higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.
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Select the correct answer.
A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
O A.
B.
100 kilometers/hour south
200 kilometers/hour
200 kilometers/hour north
O C.
O D. 100 kilometers/hour
The average velocity of the car is 100 kilometers/hour south. This means that, on average, the car is traveling 100 kilometers per hour in the south direction relative to its starting point.
To determine the average velocity of the car, we need to calculate the displacement and divide it by the time taken. Velocity is defined as the rate of change of displacement with respect to time.
In this case, the car is traveling south, and its displacement is 200 kilometers from its starting point after 2 hours.
The average velocity is given by the formula:
Average velocity = Displacement / Time
The displacement is 200 kilometers south, and the time is 2 hours. Therefore, we have:
Average velocity = 200 kilometers south / 2 hours
Simplifying the calculation:
Average velocity = 100 kilometers/hour south
Hence, the correct answer is B. 100 kilometers/hour south. This indicates that the car's average velocity is 100 kilometers per hour towards the south direction.
It's important to note that velocity is a vector quantity and includes both magnitude (speed) and direction. In this case, the direction is specified as south, which indicates that the car is moving towards the south relative to its starting point.
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The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colors) to the shortest wavelengths (the deepest violet colors) our eyes can detect. Wavelengths near 655 nm are perceived as red. Those near 515 nm are green and those near 475 nm are blue. Calculate the frequency of light (in Hz) with a wavelength of 655 nm, 515 nm, and 475 nm.
The frequency of light with a wavelength of 655 nm is[tex]4.57 x 10^14 Hz[/tex] and 515 nm is [tex]5.82 x 10^14[/tex] Hz and 475 nm is[tex]6.31 x 10^14 Hz[/tex]
The equation that links the speed of light to wavelength and frequency is
c = λν
Where, c = speed of lightλ = wavelengthν = frequency c is a constant of 2.998 x 10^8 m/s.
Calculating the frequency of light with a wavelength of
655 nm:λ = 655 nm = [tex]6.55 x 10^-7m[/tex]
Using the above equation, we get
c = λνν = c/λ = [tex](2.998 x 10^8 m/s)/(6.55 x 10^-7m)ν = 4.57 x 10^14 Hz[/tex]
Therefore, the frequency of light with a wavelength of 655 nm is 4.57 x [tex]10^14 Hz.[/tex]
Calculating the frequency of light with a wavelength of 515 nm:λ = 515 nm = [tex]5.15 x 10^-7m[/tex]
Using the above equation, we get
c = λνν = c/λ =[tex](2.998 x 10^8 m/s)/(5.15 x 10^-7m)ν = 5.82 x 10^14 Hz[/tex]
Therefore, the frequency of light with a wavelength of 515 nm is 5.82 x [tex]10^14 Hz[/tex].
Calculating the frequency of light with a wavelength of 475 nm:λ = 475 nm = [tex]4.75 x 10^-7[/tex]m Using the above equation, we get
c = λνν = c/λ = [tex](2.998 x 10^8 m/s)/(4.75 x 10^-7m)ν = 6.31 x 10^14 Hz[/tex]
Therefore, the frequency of light with a wavelength of 475 nm is 6.31 x [tex]10^14 Hz.[/tex]
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The boiling point of nitrogen is -196°C. What is that on the absolute temperature scale?
Answer:
The boiling point of nitrogen on the absolute temperature scale is 77.15 K
Explanation:
Temperature in Kelvin(Absolute temperature) = Temperature in Celcius + 273.15.
A cup of coffee cools from 120-115°f in 0.5 minutes in a room of 65°f. how long will it take to cool from 105°f _90°f in the same room use newton's law of Cooling
