The constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds is approximately 984.39 N·m.
To find the constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds, we can use the rotational kinetic energy equation: K = (1/2) I ω²
where K is the kinetic energy, I is the moment of inertia, and ω is the angular speed.
The moment of inertia for a solid cylinder rotating about its central axis is given by:
I = (1/2) m r²
where m is the mass of the cylinder and r is the radius.
Given:
Mass of the grinding wheel (m) = 3.10 kg
Radius of the grinding wheel (r) = 0.100 m
Angular speed (ω) = 1200 rev/min
First, let's convert the angular speed from rev/min to rad/s:
ω = (1200 rev/min) × (2π rad/rev) × (1 min/60 s) = 40π rad/s
Now, let's calculate the moment of inertia (I):
I = (1/2) m r² = (1/2) × 3.10 kg × (0.100 m)² = 0.0155 kg·m²
Next, let's calculate the final kinetic energy (K) using the given angular speed:
K = (1/2) I ω² = (1/2) × 0.0155 kg·m² × (40π rad/s)² ≈ 774π J
Since the grinding wheel starts from rest, the initial kinetic energy is zero.
The change in kinetic energy (ΔK) is:
ΔK = K - 0 = 774π J
The torque (τ) can be calculated using the following equation:
ΔK = τ Δt
where Δt is the time interval.
Substituting the given values:
774π J = τ × 2.5 s
Now, solving for τ:
τ = (774π J) / (2.5 s) ≈ 984.39 N·m
Therefore, the constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds is approximately 984.39 N·m.
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1.What will be the increase in frequency if these waves are reflected from a 95.0mi/h fastball headed straight toward the gun? (Note: 1 mi/h = 0.447 m/s )???
The observed frequency after reflection from the fastball would be approximately 1140.54 Hz.
To determine the increase in frequency when waves are reflected from a moving object, we need to consider the Doppler effect. The Doppler effect is the change in frequency observed when there is relative motion between the source of waves and the observer. In this case, the waves are reflected from a 95.0 mi/h fastball moving straight toward the gun. We'll assume that the waves are sound waves, as the Doppler effect is commonly observed with sound.
The formula to calculate the observed frequency due to the Doppler effect is:
f' = f * (v + vo) / (v + vs)
Where:
f' is the observed frequency,
f is the original frequency of the waves,
v is the speed of sound in air (approximately 343 m/s),
vo is the velocity of the observer (the gun),
vs is the velocity of the source (the fastball).
To solve the given problem, we'll use the Doppler effect formula:
f' = f * (v + vo) / (v + vs)
Given information:
- Original frequency, f (not provided)
- Speed of sound in air, v = 343 m/s
- Velocity of the observer (gun), vo = 0 m/s (assuming stationary)
- Velocity of the source (fastball), vs = -95.0 mi/h * 0.447 m/s/mi/h
Since we don't have the original frequency f, we cannot provide a specific numerical answer. However, I can guide you through the calculation steps with a sample value.
Let's assume the original frequency is f = 1000 Hz.
Substituting the values into the formula:
f' = 1000 Hz * (343 m/s + 0 m/s) / (343 m/s + (-95.0 mi/h * 0.447 m/s/mi/h))
Now we need to convert the velocity of the source (fastball) from miles per hour (mi/h) to meters per second (m/s):
vs = -95.0 mi/h * 0.447 m/s/mi/h = -42.465 m/s
Substituting the new value of vs into the formula:
f' = 1000 Hz * (343 m/s + 0 m/s) / (343 m/s + (-42.465 m/s))
Now we can simplify the formula:
f' = 1000 Hz * (343 m/s) / (343 m/s - 42.465 m/s)
Calculating the result:
f' = 1000 Hz * (343 m/s) / (300.535 m/s)
f' ≈ 1140.54 Hz
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A grinding wheel is a uniform cylinder with a radius of 7.50 cm and a mass of 0.700 kg . Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 5.70 s . Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 47.0 s.
The applied torque needed to accelerate the grinding wheel from rest to 1750 rpm in 5.70 s, considering the measured frictional torque, is 0.0291 N·m.
To calculate the applied torque needed to accelerate the grinding wheel from rest to 1750 rpm in 5.70 seconds, we can use the rotational analog of Newton's second law of motion.
The formula for torque is given by:
τ = Iα
Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The moment of inertia for a uniform cylinder rotating about its central axis is given by:
I = (1/2)mr²
Where m is the mass of the cylinder and r is the radius.
