The maximum bending moment developed in the beam is 3750000 N-mm. The overall stress is 4.84 MPa.
The maximum bending moment developed in a beam is equal to the force applied to the beam multiplied by the distance from the point of application of the force to the nearest support.
In this case, the force is 5000 N and the distance from the point of application of the force to the nearest support is 1500 mm. Therefore, the maximum bending moment is:
Mmax = PL/4 = 5000 N * 1500 mm / 4 = 3750000 N-mm
The overall stress is equal to the maximum bending moment divided by the moment of inertia of the beam cross-section. The moment of inertia of the beam cross-section is calculated using the following formula:
I = b * h^3 / 12
where:
b is the width of the beam in mm
h is the height of the beam in mm
In this case, the width of the beam is 127 mm and the height of the beam is 254 mm. Therefore, the moment of inertia is:
I = 127 mm * 254 mm^3 / 12 = 4562517 mm^4
Plugging in the known values, we get the following overall stress:
f = Mmax (h/2) / I = 3750000 N-mm * (254 mm / 2) / 4562517 mm^4 = 4.84 MPa
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1.) There is a seesaw with a pivot at the center of the seesaw. If the Tom weights 100 kg and sits on one end of the 5 meters on one end of the pivot, how far (from Tom) does Sarah have to sit on the other end of the pivot if she weights 150 kg to keep the seesaw at static equilibrium? (Assume that mass of the seesaw and the mass of the pivot are negligible.)
Sarah needs to sit 7.5 meters from Tom to keep the seesaw at static equilibrium.
For the seesaw to be in static equilibrium, the torques on each side of the pivot must be equal. The torque is calculated by multiplying the force by the distance from the pivot.
Tom's weight is 100 kg and he is sitting 5 meters from the pivot. This means that his torque is 500 N * 5 m = 2500 N m.
Sarah's weight is 150 kg and she needs to sit at a distance such that her torque is equal to Tom's torque. This means that she needs to sit 7.5 meters from the pivot.
Here is the calculation for the distance Sarah needs to sit:
d = 2500 N m / 150 kg = 16.67 m
This is slightly more than 7.5 meters because Sarah's weight is greater than Tom's weight.
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23. A crane raises a 90 kg piano from the ground to a balcony
that is 12 m above the ground (the piano starts and ends
motionless). How much work was done by the crane on the piano?
(a) 10,600 J (b) 1
The work done by the crane on the piano is approximately 10,584 J.
To calculate the work done by the crane on the piano, we need to determine the change in gravitational potential energy of the piano as it is raised from the ground to the balcony.
The gravitational potential energy (PE) is given by the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity, and h is the change in height.
Given:
m = 90 kg
g = 9.8 m/s^2 (approximate value)
h = 12 m
Substituting these values into the formula:
PE = (90 kg) * (9.8 m/s^2) * (12 m)
PE = 10,584 J (rounded to the nearest whole number)
Therefore, the work done by the crane on the piano is approximately 10,584 J.
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I need explanation of both questions
What is the effect of increasing the tension in the vibrating string to the frequency if linear mass density & vibrating length are held constant?
What is the effect of increasing the linear mass density of the vibrating string to the frequency if tension & vibrating length are held constant?
Increasing the tension in a vibrating string while keeping the linear mass density and vibrating length constant has the effect of increasing the frequency of the string's vibrations.
On the other hand, increasing the linear mass density of the vibrating string while keeping the tension and vibrating length constant has the effect of decreasing the frequency of the string's vibrations.The frequency of vibration in a string is determined by several factors, including the tension in the string, the linear mass density (mass per unit length) of the string, and the vibrating length of the string.
When the tension in the string is increased while the linear mass density and vibrating length are held constant, the frequency of vibration increases. This is because the increased tension results in a higher restoring force acting on the string, causing it to vibrate at a higher frequency.On the other hand, when the linear mass density of the string is increased while the tension and vibrating length are held constant, the frequency of vibration decreases. This is because the increased linear mass density increases the inertia of the string, making it more resistant to motion and reducing the frequency at which it vibrates.
Increasing the tension in a vibrating string increases the frequency of vibration, while increasing the linear mass density decreases the frequency of vibration, assuming the vibrating length and other factors remain constant.
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If the amplitude of a sound wave is increased, there is an increase in the:
1: loudness of the sound
2: pitch of the sound
3: velocity of the wave
4: energy of the wave
5: wavelength of the wave
The phrase(s) that make the statement true are _____ and _____. Put the numbers of the phrases in any order.
5. The third harmonic in an open tube is a wave that is 1.5 wavelengths long.
True or False
The correct options are 1 and 4. If the amplitude of a sound wave is increased, there is an increase in the loudness of the sound, the energy of the wave.
The loudness of sound is the degree of sound volume.
Amplitude determines the amount of energy produced by sound. Hence, increasing the amplitude of a sound wave increases the loudness of the sound.
The energy of a wave is determined by the amplitude of the wave.
Therefore, when the amplitude of a wave is increased, the energy of the wave is also increased.
Hence, increasing the amplitude of a sound wave increases the energy of the wave.
The third harmonic in an open tube is a wave that is 3/2 or 1.5 wavelengths long.
Hence, the given statement is True.
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Two blocks with mass M1 and M2 are sitting on a frictionless horizontal floor. They are
connected by means of a rope with mass M. You can neglect any sagging of the rope, and treat
it as perfectly taut and horizontally. A horizontal pulling force with magnitude P is exerted on
block M1. Calculate the tension in the front of the rope and in the back of the rope. ALSO state
what these tensions would become when the mass of the rope would be negligible.
The tension in the front of the rope is P + M2g, and the tension in the back of the rope is P + M2g.
In summary, when a horizontal pulling force P is exerted on block M1, the tension in the front and back of the rope can be calculated. The tension in the front of the rope is equal to the applied force P plus the weight of block M2 (M2g), while the tension in the back of the rope is also equal to P plus M2g.
To explain further, when the pulling force P is applied to block M1, an equal and opposite force is transmitted through the rope to block M2. The tension in the rope is the force experienced by both blocks.
In the front of the rope, the tension is equal to the pulling force P plus the weight of block M2, which is M2g. Similarly, in the back of the rope, the tension is also equal to P plus M2g.
When the mass of the rope is negligible, the tensions in the front and back of the rope would simply become equal to the applied force P. In this case, the weight of the rope would no longer contribute to the tensions since it is negligible compared to the masses of the blocks.
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A large, open-topped water tank is being filled from above by a 1.0-cm-diameter hose. The water in the hose has a uniform speed of 13 cm/s. Meanwhile, the tank springs a leak at the bottom. The hole has a diameter of 0.70 cm. Determine the equilibrium level heq of the water in the tank, measured relative to the bottom, if water continues flowing into the tank at the same rate.
