Two circuits are considered equivalent when their output values are the same for all possible input combinations.
A circuit equivalent to another is one that meets the same conditions, (eg same current), under a different configuration.
The equivalent circuit made in this way is not the same as the original one, but if the total voltages, the total currents and the total resistance of the circuit will be equal, which will be the equivalent.The concept of equivalent circuit is used when saying, for example, that a real generator is equivalent to an ideal one with its internal resistance in series.Therefore, we can conclude that two circuits are considered equivalent when their output values are the same for all possible input combinations.
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Final answer:
Two circuits are considered equivalent if they produce the same output for all possible input combinations, as can be demonstrated using a truth table in digital logic. The correct option is C.
Explanation:
Two circuits are considered equivalent when their output values are the same for all possible input combinations. This means that regardless of the input values, both circuits will produce the same output across all scenarios. For instance, in digital logic, two circuits using different logic gates such as AND, OR, NAND, or XOR would be equivalent if they yield the same output for each combination of inputs, as represented in a truth table. An example of this is the use of De Morgan's laws where an AND gate followed by a NOT gate can be equivalent to a NAND gate, producing the same outputs. Understanding these principles is essential for fields like electronics, computer engineering, and programming. Hence, Option C is correct.
A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 42 m/s. The third piece has three times the mass of the other two pieces. Randomized Variables V = 42 m/s show answer No Attempt 33% Part (a) Write an expression for the speed of the larger piece, that is in terms of only the variable V.
Answer:
Speed of larger piece is [tex]\dfrac{V\sqrt{2}}{3}[/tex]
Explanation:
We apply the principle of conservation of momentum.
The watermelon is initially at rest. The initial momentum = 0 kg m/s in all directions.
After the collision,
Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = [tex]mV + 3mv_{ly}[/tex]
Here, [tex]v_{ly}[/tex] is the y-component of velocity of larger piece.
This is equal to 0, since the initial momentum is 0.
[tex]v_{ly}=\dfrac{V}{3}[/tex]
Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = [tex]mV + 3mv_{lx}[/tex]
Here, [tex]v_{lx}[/tex] is the x-component of velocity of larger piece.
This is also equal to 0, since the initial momentum is 0.
[tex]v_{lx}=\dfrac{V}{3}[/tex]
The velocity of the larger piece, [tex]v_l[/tex], is the resultant of [tex]v_{lx}[/tex] and [tex]v_{ly}[/tex]. Since they are mutually perpendicular,
[tex]v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}[/tex]
[tex]v_l = \dfrac{V\sqrt{2}}{3}[/tex]
(1 point) Find a linearly independent set of vectors that spans the same subspace of R3R3 as that spanned by the vectors ⎡⎣⎢3−1−2⎤⎦⎥, ⎡⎣⎢−923⎤⎦⎥, ⎡⎣⎢−30−1⎤⎦⎥. [3−1−2], [−923], [−30−1]. A linearly independent spanning set for the subspace is:
Answer:
Explanation:
[tex]A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right][/tex]
This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of [tex]R^4[/tex] as all three vectors.
Therefore, the linearly independent spanning set for the subspace is
[tex]\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right][/tex]
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 cm2 . If a voltage of 145 V is applied to the arrangement, determine the voltage across the 2.5 cm2 wire.
Answer:
22.1 V
Explanation:
We are given that
[tex]A_1=0.7 cm^2=0.7\times 10^{-4} m^2[/tex]
[tex]A_2=2.5 cm^2=2.5\times 10^{-4} m^2[/tex]
[tex]A_3=2.2 cm^2=2.2\times 10^{-4} m^2[/tex]
[tex]A_4=3 cm^2=3\times 10^{-4} m^2[/tex]
Using [tex] 1cm^2=10^{-4} m^2[/tex]
We know that
[tex]R=\frac{\rho l}{A}[/tex]
In series
[tex]R=R_1+R_2+R_3+R_4[/tex]
[tex]R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}[/tex]
[tex]R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}[/tex]
Substitute the values
[tex]R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})[/tex]
[tex]R=\rho l(2.62\times 10^4)[/tex]
[tex]V=145 V[/tex]
[tex]I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}[/tex]
Voltage across the 2.5 square cm wire=[tex]IR=I\times \frac{\rho l}{A_2}[/tex]
Voltage across the 2.5 square cm wire=[tex]\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V[/tex]
Voltage across the 2.5 square cm wire=22.1 V
In an arcade game a 0.117 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releasing it. If the spring has a spring constant of 194 N/m and is compressed from its equilibrium position by 7 cm, find th
Find the speed with which the disk slides across the surface
Answer:
[tex]\boxed{2.85 m/s}[/tex]
Explanation:
The Potential energy of spring is transformed to kinetic energy hence
[tex]0.5kx^{2}=0.5mv^{2}\\kx^{2}=mv^{2}\\v=\sqrt{\frac {kx^{2}}{m}}[/tex]
Here k is the spring constant, x is the extension of spring, v is the velocity of disk and m is the mass of the disk.
