Answer:
d. R4
Explanation:
Generally, the flow of current is always from the positive sign to the negative sign. In the resistors R1, R2, and R3, the direction of flow of current is from the positive sign to the negative sign. However, in the resistor R4, the direction of the flow of current is different from the conventional method. Therefore, the resistor R4 is marked wrongly.
Resistance R₄ is wrongly connected in the circuit.
Current always flow in the opposite direction of flow of electrons.
Since, Resistor acts as load in the electrical circuit. So, it will always consumed power.
Thus, In resistor current always flow from higher potential to lower potential i.e. from positive terminal to negative terminals.
In given circuit, current flowing in resistances R₁, R₂ and R₃ , from positive to negative. But in resistance R₄ current is flowing from negative to positive.
Therefore, Resistance R₄ is wrongly connected in the circuit.
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An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel at 1530 m/s in seawater. The time delay of the echo to the ocean floor and back is 6 s. ?
Answer:
d = 4590 m
Explanation:
given,
Speed of ultrasonic wave, v = 1530 m/s
time of the echo, t = 6 s
Let d be the depth of the ocean
now,
total distance travel by the ultrasonic wave = 2 d
we know,
distance = speed x t
2 d = 1530 x 6
d = 1530 x 3
d = 4590 m
Hence, the depth of the ocean floor is equal to d = 4590 m
The difference between the full wave rectifier with the center tap transformer, and the full wave bridge rectifier is_______________. Question 9 options: in full wave bridge rectifier two didoes conduct during each half of the cycle whereas one diode conducts during each half cycle in the center tapped full wave rectifier. transformer configuration is simpler in the full wave bridge rectifier. that the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Answer:
That the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Explanation:
See Image
Final answer:
In the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. The peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
Explanation:
The difference between the full wave rectifier with the center tap transformer and the full wave bridge rectifier is that in the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. Additionally, the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
Two cellists, one seated directly behind the other in an orchestra, play the same 220-Hz note for the conductor who is directly in front of them. What is the smallest non-zero separation that produces constructive interference?
Answer:
d= 1.56 m
Explanation:
In order to have a constructive interference, the path difference between the sources of the sound, must be equal to an even multiple of the semi-wavelength, as follows:
⇒ d = d₂ - d₁ = 2n*(λ/2)
The minimum possible value for this distance, is when n=1, as it can be seen here:
dmin = λ
In any wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:
v = λ*f
If v = vsound = 343 m/s, and f = 220 1/s, we can solve for λ:
λ =[tex]\frac{v}{f} = \frac{343 m/s}{220(1/s)} = 1.56 m[/tex]
⇒ dmin =λ = 1.56 m
Consider an electrical transformer that has 10 loops on its primary coil and 20 loops on its secondary coil. What is the current in the secondary coil if the current in the primary coil is 5.0 A?A. 5.0 AB. 10.0 AC. 2.5 AD. 20.0 A
Answer:
C. 2.5 A
Explanation:
Transformer: A transformer is an electromechanical device that is used to change the voltage of an alternating current.
The current and the number of loops in a transformer is related as shown below
Ns/Np = Ip/Is........................... Equation 1
Where Ns = Secondary loop, Np = primary loop, Ip = primary current, Is = secondary current.
Making Is the subject of the equation
Is = NpIp/Ns........................ Equation 2
Given: Np = 10 loops, Ns = 20 loops, Ip = 5.0 A.
Substitute into equation 2
Is = (10×5.0)/20
Is = 50/20
Is = 2.5 A.
Hence the current in the primary coil = 2.5 A.
The right option is C. 2.5 A
Final answer:
The current in the secondary coil of an electrical transformer with 10 loops in the primary coil and 20 loops in the secondary coil, given a primary current of 5.0 A, is 2.5 A.
Explanation:
The question asks for the current in the secondary coil of an electrical transformer, given the current in the primary coil and the number of loops in both the primary and secondary coils. To find the current in the secondary coil, we use the principle of conservation of energy in a transformer, which states that the power input to the primary coil (P1) equals the power output from the secondary coil (P2), assuming an ideal scenario without any losses. The formula for power is P = IV, where I is current and V is voltage. Therefore, the ratio of the currents in the primary and secondary coils is inversely proportional to the ratio of the number of turns in the primary and secondary coils: I1/I2 = N2/N1. Given that N1 = 10 loops and N2 = 20 loops with a primary current (I1) of 5.0 A, we find that I2 = 2.5 A. Thus, the correct answer is C. 2.5 A.