The equilibrium level (heq) of the water in the tank, measured relative to the bottom, is approximately 1.68 cm.
1. Calculate the cross-sectional area of the hose:
A_in = π × (0.5 cm)^2
= 0.785 cm^2
2. Calculate the cross-sectional area of the leak:
A_out = π × (0.35 cm)^2
= 0.385 cm^2
3. Calculate the velocity of the water leaving the tank:
v_out = (A_in × v_in) / A_out
= (0.785 cm^2 × 13 cm/s) / 0.385 cm^2
≈ 26.24 cm/s
4. Calculate the equilibrium level of the water in the tank:
heq = (Q_in / A_out) / v_out
= (A_in × v_in) / (A_out × v_out)
= (0.785 cm^2 × 13 cm/s) / (0.385 cm^2 × 26.24 cm/s)
≈ 1.68 cm
Therefore, the equilibrium level (heq) of the water in the tank, measured relative to the bottom, is approximately 1.68 cm.
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1. a heavy object is lifted from the ground at a constant speed of 1.2 m/s for 2.5s and then it is dropped. At what speed does the heavy object hit the ground?
2. A 1.00x10^3 kg object is raised vertically at a constant velocity of 4.00 m/s by a crane. What is the power output of the crane is the object was raised 8.0 m from the ground?
1. The heavy object hits the ground with a speed of approximately 24 m/s.
2. The power output of the crane is 3.2 × 10⁴ W.
1. To determine the speed at which the heavy object hits the ground, we need to consider the two phases of its motion: lifting and dropping.
- Lifting phase: The object is lifted at a constant speed of 1.2 m/s for 2.5 seconds. During this phase, the object's velocity remains constant, so there is no change in speed.
- Dropping phase: After being dropped, the object falls freely under the influence of gravity. Assuming no air resistance, the object's speed increases due to the acceleration of gravity, which is approximately 9.8 m/s².
To find the speed when the object hits the ground, we can use the equation for free fall:
v = u + gt
where v is the final velocity, u is the initial velocity (0 m/s in this case since the object is dropped), g is the acceleration due to gravity, and t is the time of falling.
Using the equation, we have:
v = 0 + (9.8 m/s²)(2.5 s) ≈ 24 m/s
Therefore, the heavy object hits the ground with a speed of approximately 24 m/s.
2. The power output of the crane can be calculated using the formula:
Power = Force × Velocity
In this case, the force is the weight of the object, which is given by:
Force = mass × acceleration due to gravity
Force = (1.00 × 10³ kg) × (9.8 m/s²) = 9.8 × 10³ N
The velocity is the constant velocity at which the object is raised, which is 4.00 m/s.
Using the formula for power, we have:
Power = (9.8 × 10³ N) × (4.00 m/s) = 3.92 × 10⁴ W
Therefore, the power output of the crane is 3.2 × 10⁴ W.
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10. An electron (charge -1.60 x 10-19 C and mass 9.11 x 10-31 kg) is initially at rest at the center of a uniformly charged ring of radius 3 mm and linear charge density ^ = -4.07 nC/m. The electron is nudged slightly and starts accelerating away from the ring along its central axis. Which of the following is closest to the electron's speed by the time it is very far from the ring? a) 5.0 x106 m/s d) 8.0 x 106 m/s c) 7.0 x 106 m/s b) 6.0 x 106 m/s e) 9.0 x 106 m/s
The electron is nudged slightly and starts accelerating away from the ring along its central axis. the electron's speed when it is very far from the ring is 0 m/s. None of the given options (a, b, c, d, or e) are closest to the correct answer.
To find the speed of the electron when it is very far from the ring, we can use the principle of conservation of energy.
The initial energy of the electron is entirely in the form of electric potential energy due to the interaction with the charged ring. As the electron moves away from the ring, this potential energy is converted into kinetic energy.
The electric potential energy between the electron and the ring is given by:
U = - (k * q * Q) / r,
where U is the electric potential energy, k is the Coulomb's constant (9 x 10^9 N·m^2/C^2), q is the charge of the electron (-1.60 x 10^-19 C), Q is the linear charge density of the ring (-4.07 x 10^-9 C/m), and r is the distance between the electron and the center of the ring.
The initial potential energy of the electron is:
U_initial = - (k * q * Q * r_initial) / r_initial,
where r_initial is the initial distance between the electron and the center of the ring. Since the electron is initially at the center of the ring, r_initial = 0.
The final kinetic energy of the electron when it is very far from the ring is:
K_final = (1/2) * m * v_final^2,
where K_final is the final kinetic energy, m is the mass of the electron (9.11 x 10^-31 kg), and v_final is the final velocity of the electron.
According to the conservation of energy, the initial potential energy is equal to the final kinetic energy:
U_initial = K_final.
Solving for v_final, we get:
v_final = sqrt((2 * U_initial) / m).
Substituting the values, we have:
v_final = sqrt((2 * (-(k * q * Q * r_initial) / r_initial)) / m).
Calculating the numerical value:
v_final = sqrt((2 * (-(9 x 10^9 N·m^2/C^2) * (-1.60 x 10^-19 C) * (-4.07 x 10^-9 C/m) * 0) / (9.11 x 10^-31 kg)).
v_final = sqrt(0) = 0 m/s.
Therefore, the electron's speed when it is very far from the ring is 0 m/s. None of the given options (a, b, c, d, or e) are closest to the correct answer.
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If this wave is traveling along the x-axis from left to right
with a displacement amplitude of 0.1 m in the y direction, find the
wave equation for y as a function of x and time t.
The wave equation for the displacement y as a function of x and time t can be expressed as y(x, t) = A sin(kx - ωt),
where A represents the displacement amplitude, k is the wave number, x is the position along the x-axis, ω is the angular frequency, and t is the time.
To derive the wave equation, we start with the general form of a sinusoidal wave, which is given by y(x, t) = A sin(kx - ωt). In this equation, A represents the displacement amplitude, which is given as 0.1 m in the y direction.
The wave equation describes the behavior of the wave as it propagates along the x-axis from left to right. The term kx represents the spatial variation of the wave, where k is the wave number that depends on the wavelength, and x is the position along the x-axis. The term ωt represents the temporal variation of the wave, where ω is the angular frequency that depends on the frequency of the wave, and t is the time.
By combining the spatial and temporal variations in the wave equation, we obtain y(x, t) = A sin(kx - ωt), which represents the displacement of the wave as a function of position and time.
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a)
Calculate the density of the moon by assuming it to be a sphere of diameter 3475 km and having a mass of 7.35 × 10^22 kg. Express your answer in g/cm3.
)
A car accelerates from zero to a speed of 36 km/h in 15 s.
i.
Calculate the acceleration of the car in m/s2.
ii.
If the acceleration is assumed to be constant, how far will the car travel in 1 minute ?
iii.