Substituting 0.117 Kg for m, 194 N/m for k and 0.07 m for x then
[tex]v=\sqrt{\frac {194\times 0.07^{2}}{0.117}}=2.850401081 m/s\approx \boxed{2.85 m/s}[/tex]
A 2.00 g air‑inflated balloon is given an excess negative charge, q 1 = − 3.75 × 10 − 8 C, by rubbing it with a blanket. It is found that a charged rod can be held above the balloon at a distance of d = 6.00 cm to make the balloon float. Assume the balloon and rod to be point charges. The Coulomb force constant is 1 / ( 4 π ϵ 0 ) = 8.99 × 10 9 N ⋅ m 2 / C 2 and the acceleration due to gravity is g = 9.81 m / s 2 .
Answer:
(+2.093 × 10⁻⁷) C
Explanation:
Coulomb's law gives the force of attraction between two charges and it is given by
F = kq₁q₂/r²
where q₁ = charge on one of the two particles under consideration = charge on the balloon = - 3.75 × 10⁻⁸ C
q₂ = charge on the other body = charge on the rod = ?
k = Coulomb's constant = 1/(4 π ϵ₀) = 8.99 × 10⁹ N⋅m²/C²
r = distance between the two charges = d = 6.00 cm = 0.06 m
But for this question, the force of attraction between the charges was enough to lift the balloon and match its weight, Hence,
F = (kq₁q₂/d²) = - mg (negative because it's an attractive force)
m = mass of balloon = 2.00 g = 0.002 kg
g = acceleration due to gravity = 9.8 m/s²
(8.99 × 10⁹ × (-3.75 × 10⁻⁸) × q₂)/(0.06²) = 0.002 × 9.8
q₂ = (-0.002 × 9.8 × 0.06²)/(8.99 × 10⁹ × (-3.75 × 10⁻⁸)
q₂ = + 2.093 × 10⁻⁷ C
When electric power plants return used water to a stream, after using it in their steam turbines and condensers, this used water can lead to ____________ pollution.
Answer:
Thermal Pollution
Explanation:
Thermal pollution is a term use to describe the final result caused from the water used as a coolant by power plants and/or industrial companies. It changes the water temperature and its quality. It is usually caused from humans or companies especially manufacturing companies - They use it for their personal needs. The end product of the water after being discarded is Thermal Pollution
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).
Find the speed of the passengers when the Ferris wheel is rotating at this rate.
Answer:
V = 5.24m/s
Explanation:
See attachment below.
Air enters a 16-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows, and it leaves the pipe at 180 kPa and 43°C. The gas constant of air is 0.287 kPa·m3/kg·K. Whats the volumetric flow rate of the inlet/outlet, mass flow rate and velocity & volume flow rate at the exit?
Explanation:
(a) We will determine the mass flow rate as follows.
m = [tex]\rho_{1} V_{1}[/tex]
= [tex]\frac{P_{1}}{RT_{1}}A_{1}v_{1}[/tex]
= [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
Putting the given values into the above formula as follows.
m = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
= [tex]\frac{200}{0.287 \times 293 K} \times \frac{(0.16)^{2}}{4} \pi \times 5[/tex]
= 0.239 kg/s
Hence, the mass flow rate of the inlet/outlet is 0.239 kg/s.
(b) Now, we will determine the final volume rate as follows.
[tex]V_{2} = \frac{m}{\rho_{2}}[/tex]
= [tex]\frac{RT_{2}m}{P_{2}}[/tex]
= [tex]\frac{0.287 \times 313 \times 0.239}{180}[/tex]
= 0.119 [tex]m^{3}/s[/tex]
And, the final velocity will be determined as follows.
[tex]v_{2} = \frac{V_{2}}{A}[/tex]
= [tex]\frac{4V_{2}}{D^{2} \times \pi}[/tex]
= [tex]\frac{4 \times 0.119}{(0.16)^{2} \times \pi}[/tex]
= 5.92 m/s
Therefore, the volumetric flow rate is 0.119 [tex]m^{3}/s[/tex] and velocity rate is 5.92 m/s.
A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it has traveled 4.20 m downward
The speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.
Explanation:To calculate the speed of the ball after it has traveled 4.20 m downward, we need to use the principles of free fall and the equations of motion. Since the ball is dropped from a height of 55.0 m, we can calculate the initial velocity using the equation v_i = sqrt(2 * g * h), where v_i is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (55.0 m). Plugging in these values, we find that the initial velocity is approximately 34.02 m/s.
Next, we can calculate the final velocity using the equation v_f = sqrt(v_i^2 + 2 * g * d), where v_f is the final velocity, v_i is the initial velocity, g is the acceleration due to gravity, and d is the distance traveled downward (4.20 m). Plugging in the values, we get v_f = sqrt((34.02 m/s)^2 + 2 * (9.8 m/s^2) * (4.20 m)) = approximately 24.04 m/s.