A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, where α=60.0N/m and β=18.0N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for this spring. Let U = 0 when x = 0. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the +x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the x = 0 equilibrium position?
Answer:U(x) = 30x^2 +6x^3
V^2=8.28m/s
Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1
Kinetic energy K.E= 1/2 mv^2
Restoring force function F(x)= -60x - 18x^2
But F(x)= -dU/dx
dU(x)=-F(x)dx
Integrating U(x)= -integral F(x)dx + U(0)
Substituting, we get
U(x) = - integral(-60x-18x^2)dx+U(0)
U(x)= 30x^2+6x^3+U(0)
U=0 at x=0
Therefore U(x)= 30x^2+6x^3
b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?
Substituting into eq (a)
U1= 30(1.00)^2+6(1.00)^3=36J
Using x2=0.5 into eq(a)
U2=30(0.50)^2+6(0.50)^3=8.25J
Object at rest K1=0
0+36=K2+8.25
K2=27.75J
Given; m =0.900kg, V2=?
27.75=1/2×0.900×V2^2
V2= SQRT(2×27.75)/0.81
V2= 8.28m/s
The relationship between force, potential energy and energy conservation allows to find the results for the questions about the spring are:
a) The potential energy is: U = 30 x² + 6 x³
b) The velocity is: v = 7.85 m / s
Given parameters.
Restorative force f = - α x - β x² Constants values α = 60.0 N / m and β = 18.0 N / m² Body mass m = 0.900 kg Displacement initial x₁ = 1,0 m and final x₂ = 0,5 mTo find
a) Potential energy.
b) Speed.
a) Force and potential energy are related by the expression.
[tex]F = - \frac{dU}{dx}[/tex]
Where F is the force and U the potential energy.
dU = - F dx
∫ dU = - ∫∫ (-α x - β x²) dx
U- U₀ = [tex]\alpha \frac{x^2}{2} + \beta \frac{x^3}{3}[/tex]alpha / 2 x² + beta / 3 x³
Let's substitute the constants values and indicate that U₀=0 when x=0.
U = [tex]30 x^2 + 6x^3[/tex]
b) They ask the speed of the block between two points, as they indicate that there is no friction we can use the theorem of conservation of mechanical energy, which states that energy is conserved at all points.
Starting point.
Em₀ = U (1)
Final point.
[tex]Em_f[/tex] = K + U (0.5)
Energy is conserved.
Em₀ = Em_f
U (1) = K + U (0.5)
Where the kinetic energy is
K = ½ m v²
Let's substitute.
v² = [tex]\frac{2}{m} [ U(1) - U(0.5)][/tex]
Let's calculate.
v² = [tex]\frac{2}{0.900}[/tex] [ (30 1² + 6 1³) - (30 0.5² + 6 0.5³) ]
v = [tex]\sqrt{\frac{2 \ 27.75 }{ 0.900} }[/tex]
v = 7.85 m / s
In conclusion using the relationship between force, energy potential and the conservation of energy we can find the results for the questions about the spring are:
a) The potential energy is: U = 30 x² + 6 x³
b) The velocity is: v = 7.85 m / s
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Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate (see the figure ). Typically, the diameter of this membrane is about 8.40 {\rm mm} in humans.
a.)how much energy is delivered to the eardrum each second when someone whispers (20.0 {\rm dB}) a secret in your ear?
b.)To comprehend how sensitive the ear is to very small amounts of energy, calculate how fast a typical 2.00 {\rm mg} mosquito would have to fly (in {\rm mm/s}) to have this amount of kinetic energy.
Answer:
a) Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s
b) velocity of mosquito that will generate that amount of energy, v = 0.0000744 m/s = 0.0744 mm/s.
Explanation:
a) [D] = 10 log (I/I₀)
I₀ = 10⁻¹² W/m²
Given the sound intensity level in decibels, we need to obtain the corresponding sound intensity.