Calculate the speed of the car after 1 minute.
The density of the moon is determined to be 3.35 g/cm³ based on its mass and volume. In the case of the car, it experiences an acceleration of 2/3 m/s², enabling it to travel a distance of 4000 m in 1 minute and achieve a speed of 200/3 m/s.
a) Density of the moon: Density is the measure of mass per unit volume of a substance. It is denoted by p. It is given as:
[tex]\[Density=\frac{Mass}{Volume}\][/tex]
Given that the diameter of the moon is 3475 km and the mass of the moon is 7.35 × 10²² kg, we need to find the density of the moon. We know that the volume of a sphere is given as:
[tex]\[V=\frac{4}{3}πr^{3}\][/tex]
Here, the diameter of the sphere is 3475 km. Therefore, the radius of the sphere will be half of it, i.e.:
[tex]\[r=\frac{3475}{2}\ km=1737.5\ km\][/tex]
Substituting the given values in the formula to get the volume, we get:
[tex]\[V=\frac{4}{3}π(1737.5)^{3}\ km^{3}\][/tex]
Converting km to cm, we get:
[tex]\[1\ km=10^{5}\ cm\]\[\Rightarrow 1\ km^{3}=(10^{5})^{3}\ cm^{3}=10^{15}\ cm^{3}\][/tex]
Therefore,[tex]\[V=\frac{4}{3}π(1737.5×10^{5})^{3}\ cm^{3}\][/tex]
Now we can find the density of the moon:
[tex]\[Density=\frac{Mass}{Volume}\]\[Density=\frac{7.35×10^{22}}{\frac{4}{3}π(1737.5×10^{5})^{3}}\ g/{cm^{3}}\][/tex]
Simplifying, we get the density of the moon as:
[tex]\[Density=3.35\ g/{cm^{3}}\][/tex]
b) Acceleration of the car
i. The initial velocity of the car is zero. The final velocity of the car is 36 km/h or 10 m/s. The time taken by the car to reach that velocity is 15 s. We can use the formula of acceleration:
[tex]\[Acceleration=\frac{Change\ in\ Velocity}{Time\ Taken}\]\[Acceleration=\frac{10-0}{15}\ m/s^{2}\][/tex]
Simplifying, we get the acceleration of the car as:
[tex]\[Acceleration=\frac{2}{3}\ m/s^{2}\][/tex]
ii. If we assume that the acceleration of the car is constant, we can use the formula of distance traveled by a uniformly accelerated body:
[tex]\[Distance\ travelled=\frac{Initial\ Velocity×Time\ Taken+\frac{1}{2}Acceleration\times(Time\ Taken)^{2}}{2}\][/tex]
Here, the initial velocity of the car is zero, the acceleration of the car is 2/3 m/s² and the time taken by the car to travel a distance of 1 minute is 60 s.
Substituting these values, we get:
[tex]\[Distance\ travelled=\frac{0\times 60+\frac{1}{2}\times \frac{2}{3}\times (60)^{2}}{2}\ m\]\[Distance\ travelled=\frac{12000}{3}=4000\ m\][/tex]
Therefore, the car will travel a distance of 4000 m in 1 minute.
iii. If we assume that the acceleration of the car is constant, we can use the formula of distance traveled by a uniformly accelerated body
[tex]:\[Distance\ travelled=\frac{Initial\ Velocity×Time\ Taken+\frac{1}{2}Acceleration\times(Time\ Taken)^{2}}{2}\][/tex]
Here, the initial velocity of the car is zero, the acceleration of the car is 2/3 m/s² and the time taken by the car to travel a distance of 1 minute is 60 s. We need to find the speed of the car after 1 minute. We know that:
[tex]\[Speed=\frac{Distance\ travelled}{Time\ Taken}\][/tex]
Substituting the values of the distance traveled and time taken, we get:
[tex]\[Speed=\frac{4000}{60}\ m/s\][/tex]
Simplifying, we get the speed of the car after 1 minute as: [tex]\[Speed=\frac{200}{3}\ m/s\][/tex]
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Determine the resonant frequency of the circuit shown.
A.500 Hz
B.159 Hz
C.32 Hz
D.235 Hz
E.112.5 Hz
The resonant frequency of a circuit can be determined using the formula f = 1 / (2π√(LC)), where f is the resonant frequency, L is the inductance, and C is the capacitance of the circuit. Given the values of L and C for the circuit shown, we can calculate the resonant frequency.
To calculate the resonant frequency of the circuit, we need to determine the values of L and C. The resonant frequency can be obtained using the formula f = 1 / (2π√(LC)), where f is the resonant frequency, L is the inductance, and C is the capacitance of the circuit.
Since the specific values of L and C for the given circuit are not provided in the question, it is not possible to calculate the resonant frequency.
Therefore, none of the options A, B, C, D, or E can be selected as the correct answer.
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Two positive charges \( \mathrm{Q} 1 \) and \( \mathrm{Q} 2 \) are separated by a distance \( r \). The charges repel each other with a force \( F \). If the magnitude of each charge is doubled and th
If the magnitude of each charge is doubled and the distance between them is halved, the new force between them will be four times the original force.
Let's denote the original charges as Q1 and Q2, and the original force as F. The electric force between two charges is given by Coulomb's law:
F = k * (Q1 * Q2) / r^2, where k is the Coulomb's constant and r is the distance between the charges.
If the magnitude of each charge is doubled (2Q1 and 2Q2) and the distance between them is halved (r/2), the new force (F') can be calculated as:
F' = k * (2Q1 * 2Q2) / (r/2)^2.
Simplifying the equation:
F' = k * (4Q1 * 4Q2) / (r/2)^2,
F' = k * (16Q1 * Q2) / (r^2/4),
F' = k * (16Q1 * Q2) * (4/r^2),
F' = 64 * k * (Q1 * Q2) / r^2.
Therefore, the new force between the charges is four times the original force: F' = 4F.
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consider the right-circular cylinder of diameter d, length l, and the areas a1, a2, and a 3 representing the base, inner, and top surfaces, respectively. calculate the net radiation heat transfer, in watt, from a1 to a3 if f12 = 0.36 (a fraction of radiation heat transfer from surface 1 to surface 2), A_1 = 0.05 m^2, T_1 = 1000 K, and T_3 = 500 K.
The net radiation heat transfer from surface 1 to surface 3 is 64.8 W.
How can we calculate the net radiation heat transfer between the surfaces of a right-circular cylinder?The net radiation heat transfer between two surfaces can be calculated using the formula:
Q_net = f12 * σ * (A_1 * T_1^4 - A_2 * T_2^4)
Here, Q_net represents the net radiation heat transfer, f12 is the fraction of radiation heat transfer from surface 1 to surface 2, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2·K^4)), A_1 and A_2 are the areas of the respective surfaces, and T_1 and T_2 are the temperatures in Kelvin.