Therefore, the speed of the ball after it has traveled 4.20 m downward is approximately 24.04 m/s.
Ballistic pendulum conservation of momentum/mechanical energy?
"Mechanical energy and momentum are conserved only when certain conditions rae met. Explain why the collision between the ball and the pendulum conserves momentum but not mechanical energy. Similarly, explain why the motion of the pendulum during its swing conserves mechanical energy but (apparently) not momentum."
I understand why during the collision kinetic energy is lost and momentum is conserved, but why would mechanical energy be conserving during the swing and not momentum?
Answer:
In the explanation the answers for the two questions are analyzed.
Explanation:
1. When a collision occurs, it can be said that the total energy is not conserved, but the momentum is conserved. Kinetic energy is converted to heat in the case of an inelastic collision, therefore the sum of all energies is the same before and after the collision. Regarding the net moment before the collision is equal to the net moment after said collision.
2. Regarding the impulse, this is not conserved in the case of an oscillating pendulum because the resulting force acting on the pendulum is not equal to zero. While the total energy would be equal to the sum of the potential energy plus the kinetic energy.
In pushing a heavy box across the floor, is the force you need to apply to start the box moving greater than, less than, or the same as the force needed to keep the box moving? On what are you basing your choice?
How do you think the force of friction is related to the weight of the box? Explain.
Answer:
Explanation:
Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .
A threshold force is needed to move the box and when box started to move kinetic friction comes into play.
Friction force is directly related to the weight of the box as the friction force is
coefficient of friction time Normal reaction .
And Normal reaction is equal to the weight of box if no force is applied.
[tex]f_r=\mu N[/tex]
[tex]N=mg[/tex]
The force to start moving a box is greater than to keep it moving due to static and kinetic friction. The force of friction is directly proportional to the weight of the box; the heavier the box, the greater the friction.
Explanation:In physics, the force required to start moving an object is often greater than the force required to keep it moving. This is due to a concept known as friction, particularly, static friction and kinetic friction. Static friction is the force that resists the initiation of sliding motion, and it's usually greater than kinetic friction, which is the force that opposes motion of an object when the object is in motion.
Now, relating friction to the weight of the box, the weight of an object is equal to its mass times the acceleration due to gravity, and it's directly proportional to the force of friction. In essence, the heavier the box is, the more the static and kinetic friction that you have to overcome to move and keep it moving.
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A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N , against a friction force of 210 N . What is the acceleration of the box? How far would the box move in 3 s , if it starts from rest?
Answer:
0.67 m/s² or 2/3 m/s²
3 m
Explanation:
Using
F-F' = ma............ Equation 1
Where F = Horizontal force applied to the box, F' = Frictional force, m = mass of the box, a = acceleration of the box.
make a the subject of the equation
a = (F-F')/m............ Equation 2
Given: F = 230 N, F' = 210 N, m = 30 kg.
substitute into equation 2
a = (230-210)/30
a = 20/30
a = 0.67 m/s² or 2/3 m/s²
The acceleration of the box = 2/3 m/s²
Using,
s = ut+1/2at²............ Equation 3
Where u = initial velocity, t = time, a = acceleration, s = distance.
Given: u = 0 m/s (from rest), t = 3 s, a = 2/3 m/s²
substitute into equation 3
s = 0(3)+1/2(2/3)(3²)
s = 3 m.
Hence the box moves 3 m
The acceleration of the 30.0-kg box is approximately 0.67 m/s². If it starts from rest, it would move around 3.02 meters in 3 seconds.
Explanation:In order to solve the problem, we'll apply the principle of Newton's second law, which states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
First, we identify the net force on the box. The box is being pulled by a force of 230 N, and there is friction acting against this pull with a force of 210 N. The net force (F) is therefore 230 N (pulling force) - 210 N (friction) = 20 N.
To find the acceleration (a), we use the formula:
a = F/m
Substituting the given values, we get:
a = 20 N / 30.0 kg = 0.67 m/s².
Now, the second part of the problem asks for the distance the box would move in 3 seconds if it starts from rest. We use the formula for distance (d) traveled under constant acceleration:
d = 0.5 * a * t²
Substituting the calculated acceleration and time (3 seconds), we find:
d = 0.5 * 0.67 m/s² * (3 s)² = 3.02 m.
So, under the given conditions, the box's acceleration is 0.67 m/s², and it would move approximately 3.02 meters in 3 seconds.
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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.30 A. When they are wired in parallel with the same battery, the current is 1.6 A. What are the values of the two resistors?
Note: I understand other people have asked the same concept with different numbers, and their questions have been answered, however I do not understand why you need to multiple R1 by R2 when the resistors are in parallel, and furthermore, why are you supposed to multiply the "in parallel" equation by (R1+R2; the in series resistors) to find R1*R2?