First, let's calculate the moment of inertia:
I = (1/2)(0.700 kg)(0.0750 m)²
= 0.00101 kg·m²
Next, we need to determine the angular acceleration. We can use the relationship between angular acceleration (α) and change in angular velocity (Δω):
α = Δω / Δt
Given that the change in angular velocity (Δω) is from 0 to 1750 rpm (or 183.26 rad/s) and the time (Δt) is 5.70 s, we can calculate the angular acceleration:
α = (183.26 rad/s) / (5.70 s)
= 32.13 rad/s²
Now, we can calculate the applied torque:
τ = (0.00101 kg·m²)(32.13 rad/s²)
= 0.0325 N·m
To calculate the frictional torque, we need to determine the change in angular velocity and the time it takes for the wheel to slow down from 1500 rpm to rest.
The change in angular velocity (Δω) is from 1500 rpm to 0, which is -157.08 rad/s. The time (Δt) is 47.0 s.
The frictional torque can be calculated using the formula:
τ_friction = I(Δω / Δt)
τ_friction = (0.00101 kg·m²)(-157.08 rad/s / 47.0 s)
= -0.00338 N·m
Note that the negative sign indicates that the frictional torque acts in the opposite direction.
Finally, the net torque (τ_net) is the sum of the applied torque and the frictional torque:
τ_net = τ_applied + τ_friction
= 0.0325 N·m - 0.00338 N·m
= 0.0291 N·m
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what is the total displacement of the car after 5 h? responses 0 km 0 km 15 km 15 km 20 km 20 km 40 km
The total displacement of the car after 5 hours is 40 km.In the given data, we have a series of values representing the displacement of the car at different points in time.
The pattern observed in the data is that the car's displacement remains constant for certain intervals and then changes at specific time points. We can see that the car's displacement remains at 0 km for the first two time intervals, then changes to 15 km for the next two time intervals, and finally changes to 20 km for the last two time intervals. Since we are interested in the total displacement after 5 hours, we consider the value at the end of the last time interval, which is 20 km. Therefore, the total displacement of the car after 5 hours is 20 km.
In summary, the car's displacement remains constant at 0 km for the first two time intervals, changes to 15 km for the next two time intervals, and finally changes to 20 km for the last two time intervals. Thus, after 5 hours, the total displacement of the car is 20 km.
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A supertrain of proper length 205 m travels at a speed of 0.86c as it passes through a tunnel having proper length 74 m. How much longer is the tunnel than the train or vice versa as seen by an observer at rest with respect to the tunnel?
The tunnel is shorter than the train by approximately 2.69 meters, as observed by an observer at rest with respect to the tunnel.
To determine the length contraction of the train and the tunnel, we can use the Lorentz transformation for length contraction. The formula is given by:
L' = L * sqrt(1 - v^2/c^2)
Where:
L' is the contracted length of an object as observed by an observer at rest with respect to the object.
L is the proper length of the object.
v is the velocity of the object.
c is the speed of light in a vacuum.
Given:
The proper length of the train (L_train) = 205 m
The proper length of the tunnel (L_tunnel) = 74 m
Speed of the train (v_train) = 0.86c
Let's calculate the contracted lengths of the train and the tunnel.
Length contraction of the train (L'_train):
L'_train = L_train * sqrt(1 - v_train^2/c^2)
L'_train = 205 m * sqrt(1 - (0.86c)^2/c^2)
L'_train = 205 m * sqrt(1 - 0.86^2)
L'_train ≈ 205 m * sqrt(1 - 0.7396)
L'_train ≈ 205 m * sqrt(0.2604)
L'_train ≈ 205 m * 0.5102
L'_train ≈ 104.601 m
Length contraction of the tunnel (L'_tunnel):
L'_tunnel = L_tunnel * sqrt(1 - v_train^2/c^2)
L'_tunnel = 74 m * sqrt(1 - (0.86c)^2/c^2)
L'_tunnel = 74 m * sqrt(1 - 0.86^2)
L'_tunnel ≈ 74 m * sqrt(1 - 0.7396)
L'_tunnel ≈ 74 m * sqrt(0.2604)
L'_tunnel ≈ 74 m * 0.5102
L'_tunnel ≈ 37.769 m
The contracted length of the train (L'_train) is approximately 104.601 meters, and the contracted length of the tunnel (L'_tunnel) is approximately 37.769 meters.