In this case, the areas are given as A_1 = 0.05 m^2, A_2 = 0.05 m^2, and A_3 = 0.05 m^2 (assuming the base, inner, and top surfaces have the same area). The temperatures are T_1 = 1000 K and T_3 = 500 K.
Substituting the given values into the formula, we have:
Q_net = 0.36 * 5.67 x 10^-8 * (0.05 * 1000^4 - 0.05 * 500^4)
≈ 64.8 W
Therefore, the net radiation heat transfer from surface 1 to surface 3 is approximately 64.8 W.
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A charge of -3.20 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y = 3.85 cm If a third charge, of 5.00 nC, is now placed at the point i = 2.95 cm, y = 3.85 cm find the r and y components of the total force exerted on this charge by the other two charges. Express answers numerically separated by a comma.
The x -component of the resultant force [tex]$F_R^x=77.88 \times 10^{-6} \mathrm{~N}$[/tex]
And y- component of the resultant force [tex]$F_R^y=-38.67 \times 10^{-6} N$[/tex]
The electric force on charge q₂ due to charge q₁ is given by as follows:
[tex]\vec{F}=\frac{1}{4 \pi \epsilon_o} \frac{q_1 q_2}{\left|\vec{r}_2-\vec{r}_1\right|^3}\left(\vec{r}_2-\vec{r}_1\right) \\\vec{F}=\left(9 \times 10^9 N m^2 / C^2\right) \times \frac{q_1 q_2}{\left|\vec{r}_2-\vec{r}_1\right|^3}\left(\vec{r}_2-\vec{r}_1\right)[/tex] ......(i)
Where;
r₁ and r₂ are position vectors of charges respectively.
ε₀ is vacuum permittivity.
In our case, we have to find a net force on a third charge due to two other charges.
First, we will determine the force on 5.00 nC due to -3.20 nC.
We have the following information
Charge q₁ = 3.20 nC
= 3.20 × 10⁻⁹ C
Charge q₃ = 5.00 nC
= 5 × 10⁻⁹ C
Position of charge q₁ is the origin = [tex]\vec{r}_1=0 \hat{i}+0 \hat{j}[/tex]
Position of charge q₃ = [tex]\quad \vec{r}_3=(x=2.90 \mathrm{~cm}, y=3.85 \mathrm{~cm})=0.029 \mathrm{~m} \hat{i}+0.0385 \mathrm{~m} \hat{j}$[/tex]
Then,
[tex]$\vec{r}_3-\vec{r}_1=(0.029 m \hat{i}+0.0385 m \hat{j})-(0 \hat{i}+0 \hat{j})=0.029 m \hat{i}+0.0385 m \hat{j}$$[/tex]
And,
[tex]$$\left|\vec{r}_3-\vec{r}_1\right|=|0.029 m \hat{i}+0.0385 m \hat{j}|=0.0482 m$$[/tex]
Plugging in these values in equation (i), we get the following;
[tex]\vec{F}_{13}=\left(9 \times 10^9 \mathrm{Nm}^2 / C^2\right) \times \frac{\left(-3.20 \times 10^{-9} C\right) \times\left(5.00 \times 10^{-9} C\right)}{(0.0482 m)^3} \times(0.029 m \hat{i}+0.0385 m \hat{j}) \\\vec{F}_{13}=-29.13 \times 10^{-6} N \hat{i}-38.67$$[/tex]
Similarly ;
We will determine the force on the third charge due to the charge of 2.00 nC.
We have the following information;
Charge q₂ = 2.00 nC
= 2 × 10⁻⁹ C
Charge q₃ = 5.00 nC
= 5 × 10⁻⁹ C
Position of charge q₂ is y = 3.85 cm
[tex]\vec{r}_2=0.0385 \mathrm{~m} \hat{j}$[/tex]
Position of charge q₃ [tex]\vec{r}_3=(x=2.90 \mathrm{~cm}, y=3.85 \mathrm{~cm})=0.029 \mathrm{~m} \hat{i}+0.0385 \mathrm{~m} \hat{j}$[/tex]
Then,
[tex]$\vec{r}_3-\vec{r}_2=(0.029 m \hat{i}+0.0385 m \hat{j})-(0.0385 m \hat{j})=0.029 m \hat{i}$$[/tex]
And
[tex]$$\left|\vec{r}_3-\vec{r}_2\right|=|0.029 m \hat{i}|=0.029 m$$[/tex]
Plugging in these values in equation (i), we get following:
[tex]$\vec{F}_{23}=\left(9 \times 10^9 \mathrm{Nm}^2 / C^2\right) \times \frac{\left(2.00 \times 10^{-9} C\right) \times\left(5.00 \times 10^{-9} C\right)}{(0.029 m)^3} \times(0.029 m \hat{i}) \\\\[/tex][tex]\vec{F}_{23}=107.01 \times 10^{-6} N \hat{i}$$[/tex]
Net Force :
[tex]$\vec{F}_R=\vec{F}_{13}+\vec{F}_{23}[/tex]
[tex]\vec{F}_R=\left(-29.13 \times 10^{-6} N \hat{i}-38.67 \times 10^{-6} N \hat{j}\right)+\left(107.01 \times 10^{-6} N \hat{i}\right)[/tex]
[tex]\vec{F}_R=77.88 \times 10^{-6} N \hat{i}-38.67 \times 10^{-6} 1$$[/tex]
Thus, the x -component of the resultant force [tex]$F_R^x=77.88 \times 10^{-6} \mathrm{~N}$[/tex]
And y- component of the resultant force [tex]$F_R^y=-38.67 \times 10^{-6} N$[/tex]
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A motor driven pump transfers 5000 litres of oil per hour through an elevation of 16 m. if the specific gravity of the oil is 0.8, what is the input power to the pump?
The input power to the pump is approximately 174.72 watts.
To calculate the input power to the pump, we can use the following formula:
Power = (Flow rate) x (Head) x (Density) x (Gravity)
Given:
Flow rate = 5000 liters/hourElevation (Head) = 16 mSpecific gravity (Density relative to water) = 0.8Gravity = 9.8 m/s^2 (acceleration due to gravity)First, we need to convert the flow rate from liters/hour to cubic meters/second since the SI unit is used for power (watts).
Flow rate = 5000 liters/hour
= (5000/1000) cubic meters/hour
= (5000/1000) / 3600 cubic meters/second
≈ 0.0014 cubic meters/second
Now, we can calculate the input power:
Power = (0.0014 cubic meters/second) x (16 m) x (0.8) x (9.8 m/s^2)
≈ 0.17472 kilowatts
≈ 174.72 watts
Therefore, the input power to the pump is approximately 174.72 watts.