Answer:
[tex]30\Omega, 10\Omega[/tex]
Explanation:
Let two resistors R1 and R2 are wired in series.
Potential difference, V=12 V
Current=I=0.3 A
We have to find the value of two resistors.
When two resistors are connected in series
[tex]R=R_1+R_2[/tex]
[tex]V=IR=I(R_1+R_2)[/tex]
Substitute the values
[tex]12=0.3(R_1+R_2)[/tex]
[tex]R_1+R_2=\frac{12}{0.3}=40[/tex]
[tex]R_1+R_2=40[/tex]..(1)
In parallel
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}[/tex]
[tex]R=\frac{R_1R_2}{R_1+R_2}[/tex]
Current in parallel, I=1.6 A
[tex]V=IR[/tex]
[tex]V=1.6(\frac{R_1R_2}{R_1+R_2})[/tex]
[tex]\frac{12}{1.6}=\frac{R_1R_2}{40}[/tex]
[tex]R_1R_2=\frac{12\times 40}{1.6}=300[/tex]
[tex]R_1-R_2=\sqrt{(R_1+R_2)^2-4R_1R_2}[/tex]
[tex]R_1-R_2=\sqrt{(40)^2-4(300)}=20[/tex]....(2)
Adding equation (1) and (2)
[tex]2R_1=60[/tex]
[tex]R_1=\frac{60}{2}=30\Omega[/tex]
Substitute the value in equation (1)
[tex]30+R_2=40[/tex]
[tex]R_2=40-30=10\Omega[/tex]
Final answer:
To find the values of two resistors based on their behavior in series and parallel circuits, one must calculate the total equivalent resistances for each configuration with Ohm's law and then solve the equations relating the individual resistances.
Explanation:
When looking to determine the values of two resistors R1 and R2 based on current measurements in both series and parallel configurations, the approach involves some electrical principles and algebra. Let's break down the steps needed to solve for the resistors' values.
Determination of resistors in series
Firstly, we calculate the equivalent resistance for the series circuit using Ohm's law (V = I × R), where V is the voltage, I is the current, and R is the resistance. With a 12 V battery and a current of 0.30 A, the equivalent resistance, Req-series, is 12 V / 0.30 A = 40 Ω.
Determination of resistors in parallel
For the parallel configuration, the current through the battery increases to 1.6 A with the same voltage of 12 V. Again, using Ohm's law, we find the equivalent parallel resistance, Req-parallel, which is 12 V / 1.6 A = 7.5 Ω.
In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of each individual resistance. This can be expressed as 1/Req-parallel = 1/R1 + 1/R2. To find R1 and R2, we need to relate the resistances in parallel to the sum of their series counterpart (R1 + R2 = Req-series).
Finding Individual Resistances
Given the equations R1 + R2 = 40 Ω and 1/R1 + 1/R2 = 1/7.5 Ω, we can simplify the latter to R1×R2/(R1 + R2) = 7.5 Ω. Substituting R1 + R2 = 40 Ω into this equation, we finally solve for the individual resistance values, R1 and R2.
Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer I0 and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of φ = 2θ.
Answer:
Explanation:
Intensity of unpolarised light = I₀
intensity after passing through first polariser = I₀ / 2
Angle between first and second polariser is φ so
intensity after passing through second polariser
= (I₀ / 2) cos²φ
Now angle between second and third polariser
= 90 - φ
intensity after passing though third polariser
= (I₀ / 2) cos²φ cos²( 90 - φ)
= (I₀ / 2) cos²φ sin²φ
= (I₀ / 8) 4cos²φ sin²φ
= (I₀ / 8) sin²2φ
The intensity after the third polarizer will be:
I₃ = (I₀/8)*sin^2(2φ)
What is the resulting intensity?
For non-polarized light that passes through any polarizer, we say that the intensity is reduced to its half.
Original intensity = I₀
After the first polarizer, the intensity will be:
I₁ = I₀/2.
Now, when it passes through a polarizer such that the difference in angles with the polarization is x, the new intensity will be:
I₂ = I₁*cos^2(x).
The angle between the second and the first polarizer is φ, then we have:
I₂ = (I₀/2)*cos^2(φ).
Now we also know that the first and the last polarizer are crossed (so there is an angle of 90°). Then if we define θ as the angle between the second and the third polarizer, we will have that:
φ + θ = 90°
then:
θ = 90° - φ
The intensity after the third polarizer will be:
I₃ = (I₀/2)*cos^2(φ)*cos^2(90° - φ)
And we know that:
cos(90° - φ) = sin(φ)
Then we can rewrite:
I₃ = (I₀/2)*cos^2(φ)*sin^2(φ)
But we want a single trigonometric function, then we use the relation:
cos^2(φ)*sin^2(φ) = sin^2(2φ)/4
And replacing that, we get:
I₃ = (I₀/2)*sin^2(2φ)/4 = (I₀/8)*sin^2(2φ)
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Explain whether a solenoid will attract paper clips if there is no current through its coils. Predict what will happen to the force between the solenoid and the paper clips if you increase the current in the coils or increased the number of turns of coil per un
Answer:
magnetic field strength increases.