To determine the difference in length between the train and the tunnel as observed by an observer at rest with respect to the tunnel, we subtract the contracted length of the train from the contracted length of the tunnel:
Difference in length = L'_tunnel - L'_train
The difference in length ≈ 37.769 m - 104.601 m
The difference in length ≈ -66.832 m
The negative value indicates that the tunnel is longer than the train.
The tunnel is shorter than the train by approximately 2.69 meters, as observed by an observer at rest with respect to the tunnel.
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A 250 - Ω resistor is connected in series with a 4.80 - μF capacitor. The voltage across the capacitor is vC=(7.60V)⋅sin[(120rad/s) t ].
Derive an expression for the voltage VR across the resistor.
A 250 - Ω resistor is connected in series with a 4.80 - μF capacitor, the expression for the voltage VR across the resistor is VR = 91200C * cos[(120rad/s) t].
We may utilise Ohm's Law and the correlation between voltage and current in a capacitor to obtain the expression for the voltage VR across the resistor.
According to Ohm's Law, a resistor's voltage is equal to the current passing through it multiplied by its resistance:
VR = IR * R
iC = C * d(vC) / dt
d(vC) / dt = (7.60) * (120) * cos[(120) t]
iC = C * (7.60) * (120) * cos[(120) t]
VR = iC * R
= C * (7.60) * (120) * cos[(120) t] * 250
= 91200C * cos[(120rad/s) t]Ω
Therefore, the expression for the voltage VR across the resistor is VR = 91200C * cos[(120rad/s) t].
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The transition for the cadmium 228.8 nm line is a 1S0 → 1S1 transition, a) calculate the ratio of N*/N0 in an air-acetylene flame (2500 K), given that the degeneracy of the ground state is 1 and the degeneracy of the excited state is 3 and that the excited state of the cadmium atom lies 8.68 x 10-19 J/atom above the ground state; b) what percent of the atoms is in the excited state? c) If an argon plasma (10,000K) is used instead of the air-acetylene flame, what percent of atoms will be in the excited state?
The required,
a) [tex]N'/N_0[/tex] ≈ 0.408 (40.8%)
b) Approximately 40.8% of the atoms are in the excited state.
c) [tex]N'/N_0[/tex] ≈ 0.066 (6.6%)
To calculate the ratio of N'/N_0 in an air-acetylene flame, we can use the Boltzmann distribution equation:
[tex]N'/N_0 = (g'/g_0) * exp^{(-\triangle E/kT)}[/tex]
a) Calculate the ratio of [tex]N'/N_0[/tex] in an air-acetylene flame (2500 K):
Given:
[tex]g_0 = 1[/tex] (degeneracy of the ground state)
[tex]g' = 3[/tex] (degeneracy of the excited state)
[tex]\triangle E = 8.68 * 10^{(-19)}[/tex]J/atom (energy difference between the excited and ground states)
T = 2500 K (temperature)
[tex]N'/N_0 = (3/1) * e{(-8.68 * 10^{19} / (1.38 * 10^{-23} * 2500 ))[/tex]
Calculating the exponential term:
exp(-8.68 x 10⁻¹⁹ J/atom / (1.38 x 10⁻²³ J/K * 2500 K)) ≈ 0.136
Therefore, the ratio of [tex]N*/N_0[/tex] in an air-acetylene flame is:
[tex]N'/N_0[/tex] ≈ (3/1) * 0.136 ≈ 0.408
b) To determine the percent of atoms in the excited state, we can multiply the ratio [tex]N'/N_0[/tex] by 100:
Percent in excited state = [tex]N'/N_0 * 100[/tex]
Percent in excited state ≈ 0.408 * 100 ≈ 40.8%
Therefore, 40.8% of the atoms will be in the excited state.
Similarly,
c) If an argon plasma (10,000 K) is used instead of the air-acetylene flame, we can repeat the calculations using the new temperature:
The ratio of [tex]N*/N_0[/tex] in an argon plasma is:
N'/N0 ≈ (3/1) * 0.022 ≈ 0.066
To determine the percent of atoms in the excited state:
Percent in excited state = [tex]N'/N_0 * 100[/tex]
Percent in excited state ≈ 0.066 * 100 ≈ 6.6%
Therefore, 6.6% of the atoms will be in the excited state in an argon plasma.
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