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calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark. (treat the stick as a thin rod.) (a) 1.1 kgm2 (b) 3.2 kgm2 (c) 4.2 kgm2 (d) 0.097 kgm2
Rounding to two decimal places, the rotational inertia of the meter stick is approximately 0.097 kgm^2. Therefore, the correct answer is (d) 0.097 kgm^2.
To calculate the rotational inertia of the meter stick, we need to use the formula for the rotational inertia of a thin rod. The formula is given by I = (1/3) * m * L^2, where I is the rotational inertia, m is the mass of the rod, and L is the length of the rod.
In this case, the mass of the meter stick is given as 0.56 kg, and the length of the stick is 1 meter. Since the axis of rotation is perpendicular to the stick and located at the 20 cm mark, we need to consider the rotational inertia of two parts: one part from the 0 cm mark to the 20 cm mark, and another part from the 20 cm mark to the 100 cm mark.
For the first part, the length is 0.2 meters and the mass is 0.2 * 0.56 = 0.112 kg. Plugging these values into the formula, we get:
I1 = (1/3) * 0.112 * (0.2)^2 = 0.00149 kgm^2.
For the second part, the length is 0.8 meters and the mass is 0.8 * 0.56 = 0.448 kg. Plugging these values into the formula, we get:
I2 = (1/3) * 0.448 * (0.8)^2 = 0.09504 kgm^2.
Finally, we add the rotational inertias of both parts to get the total rotational inertia:
I_total = I1 + I2 = 0.00149 + 0.09504 = 0.09653 kgm^2.
Rounding to two decimal places, the rotational inertia of the meter stick is approximately 0.097 kgm^2. Therefore, the correct answer is (d) 0.097 kgm^2.
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Inside a uniform 0.2-T magnetic field, a beam of charged particles moves in a circle with radius 0.3 m. How fast must the particles be moving? (in m/s) (charge-to-mass ratio of each particle =158 ) Your Answer: Answer
The particles must be moving at approximately 9.48 m/s to maintain a circular path inside the 0.2-T magnetic field.
Explanation:
To find the speed of the charged particles moving in a circle inside a magnetic field, we can use the equation for the centripetal force and the equation for the magnetic force.
The centripetal force required to keep an object moving in a circle is given by:
F_c = (m * v^2) / r,
where F_c is the centripetal force, m is the mass of the particle, v is the velocity of the particle, and r is the radius of the circle.
The magnetic force experienced by a charged particle moving in a magnetic field is given by:
F_m = q * v * B,
where F_m is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
Since the charged particle moves in a circle, the centripetal force is provided by the magnetic force:
F_c = F_m.
Equating the two forces, we have:
(m * v^2) / r = q * v * B.
Rearranging the equation, we can solve for the velocity v:
v = (q * B * r) / m.
Given:
B = 0.2 T (magnetic field strength)
r = 0.3 m (radius of the circle)
q/m = 158 (charge-to-mass ratio of each particle)
Substituting the given values into the equation, we get:
v = (158 * 0.2 * 0.3) / 1.
Calculating the result:
v = 9.48 m/s.
Therefore, the particles must be moving at approximately 9.48 m/s to maintain a circular path inside the 0.2-T magnetic field.
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In the figure, the rod moves to the right with a speed of 1.8 m/s and has a resistance of 2.6 N .(Figure 1) The rail separation is l = 27.0 cm . The magnetic field is 0.33 T, and the resistance of the U-shaped conductor is 25.5 12 at a given instant. Figure 1 of 1 dA B (outward) v dt Part A Calculate the induced emf. Express your answer to two significant figures and include the appropriate units. I MÅ ? moving along rails in a uniform magnetic field Units Submit Request Answer Part B Calculate the current in the U-shaped conductor. Express your answer to two significant figures and include the appropriate units. 01 MÅ ? I = Value Units Part C Calculate the external force needed to keep the rod's velocity constant at that instant. Express your answer to two significant figures and include the appropriate units. MÅ 0! ? F = Value Units Submit Request Answer
According to the given information, the external force needed to keep the rod's velocity constant is 0.0005 N.
According to the given information.Given:
Speed of rod, v = 1.8 m/s
Resistance, R = 2.6 N
Distance between the rails, l = 27.0 cm = 0.27 m
Magnetic field, B = 0.33 T
Resistance of the U-shaped conductor, R' = 25.5 ΩPart A
The induced emf can be calculated by using the formula given below: emf = Bvl
where, B = Magnetic field
v = Velocity of ro
dl = Distance between the rails
Substituting the given values, emf = (0.33 T)(1.8 m/s)(0.27 m)
emf = 0.16146 V ≈ 0.16 V
Therefore, the induced emf is 0.16 V.
Part BThe current in the U-shaped conductor can be calculated by using the formula given below: I = emf/R'
where, emf = Induced emf
R' = Resistance of the U-shaped conductor
Substituting the given values, I = (0.16 V)/(25.5 Ω)I = 0.00627 A ≈ 0.006 A
Therefore, the current in the U-shaped conductor is 0.006
A.
Part CThe external force needed to keep the rod's velocity constant can be calculated by using the formula given below: F = BIl where, B = Magnetic field
I = Current
l = Length of the conductor
Substituting the given values,
F = (0.33 T)(0.006 A)(0.27 m)F = 0.0005346 N ≈ 0.0005 N
Therefore, the external force needed to keep the rod's velocity constant is 0.0005 N.
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The voltage and Power ratings of a Particular light bulb, which are It's normal operating values are lov and 60w. Assume the resistance of the of with ating Conditions. If the light bulb is operated with a Current that is 50% of the current. rating Idrawn by the bulb? of the bulb, what is the actual Power
The actual power consumed is 30W when the light bulb is worked with a current that is 50% of its current rating using Ohm's Law.
Normal operating value = 60W
Bulb operation = 50% of current.
The relation between voltage, current, and resistance is given by Ohm's Law.
V = I * R.
R = V / I
The formula used for calculating the power rating in normal operating conditions is:
P_0 = V_0 * I_0
The actual current drawn by the bulb I_actual is:
V_0 = I_actual * R
R = V_0 / I_actual
P_actual = V_0 * I_actual
Substituting the values we get:
P_actual = V_0 * I_actual = V_0 * (0.5 * I_0)
60W = V_0 * I_0
V_0 = 60W / I_0
P_actual = (60W / I_0) * (0.5 * I_0) = 30W
Therefore, we can conclude that the actual power consumed is 30W.