Explanation:
No, the solenoid will not attract any magnetic substance if it does not carries a direct electric current.
The formula for the magnetic field strength of the solenoid is given as:
[tex]B=\mu.n.I[/tex]
where:
[tex]\mu=[/tex] permeability of free space
[tex]n=[/tex] no. of coil turns in the solenoid
[tex]I=[/tex] current in the coil
So, when the current & no. of coil turns in the coil are increased then the magnetic field strength also increases.
fter driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average mass of 69 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed
Answer:
The spring compressed is 0.44 m.
Explanation:
Given that,
Number of person = 25
Average Mass of each person [tex]m_{p}= 25\times69 = 1725\ kg[/tex]
Mass of goat [tex]m_{g}= 3\times15 = 45\ kg[/tex]
Mass of chicken [tex]m_{c}= 5\times3 = 15\ kg[/tex]
Mass of bananas = 25 kg
If the spring constant is [tex]4\times10^{4}\ N/m[/tex]
We need to calculate the total mass
[tex]M=m_{p}+m_{g}+m_{c}+m_{b}[/tex]
Put the value in the formula
[tex]M=1725+45+15+25[/tex]
[tex]M=1810\ kg[/tex]
The weight of all these things is
[tex]W=Mg[/tex]
[tex]W=1810\times9.8[/tex]
[tex]W=17738\ N[/tex]
We need to calculate the distance
Using formula of restoring force
[tex]F=kx[/tex]
[tex]x=\dfrac{F}{k}[/tex]
Put the value into the formula
[tex]x=\dfrac{17738}{4\times10^{4}}[/tex]
[tex]x=0.44\ m[/tex]
Hence, The spring compressed is 0.44 m.
A "mechanical wave" occurs in a physical medium because of some restoring force acting on the medium, and can be described by an amplitude, an oscillation frequency, and an amplitude.
You can easily decrease the wavelength of a mechanical wave by half by doubling the frequency ONLY if which of the following is true? Select all that apply
(A) The ratio between the restoring force and the density of the medium remain constant
(B) The amplitude is doubled
(C) The amplitude is reduced by 1/2
(D) The same medium is used at the same temperature
Answer:
(A) The ratio between the restoring force and the density of the medium remain constant
Explanation:
The ratio between the restoring force and the density of the medium is equal to the square of the velocity of the wave.
[tex]v = \sqrt{\frac{F}{\mu}}[/tex]
The general formula that relates the displacement and velocity (x = vt) can be written in wave mechanics such that
[tex]v = \lambda f[/tex]
where f is the frequency, λ is the wavelength, and v is the velocity of the wave.
According to this equation, in order to halve the wavelength by doubling the frequency, the velocity should be constant. Therefore, the correct answer is (A).
The wavelength of a mechanical wave can be decreased by half by doubling the frequency ONLY if; Choice (A) The ratio between the restoring force and the density of the medium remain constant and Choice D: The same medium is used at the same temperature.
Discussion:
The speed of the mechanical wave is dependent on the ratio of the restoring force and the density of the medium.
Additionally, when the same medium is used at the same temperature; the density of the medium remains constant.
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Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.
(a) How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.
(b) If Kate just touches the surface of the river on her first downward trip (i.e. before the first bounce), what is the spring constant k?
Final answer:
To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x.
Explanation:
(a) To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. At this point, her net force will be zero. The gravitational force can be calculated as mg, where m is her mass and g is the acceleration due to gravity. The spring force can be calculated using Hooke's Law: Fs = -kx, where k is the spring constant and x is the displacement of the cord from its equilibrium position. Equating the gravitational force and the spring force and solving for x will give us the distance below the bridge where Kate will be hanging.
(b) To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x. Solving for k will give us the spring constant of the bungee cord.
A 14-kg block rests on a level frictionless surface and is attached by a light string to a 5.0-kg hanging mass where the string passes over a massless frictionless pulley. If g=9.8 m/s2 what is the tension in the connecting string?
Answer:
Explanation:
Let T be the tension in the string .
For hanging mass
net force = m₁g - T
m₁g - T = m₁a
For mass placed on horizontal plane
T = m₂a
m₁g - m₂a = m₁a
m₁g = a ( m₂ +m₁)
a = m₁g / ( m₂ +m₁)
= 5 x 9.8 / 19
= 2.579 m /s²
m₁g - T = m₁a
T = m₁g - m₁a
= m₁ ( g - a )
= 5 ( 9.8 - 2.579)
= 36.10 N
A bat strikes a 0.145 kgkg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/sm/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘30∘ above horizontal with a speed of 38.0 m/sm/s. The ball and bat are in contact for 1.75 msms. find the horizontal and vertical components of the average force on the ball.