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A 2 M resistor is connected in series with a 2.5 µF capacitor and a 6 V battery of negligible internal resistance. The capacitor is initially uncharged. After a time t = ↑ = RC, find each of the following. (a) the charge on the capacitor 9.48 HC (b) the rate at which the charge is increasing 1.90 X HC/s (c) the current HC/S (d) the power supplied by the battery μW (e) the power dissipated in the resistor μW (f) the rate at which the energy stored in the capacitor is increasing. μW
The rate at which the energy stored in the capacitor is increasing. = μW
We know that;
Charging of a capacitor is given as:q = Q(1 - e- t/RC)
Where, q = charge on capacitor at time t
Q = Final charge on the capacitor
R = Resistance
C = Capacitance
t = time after which the capacitor is charged
On solving this formula, we get;
Q = C X VC X V = Q/C = 6 V / 2.5 µF = 2.4 X 10-6 C
Other data in the question is:
R = 2 MΩC = 2.5 µFV = 6 V(
The charge on the capacitor:
q = Q(1 - e- t/RC)q = 2.4 X 10-6 C (1 - e- 1)q = 9.48 X 10-6 C
The rate at which the charge is increasing:
When t = RC; q = Q(1 - e- 1) = 0.632QdQ/dt = I = V/RI = 6/2 X 106 = 3 X 10-6 Adq/dt = d/dt(Q(1 - e-t/RC))= I (1 - e-t/RC) + Q (1 - e-t/RC) (-1/RC) (d/dt)(t/RC)q = Q(1 - e- t/RC)dq/dt = I (1 - e- t/RC)dq/dt = (3 X 10-6 A)(1 - e- 1) = 1.9 X 10-6 A
the current: Current flowing through the circuit is given by; I = V/R = 6/2 X 106 = 3 X 10-6 A
the power supplied by the battery: Power supplied by the battery can be given as:
P = VI = (6 V)(3 X 10-6 A) = 18 X 10-6 μW
the power dissipated in the resistor: The power dissipated in the resistor can be given as; P = I2 R = (3 X 10-6 A)2 (2 X 106 Ω) = 18 X 10-6 μW
the rate at which the energy stored in the capacitor is increasing: The rate at which the energy stored in the capacitor is increasing is given as;dW/dt = dq/dt X VdW/dt = (1.9 X 10-6 A)(6 V) = 11.4 X 10-6 μW
Given in the question that, a 2 M resistor is connected in series with a 2.5 µF capacitor and a 6 V battery of negligible internal resistance. The capacitor is initially uncharged. We are to find various values based on this. Charging of a capacitor is given as;q = Q(1 - e-t/RC)Where, q = charge on capacitor at time t
Q = Final charge on the capacitor
R = Resistance
C = Capacitance
t = time after which the capacitor is charged
We have;R = 2 MΩC = 2.5 µFV = 6 VTo find Q, we have;Q = C X VQ = 2.4 X 10-6 C
Other values that we need to find are
The charge on the capacitor:q = 2.4 X 10-6 C (1 - e- 1)q = 9.48 X 10-6 C
The rate at which the charge is increasing:dq/dt = I (1 - e- t/RC)dq/dt = (3 X 10-6 A)(1 - e- 1) = 1.9 X 10-6 A
The current: Current flowing through the circuit is given by; I = V/R = 6/2 X 106 = 3 X 10-6 A
The power supplied by the battery: Power supplied by the battery can be given as:
P = VI = (6 V)(3 X 10-6 A) = 18 X 10-6 μW
The power dissipated in the resistor: Power dissipated in the resistor can be given as; P = I2 R = (3 X 10-6 A)2 (2 X 106 Ω) = 18 X 10-6 μW
The rate at which the energy stored in the capacitor is increasing: The rate at which the energy stored in the capacitor is increasing is given as;dW/dt = dq/dt X VdW/dt = (1.9 X 10-6 A)(6 V) = 11.4 X 10-6 μW
On calculating and putting the values in the formulas of various given entities, the values that are calculated are
The charge on the capacitor = 9.48 HC
The rate at which the charge is increasing = 1.90 X HC/s
The current = HC/S
The power supplied by the battery = μW
The power dissipated in the resistor = μW
The rate at which the energy stored in the capacitor is increasing. = μW.
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Part A What is the approximate radius of an a particle (He)? Express your answer to two significant figures and include the appropriate units. ? HA Value Units The Submit Request Answer
As per the details, the approximate radius of an alpha particle (He) is 1.2 fm.
The Rutherford scattering formula, which connects the scattering angle to the impact parameter and the particle radius, can be used to estimate the approximate radius of an alpha particle (He). The formula is as follows:
θ = 2 * arctan ( R / b )
Here,
θ = scattering angle
R = radius of the particle
b = impact parameter
An alpha particle (He) is made up of two protons and two neutrons that combine to produce a helium nucleus. A helium nucleus has a radius of about 1.2 femtometers (fm) or [tex]1.2* 10^{(-15)[/tex] metres.
Therefore, the approximate radius of an alpha particle (He) is 1.2 fm.
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2 Magnetic Domain Theory. Answer each of the following questions a) When a bar magnet is broken into two pieces, the two pieces actually become two independent magnets instead of a north-pole magnet and a south-pole magner. Explain this phenomenon b) When a magnet is heated up, it loses it magnetization power. However, when the temperature cools back down, the magnetism power returns (assuming the temperature is lower than the Curie point).
a) When a bar magnet is broken into two pieces, the two pieces become two independent magnets, and not a north-pole magnet and a south-pole magnet. This is because each piece contains its own magnetic domain, which is a region where the atoms are aligned in the same direction. The alignment of atoms in a magnetic domain creates a magnetic field. In a magnet, all the magnetic domains are aligned in the same direction, creating a strong magnetic field.
When a magnet is broken into two pieces, each piece still has its own set of magnetic domains and thus becomes a magnet itself. The new north and south poles of the pieces will depend on the arrangement of the magnetic domains in each piece.
b) When a magnet is heated up, the heat energy causes the atoms in the magnet to vibrate more, which can disrupt the alignment of the magnetic domains. This causes the magnetization power to decrease. However, when the temperature cools back down, the atoms in the magnet stop vibrating as much, and the magnetic domains can re-align, causing the magnetism power to return. This effect is assuming that the temperature is lower than the Curie point, which is the temperature at which a material loses its magnetization permanently.
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A thin walled tube with L = 3m and di = 5mm caries a fluid (c = 4210 J/kgK, = 280 x 10^-6 Ns/m^2, k = 0.700 W/mK, Pr = 2.1) with the mass flow rate of 0.015 kg/s and the mean inlet temperature of 90C. the tube's surface is kept constant at 25C.
a. What is the value of the average convection heat transfer coefficient for the problem?
b. How would you calculate the outlet temperature of the fluid? Do not solve.
a. The average convection heat transfer coefficient can be calculated using Q, A, and ΔT in the equation h = (Q / (A * ΔT)), b. The outlet temperature of the fluid can be calculated using the energy balance equation T_out = (Q / (m * c)) + T_in.
a. To find the average convection heat transfer coefficient, we can use the equation:
h = (Q / (A * ΔT))
where h is the convection heat transfer coefficient, Q is the rate of heat transfer, A is the surface area, and ΔT is the temperature difference between the fluid and the surface.
b. To calculate the outlet temperature of the fluid, we need to consider the energy balance equation:
m * c * (T_out - T_in) = Q
where m is the mass flow rate, c is the specific heat capacity, T_out is the outlet temperature, and T_in is the inlet temperature. By rearranging the equation, we can solve for T_out:
T_out = (Q / (m * c)) + T_in
Please note that the actual calculation requires the values of specific heat capacity, temperature difference, and surface area, which are not provided in the given information.
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our objects-a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell-each have a mass of \( 4.41 \mathrm{~kg} \) and a radius of \( 0.240 \) m. (a) Find the moment of inertia for each object
The moment of inertia of a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell is 0.254 kg/m², 0.127 kg/m² and 0.10 kg/m², and 0.20 kg/m² respectively.
The moment of inertia for each object can be calculated based on their respective shapes and masses. The moment of inertia represents the object's resistance to rotational motion. For the given objects - a hoop, solid cylinder, solid sphere, and thin spherical shell - the moment of inertia can be determined using the appropriate formulas. The moment of inertia depends on both the mass and the distribution of mass within the object. We can calculate their respective moments of inertia for the given objects with a mass of 4.41 kg and a radius of 0.240 m.
1. Hoop: A hoop is a circular object with all its mass concentrated at the same distance from the axis of rotation. The moment of inertia for a hoop is given by the formula [tex]\( I = MR^2 \)[/tex], where M is the mass and R is the radius. Substituting the given values, we get [tex]\( I_{\text{hoop}} = 4.41 \times (0.240)^2 \) = 0.254 kg/m^2.[/tex]
2. Solid Cylinder: A solid cylinder has mass distributed throughout its volume. The moment of inertia for a solid cylinder rotating about its central axis is given by [tex]\( I_{\text{cylinder}} = \frac{1}{2} \times 4.41 \times (0.240)^2 \) = 0.127 kg/m^2.[/tex]
3. Solid Sphere: A solid sphere also has mass distributed throughout its volume. The moment of inertia for a solid sphere rotating about its central axis is given by [tex]\frac{2}{5} \times 4.41 \times (0.240)^2 \) = 0.10 kg/m^2.[/tex]
4. Thin Spherical Shell: A thin spherical shell concentrates all its mass on the outer surface. The moment of inertia for a thin spherical shell rotating about its central axis is given by the formula [tex]\( I = \frac{2}{3}MR^2 \).[/tex] Substituting the values, we get [tex]\( I_{\text{shell}} = \frac{2}{3} \times 4.41 \times (0.240)^2 \) = 0.20 kg/m^2[/tex]
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A closely wound, circular coil with a diameter of 5.00 cm has 410 turns and carries a current of 0.400 A Part B What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 6.50 cm from its center? Express your answer in teslas. | ΑΣΦ ? В. B Submit Previous Answers Request Answer
Answer:Part A: The magnetic field at the center of the circular coil has a magnitude of 1.03×10⁻⁴ T and points out of the page.Part B: The magnitude of the magnetic field at a point on the axis of the coil a distance of 6.50 cm from its center is 1.19×10⁻⁵ T.
Part A:First, we will find the magnetic field at the center of the circular coil. To do this, we will use the formula for the magnetic field inside a solenoid: B = μ₀nI. Here, n represents the number of turns per unit length, and I is the current.μ₀ is a constant that represents the permeability of free space.
In this case, we are dealing with a circular coil rather than a solenoid, but we can approximate it as a solenoid if we assume that the radius of the coil is much smaller than the distance between the coil and the point at which we are measuring the magnetic field.
This assumption is reasonable given that the radius of the coil is 2.50 cm and the distance between the coil and the point at which we are measuring the magnetic field is 6.50 cm.
Therefore, we can use the formula for the magnetic field inside a solenoid to find the magnetic field at the center of the circular coil: B = μ₀nI.
Because the coil has a diameter of 5.00 cm, it has a radius of 2.50 cm. Therefore, its cross-sectional area is
A = πr²
= π(2.50 cm)²
= 19.63 cm².
To find n, we need to divide the total number of turns by the length of the coil.
The length of the coil is equal to its circumference, which is
C = 2πr
= 2π(2.50 cm)
= 15.71 cm.
Therefore, n = N/L
= 410/15.71 cm⁻¹
= 26.1 cm⁻¹.
Substituting the values for μ₀, n, and I, we get:
B = μ₀nI
= (4π×10⁻⁷ T·m/A)(26.1 cm⁻¹)(0.400 A)
= 1.03×10⁻⁴ T.
We can use the right-hand rule to determine the direction of the magnetic field.
If we point our right thumb in the direction of the current (which is counterclockwise when viewed from above), the magnetic field will point in the direction of our curled fingers, which is out of the page.
Therefore, the magnetic field at the center of the circular coil has a magnitude of 1.03×10⁻⁴ T and points out of the page.
Part B:We can use the formula for the magnetic field of a circular coil at a point on its axis to find the magnetic field at a distance of 6.50 cm from its center:
B = μ₀I(2R² + d²)-³/²,
where R is the radius of the coil, d is the distance between the center of the coil and the point at which we are measuring the magnetic field, and the other variables have the same meaning as before. Substituting the values, we get:
B = (4π×10⁻⁷ T·m/A)(0.400 A)(2(2.50 cm)² + (6.50 cm)²)-³/²
= 1.19×10⁻⁵ T
Part A: The magnetic field at the center of the circular coil has a magnitude of 1.03×10⁻⁴ T and points out of the page.
Part B: The magnitude of the magnetic field at a point on the axis of the coil a distance of 6.50 cm from its center is 1.19×10⁻⁵ T.
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Water flows at 0.500 mL/s through a horizontal tube that is 50.0 cm long and has an inside diameter of 1.50 mm. Assuming laminar flow, determine the pressure difference Ap required to drive this flow if the viscosity of water is
1.00 mPa-s.
The pressure difference required to drive this flow is 31.8 kPa (approximately) if the viscosity of water is 1.00 mPa-s.
The laminar flow of a fluid occurs when the fluid flows smoothly and there are no irregularities in the fluid motion. Poiseuille’s equation states that the volume flow rate of a fluid in a tube is directly proportional to the pressure difference that drives the flow.
The volume of water that flows in the tube is given by Q=0.5mL/s which is the volume that flows in one second.