Answer with Explanation:
We are given that
Mass of ball=m=0.145 kg
Initially horizontal velocity of ball,ux=50 m/s
[tex]\theta=30^{\circ}[/tex]
Final velocity of ball,v=38m/s
Time ,t=1.75 ms=[tex]1.75\times 10^{-3} s[/tex]
[tex]1 ms=10^{-3} s[/tex]
Horizontal component of average force, [tex]F_x=\frac{m(vcos\theta-u_x)}{t}[/tex]
Using the formula
Horizontal component of average force, [tex]F_x=\frac{0.145(-38cos30-50)}{1.75\times 10^{-3}}=-6.9\times 10^3[/tex]N
Vertical component of average force, [tex]F_y=\frac{m(vsin\theta-u_y)}{t}[/tex]
Vertical component of average force,[tex]F_y=\frac{0.145(38sin30-0}{1.75\times 10^{-3}}=1.6\times 10^3 N[/tex]
An airplane is flying at 150 ft/s at an altitude of 2000 ft in a direction that will take it directly over an observer at ground level. Find the rate of change of the angle of elevation between the observer and the plane when the plane is directly over a point on the ground that is 2000 ft from the observer.
Answer:
Explanation:
Let x be the horizontal distance of airplane . angle of elevation of airplane from observer = θ , altitude of airplane = 2000 ft ,
from the construction θ = 45 degree. , x = 2000 ft .
Tanθ = 2000 / x
sec²θ dθ / dt = (2000 / x²) dx / dt
dθ / dt = 2000 /(sec²θ x²) x dx / dt
dx / dt = 150 ft /s
dθ / dt = 2000x 150 /(sec²θ x²)
= 300000 / sec²45 x 2000²
= .15 degree/ s
The rate of change of the angle of elevation when an airplane (traveling at 150 ft/s) is over a point 2000 ft from the observer is approximately -2.15° per second.
Explanation:This problem falls under the department of mathematics known as trigonometry specifically, its applications in real-world scenarios. In this case, we're looking to find the rate of change of the angle of elevation when the airplane is at a certain point.
When the airplane is directly over a point 2000 ft from the observer, it forms a right triangle with the observers' location and the point over which it is flying. The hypotenuse of this triangle is the line between observer and airplane.
The angle of elevation, θ, from the observer to the plane changes over time as the airplane moves overhead so its rate of change(dθ/dt) is what we need to find. We use the trigonometric relation tangent (tan), which in this case equals to the altitude (opposite side, 2000 ft) over the horizontal distance between the observer and the plane (adjacent side, 2000 ft).
The relation is tan θ = opposite/adjacent = 2000/2000 = 1, thus θ = 45 degrees. With the plane's speed (150 ft/s), this changes the horizontal distance over time, so we differentiate tan θ giving us sec² θ*dθ/dt = -150/2000² = -0.0375 radians/sec, by applying the chain rule and remembering sec² θ is 1/cos² θ.
In degrees per second, this is approximately -2.15°/s, so that is the rate of change of the angle of elevation when the airplane is directly over a point 2000 ft from the observer.
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At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field at location < 0.03, 0.05, 0 > m, due to the moving proton. What is the vector r?
Answer:
[tex]9.7\times 10^{-5} T[/tex]
Explanation:
Velocity =[tex]5\times 10^4i-2\times 10^4j[/tex]
r=[tex]0.03i+0.05j[/tex]
r=[tex]\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058[/tex]
v=[tex]\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}[/tex]
We know that
[tex]B=\frac{mv}{qr}[/tex]
Where q=[tex]1.6\times 10^{-19} C[/tex]
Mass of proton=[tex]1.67\times 10^{-27} kg[/tex]
Using the formula
[tex]B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}[/tex]
[tex]B=9.7\times 10^{-5} T[/tex]
The vector r represents the position at which the magnetic field is being calculated. The formula used to calculate the magnetic field is the Biot-Savart law. The given values can be plugged into the formula to find the magnitude and direction of the magnetic field.
Explanation:The vector r represents the position of the point where we want to calculate the magnetic field. In this case, r = < 0.03, 0.05, 0 > m.
To calculate the magnetic field at this point, we can use the Biot-Savart law. The Biot-Savart law states that the magnetic field at a point due to a moving charge is given by B = (μ₀/4π) imes ((qv)×r)/r³, where B is the magnetic field, μ₀ is the permeability of free space, q is the charge, v is the velocity of the charge, and r is the displacement vector from the charge to the point where the magnetic field is being calculated.
Plugging in the values given, we have q = +e, v = <5e4, -2e4, 0> m/s, and r = <0.03, 0.05, 0> m. The magnitude of the magnetic field can be calculated using the formula B = (μ₀/4π) imes ((qv)×r)/r³ and the direction can be determined using the right-hand rule.
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A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150 N. If the coefficient of friction between box and floor is 0.300, find the following.(a) the work done by the applied force
J
(b) the increase in internal energy in the box-floor system due to friction
J
(c) the work done by the normal force
J
(d) the work done by the gravitational force
J
(e) the change in kinetic energy of the box
J
(f) the final speed of the box
m/s
Answer:
a) 607.5 J
b) 160.531875 J
c) 0 J
d) 0 J
e) 2.925 m\s
Explanation:
The given data :-
Mass of the box ( m ) = 37.5 kg.Displacement made by box ( x ) = 4.05 m.Horizontal force ( F ) = 150 N.The co-efficient of friction between box and floor ( μ ) = 0.3Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875Solution:-
a) The work done by applied force ( W )
W = force applied × displacement = 150 × 4.05 = 607.5 J
b) The increase in internal energy in the box-floor system due to friction.
Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N
Change in internal energy = change in kinetic energy.
ΔU = ( K.E )₂ - ( K.E )₁
Since the initial velocity is zero so the ( K.E )₁ = 0
ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
c) The work done by the normal force .
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
d) The work done by the gravitational force.
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
e) The change in kinetic energy of the box
( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
f) The final speed of the box
( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²
v² = 8.56
v = 2.925 m\s.
An electron is traveling horizontally toward the north in a uniform magnetic field that is directed vertically downward. In what direction does the magnetic force act on the electron? An electron is traveling horizontally toward the north in a uniform magnetic field that is directed vertically downward. In what direction does the magnetic force act on the electron? upward south downward west east north
Answer:
east direction.
Explanation:
Given, that the electron is travelling in the north direction in the horizontal direction whereas the magnetic field applied is in the vertically downward direction.
Using the Maxwell's left hand rule, we point the index finger in the direction of magnetic field while the perpendicular middle finger points in the direction of motion of the positive charge and the thumb points in the direction of the force. During this position the angle between all the three fingers must be mutually perpendicular to each other.
Therefore we find that here in this case the magnetic force on the electron acts in the east direction.
The first finger points in the direction of the magnetic field and the middle finger in the direction of the electric current. In the east direction the magnetic force act on the electron.
What is the left-hand rule?The thumb will point in the direction of the force on the conductor if the thumb and first two fingers of the left hand are arranged at right angles to each other on a conductor and the hand is oriented.
So that the first finger points in the direction of the magnetic field and the middle finger in the direction of the electric current.
Given that the electron travels in a horizontal north-south path, whereas the magnetic field applied is vertically downward.
We point the index finger in the direction of the magnetic field, the perpendicular middle finger in the direction of positive charge motion, and the thumb in the direction of force using Maxwell's left-hand rule.
Hence in the east direction the magnetic force act on the electron.
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A major artery with a cross-sectional area of 1.00cm2 branches into 18 smaller arteries, each with an average cross-sectional area of 0.400cm2. By what factor is the average velocity of the blood reduced when it passes into these branches
Answer:
The velocity in the smaller arteries will be reduced by a factor of 0.139
Explanation:
The flow rate of blood is going to stay the same when it is transferred from the major artery to the smaller ones.
flow rate = Velocity * Area
Since the flow rate remains constant, we have:
Flow rate in major artery = combined flow rate in smaller arteries
Velocity in Major artery * 1.00 = Velocity in smaller artery * (0.4 * 18)
[tex]V_M * 1 = V_S * (18*0.4)[/tex]
[tex]\frac{V_S}{V_M}=\frac{1}{18*0.4}[/tex]
[tex]\frac{V_S}{V_M}= 0.139[/tex]
Thus, the velocity in the smaller arteries will be reduced by a factor of 0.139
(8%) Problem 3: Sound in water travels at a velocity governed by the relation v = √(B/rho) where B is the bulk modulus and rho is the density. For salt water, take B = 2.38 × 109 Pa and rho = 1046 kg/m3. A whale sends out a high frequency (9 kHz) song to another whale 1.0 km away. How much time, in seconds, does it take for the sound to travel between the whales, t?
Answer:
the time required for the sound to travel between the whales 0.66 S.
Explanation:
As given in the problem, the velocity of sound wave ([tex]v[/tex]) is governed by the equation
[tex]v = \sqrt{\dfrac{B}{\rho}}[/tex]
Given, [tex]B = 2.38 \times 10^{9} Pa[/tex] and [tex]\rho = 1046 Kg m^{-3}[/tex]
So for salt water, the velocity of sound wave ([tex]v_{s}[/tex]) can be written as
[tex]v_{s} = \sqrt{\dfrac{2.38 \times 10^{9}}{1046}} ms^{-1} = 1.508 \times 10^{3} ms^{-1}[/tex]
As the whales are d = 1 Km or 1000 m apart from each other, so the time ([tex]t[/tex]) required for the sound wave to travel this distance is given by
[tex]t = \dfrac{d}{v_{s}} = \dfrac{1000 m}{1.508 \times 10^{3}} = 0.66 s[/tex]
To find the time it takes for the sound to travel between two whales in water, use the equation v = √(B/ρ) where v is the velocity, B is the bulk modulus, and ρ is the density. Rearrange the equation to solve for time, t, using the formula t = distance / velocity.