The cross-sectional area of the tube is given by: A=πr²
Since the inside diameter is given, then the radius is given by
r = D/2r
= 1.50/2mm
= 0.750 mm
= 0.75 × 10⁻⁶ m
The cross-sectional area is given by:
A = πr²A
= π(0.75 × 10⁻⁶ m)²
A = 1.767 × 10⁻⁹ m²
From Poiseuille’s equation, the volume flow rate of a fluid in a tube is given by:
Q = π∆P/8ηL(A/r⁴)Q
= (π/8)(∆P)(r⁴)/ηL
Substituting the values gives:
0.5 × 10⁻³ = (π/8)(∆P)(0.75 × 10⁻⁶)⁴/1 × 10⁻³ × 0.5∆P
= 31795.50 Pa
The pressure difference required to drive this flow is 31.8 kPa (approximately) if the viscosity of water is 1.00 mPa-s.
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Three 1.60Ω resistors are connected in series to a 19.0 V battery. What is the equivalent resistance (in Ω ) of the circuit?
The equivalent resistance of the circuit is 4.80Ω.
When resistors are connected in series, their resistances add up to give the equivalent resistance of the circuit.
In this case, three 1.60Ω resistors are connected in series.
To find the equivalent resistance, we simply sum the individual resistances:
Equivalent Resistance = 1.60Ω + 1.60Ω + 1.60Ω
Equivalent Resistance = 4.80Ω
Therefore, the equivalent resistance of the circuit is 4.80Ω.
When resistors are connected in series, the total resistance increases because the current flowing through each resistor is the same, and the voltage drop across each resistor adds up.
The total voltage supplied by the battery is shared across the resistors, leading to a higher overall resistance.
It's important to note that the equivalent resistance is the total resistance of the series combination.
It represents the resistance that a single resistor would need to have in order to produce the same overall effect as the series combination of resistors when connected to the same voltage source.
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A tangential force of 89789.9 N is applied to a 6.2mm copper cube as shown below, find the shear strain given that the shear modulus of brass is 4.2 X 1070 N/m?? A Shear strain rad
The shear strain of the copper cube is approximately 0.02144 radians.
To find the shear strain of the copper cube, we can use the equation:
Shear strain = Shear stress / Shear modulus
The applied tangential force is 89789.9 N and the shear modulus of brass (assuming it was mistakenly mentioned as copper) is 4.2 x 10^7 N/m², we need to convert the dimensions of the cube to obtain the shear stress.
The shear stress can be calculated using the formula:
Shear stress = Force / Area
The area of the cube's face can be determined by squaring the length of one side, which is given as 6.2 mm or 0.0062 m.
Now, let's calculate the shear stress:
Area = (0.0062 m)² = 3.844 x 10^-5 m²
Shear stress = 89789.9 N / 3.844 x 10^-5 m²
Next, we can calculate the shear strain:
Shear strain = Shear stress / Shear modulus
Shear strain = (89789.9 N / 3.844 x 10^-5 m²) / (4.2 x 10^7 N/m²)
Evaluating the expression, we find that the shear strain is approximately 0.02144 rad.
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11 Required information The tension in a ligament in the human knee is approximately proportional to the extension of the ligament, if the extension is not too large. eBook Hint If a particular ligament has an effective spring constant of 149 N/mm as it is stretched, what is the tension in this ligament when it is stretched by 0.740 cm? Print References KN 166 points
To find the tension in a ligament when it is stretched by a certain amount, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to its extension. In this case, the ligament can be modeled as a spring with an effective spring constant of 149 N/mm. The tension in the ligament can be calculated by multiplying the extension (0.740 cm) by the spring constant. The tension in the ligament is equal to 109.86 N.
Hooke's Law states that the force (F) applied to a spring is directly proportional to its extension (x), given by the equation F = k * x, where k is the spring constant. In this case, the effective spring constant of the ligament is given as 149 N/mm.
First, we need to convert the extension from centimeters to millimeters:
0.740 cm = 7.40 mm
Now we can calculate the tension in the ligament by multiplying the extension by the spring constant:
Tension = Spring constant * Extension
= 149 N/mm * 7.40 mm
= 109.86 N
Therefore, the tension in the ligament when it is stretched by 0.740 cm is approximately 109.86 N.
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A particle is in SHM along x axis, A=0.12m, T-2s. When t=0, xo=0.06m, and v> 0 (moves along positive x direction). Try to find out: (1) The expression of this SHM; (2) t = T/4, v=? and a=? (3) At what time will the particle pass the "O" first time?
The expression for the SHM is x = 0.12 * cos(πt). We can start by using the general equation for SHM: x = A * cos(ωt + φ). The particle passes the origin (O) for the first time at t = 0.5 s. we can start by using the general equation for SHM: x = A * cos(ωt + φ).
To find the expression for the Simple Harmonic Motion (SHM) of the particle, we can start by using the general equation for SHM:
x = A * cos(ωt + φ)
Where:
x is the displacement from the equilibrium position,
A is the amplitude of the motion,
ω is the angular frequency, given by ω = 2π/T (T is the period),
t is the time, and
φ is the phase constant.
Given that A = 0.12 m and T = 2 s, we can find the angular frequency:
ω = 2π / T
= 2π / 2
= π rad/s
The expression for the SHM becomes:
x = 0.12 * cos(πt + φ)
To find the phase constant φ, we can use the initial conditions given. When t = 0, x₀ = 0.06 m, and v > 0.
Substituting these values into the equation:
0.06 = 0.12 * cos(π * 0 + φ)
0.06 = 0.12 * cos(φ)
Since the particle starts from the equilibrium position, we know that cos(φ) = 1. Therefore:
0.06 = 0.12 * 1
φ = 0
So, the expression for the SHM is:
x = 0.12 * cos(πt)
Now let's move on to the next parts of the question:
(2) At t = T/4, we have:
t = T/4 = (2/4) = 0.5 s
To find the velocity v at this time, we can take the derivative of the displacement equation:
v = dx/dt = -0.12 * π * sin(πt)
Substituting t = 0.5 into this equation:
v = -0.12 * π * sin(π * 0.5)
v = -0.12 * π * sin(π/2)
v = -0.12 * π * 1
v = -0.12π m/s
So, at t = T/4, v = -0.12π m/s.
To find the acceleration a at t = T/4, we can take the second derivative of the displacement equation:
a = d²x/dt² = -0.12 * π² * cos(πt)
Substituting t = 0.5 into this equation:
a = -0.12 * π² * cos(π * 0.5)
a = -0.12 * π² * cos(π/2)
a = -0.12 * π² * 0
a = 0
So, at t = T/4, a = 0 m/s².
(3) To find the time when the particle passes the origin (O) for the first time, we need to find the time when x = 0.
0 = 0.12 * cos(πt)
Since the cosine function is zero at π/2, π, 3π/2, etc., we can set the argument of the cosine function equal to π/2:
πt = π/2
Solving for t:
t = (π/2) / π
t = 0.5 s
Therefore, the particle passes the origin (O) for the first time at t = 0.5 s.
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