Explanation:To find the time it takes for the sound to travel between the whales, we can use the equation v = √(B/ρ), where v is the velocity, B is the bulk modulus, and ρ is the density. In this case, B = 2.38 × 109 Pa and ρ = 1046 kg/m3. We can rearrange the equation to solve for the time, t: t = d/v, where d is the distance between the whales. In this case, d = 1.0 km = 1000 m. Plugging in the values, we get t = 1000 / √(2.38 × 109 / 1046). Simplifying this expression gives us the time it takes for the sound to travel between the whales.
The power rating of a 400-Ω resistor is 0.800 W.(a) What is the maximum voltage that can be applied across this resistor without damaging it? Use three significant figures in your answer.
Answer:
[tex]V=17.9\ Volt[/tex]
Explanation:
Joule's Law in Electricity
The Joule's law allows us to calculate the power dissipated in a resistor of resistance R through which goes a current I.
[tex]P=I^2R[/tex]
The relation between the voltage and the current is given by Ohm's law:
[tex]V=RI[/tex]
Solving for I and replacing int the first equation
[tex]\displaystyle P=\frac{V^2}{R}[/tex]
Solving for V
[tex]V=\sqrt{PR}[/tex]
[tex]V=\sqrt{0.8\cdot 400}=17.9[/tex]
[tex]\boxed{V=17.9\ Volt}[/tex]
Maximum voltage for given power rating, applied across this resistor without damaging it is 17.9 volts.
What is the Ohm's law?Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it. Thus,
[tex]V\propto I[/tex]
Here, (V) is the potential difference and (I) is the current.
It can be written as,
[tex]V=IR[/tex]
Here, (R) is the resistance of the circuit.
Given information-
The value of resistance is 400-Ω.
The value of power rating is 0.800 W.
By the Joule's law, the power of a circuit is equal to the product of the square of the current flowing in it and the resistance. It can be given as,
[tex]P=I^2\times R\\I=\sqrt{\dfrac{P}{R}}[/tex]
Put this value of current in ohm's law as,
[tex]V=\sqrt{\dfrac{P}{R}}\times R\\V=\sqrt{PR}[/tex]
Put the value of power and current in the above formula,
[tex]V=\sqrt{0.800\times 400}\\V=17.9\rm Volts[/tex]
'
Thus the maximum voltage that can be applied across this resistor without damaging it is 17.9 volts.
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how is Newton’s three laws of motion are used in testing the safety of our automobiles.
Answer
Newton's third law is used for the impact test on the vehicle.
In the frontal collision of the vehicle, the impact of the vehicle is made against the concrete wall then the impact on the passengers and driver is calculated.
In the impact test time of the opening of the airbag is also calculated.
Similar tests also take place when the impact is sideways.
The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10−18 JJ of kinetic energy in this distance?
The solution is in the attachment
Answer:
[tex]a=1.1*10^{18}\frac{m}{s^2}[/tex]
Explanation:
We use the following kinematic formula to calculate the acceleration:
[tex]v_f^2=v_0^2+2ax[/tex]
The kinetic energy is defined as:
[tex]\Delta K=\frac{m(v_f^2-v_0^2)}{2}\\v_f^2-v_0^2=\frac{2\Delta K}{m}[/tex]
Replacing this in the acceleration formula and solving for a:
[tex]\frac{2\Delta K}{m}=2ax\\a=\frac{\Delta K}{mx}\\a=\frac{2*10^{-18}J}{(9.1*10^{-31}kg)(2*10^{-6}m)}\\a=1.1*10^{18}\frac{m}{s^2}[/tex]
A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.
What is the strength of the electric field?
To solve this problem we will apply the concepts related to the potential energy in the molecules and obtain its electric field through the relationship given by the dipole moment. The change in potential energy from one state to another is given by,
[tex]U_2-U_1 = 1.4*10^{-21} J[/tex]
From this difference we can identify that [tex]U_1[/tex] is equivalent to the potential energy when it is perpendicular to the electric field. At the same time, the potential energy [tex]U_2[/tex] would be equivalent when it is aligned with the electric field.
From there the relationship between energy, the dipole moment and the electric field would be subject to
[tex]U_2-U_1 = pE[/tex]
Here,
[tex]p = \text{Dipole moment} = 6.2*10^{-30} C \cdot m[/tex]
Rearranging to find the electric field,
[tex]E = \frac{(U_2-U_1)}{p}[/tex]
[tex]E = \frac{(1.4*10^{-21})}{(6.2*10^{-30})}[/tex]
[tex]E =2.25*10^8 N/C[/tex]
Therefore the electric field is [tex]2.25*10^8N/C[/tex]