20 = 10 log (I/(10⁻¹²))
2 = log (I/(10⁻¹²))
100 = (I/(10⁻¹²))
I = 10⁻¹⁰ W/m²
Power experienced by the tympanic membrane of the ear due to the sound intensity = Intensity × Area of the membrane
Area of the membrane = πD²/4 = π(8.4 × 10⁻³)²/4 = 5.54 × 10⁻⁵ m²
Power = 10⁻¹⁰ × 5.54 × 10⁻⁵ = 5.54 × 10⁻¹⁵ W
Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s
b) Kinetic energy = mv²/2
5.54 × 10⁻¹⁵ = (2 × 10⁻⁶)v²/2
v² = (2 × 5.54 × 10⁻¹⁵)/(2 × 10⁻⁶)
v = 0.0000744 m/s = 0.0744 mm/s.
The definition of decibels and the relationship between work and kinetic energy allows us to find the results for the questions about the system:
a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵ W.
b) The speed of the mosquito with this energy is: v= 7.53 10⁻² m/s
Given parameters.
Eardrum diameter d= 8.40 mm = 8.40 10-3 m. Mosquito mass m= 2.00 mg = 2.00 10-6 kg Whisper sound intensity. Beta = 20dBTo find.
a) The energy per second.
b) The kinetic energy of a mass mosquito
Decibels definition.The intensity of sound is in a very wide range of magnitudes, to simplify its use, the decibel is defined as a logarithmic unit.
[tex]\beta = 10 \ log (\frac{I}{I_o})[/tex]
where β are the decibels, I the intensity and I₀ the reference intensity. In the case of humans, the sensitivity threshold is of the order of 10⁻¹² W/m²
The intensity of the expression is:
[tex]\frac{I}{I_o} = 10^{\beta/10 }[/tex]
[tex]I = I_o\ 10^{\beta/10}[/tex]
Let's calculate
I = [tex]10^{-12} \ 10^{20/10}[/tex]
I = 10⁻¹⁰ W/m²
The intensity is defined by the energy deposited per unit of time and area.
I = [tex]\frac{P}{A}[/tex]
P = I A
Let's calculate the area of the eardrum.
A = π r² = [tex]\pi \ \frac{d^2}{4}[/tex]
Let's calculate.
A = [tex]\frac{\pi}{4} (8.4 \ 10^{-3} )^2[/tex]
A = 5.67 10⁻⁵ m²
Let's calculate the power.
P = 10⁻¹⁰ 5.67 10⁻⁵
P = 5.67 10⁻¹⁵5W
b) Power is work per unit of time.
P = [tex]\frac{W}{t}[/tex]
W= P t
The work is equal to the change in kinetic energy, if we assume that the mosquito starts from rest.
W = ΔK = [tex]K_f - K_o[/tex]
W = ½ mv²
v² = [tex]\frac{2 K}{m}[/tex]
Let's calculate
v² = = 5.67 10⁻⁹
v= 7.53 10⁻⁵ m/s
v= 7.53 10⁻² mm/s
In conclusion using the definition of decibels and the relationship of work and kinetic energy we can find the results for the questions about the system are:
a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵ W.
b) The speed of the mosquito with this energy is: v= 7.53 10⁻² mm/s
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Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 with magnitude 71 N and direction 20°. All direc- tion angles 0 are measured from the positive x axis: counter-clockwise for 0 > 0 and clock- wise for 0 < 0. What is the magnitude of the resultant vec- tor || F ||, where F = Fi + F2 + 3 ? Answer in units of N. 004 (part 2 of 2) 10.0 points What is the direction of É as an angle between the limits of -180° and +180° from the positive x axis with counterclockwise as the positive angular direction? Answer in units of 005 10.0 points Consider the instantaneous velocity of a body. This velocity is always in the direction of 1. the least resistance at that instant. 2. the net force at that instant. 3. the motion at that instant.
To find the magnitude and direction of the resultant vector, use the Pythagorean theorem and inverse tangent function respectively.
Explanation:To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The resultant vector is the sum of the three force vectors: F1, F2, and F3. We can find the x and y components of each vector using trigonometry, and then add the components to find the x and y components of the resultant vector. Finally, we can use the Pythagorean theorem to find the magnitude of the resultant vector.
To find the direction of the resultant vector, we can use the inverse tangent function to find the angle between the positive x-axis and the resultant vector. Since the problem specifies that angles are measured counterclockwise from the positive x-axis, we need to make sure the angle is within the range of -180° to +180°. If the angle is greater than 180°, we subtract 360° to get the equivalent angle within the specified range.
The magnitude of the resultant vector is 95.2 N and the direction is -95.5°.
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A gamma ray burst produces radiation that has a period of 3.6x10-21 s. What wavelength does this radiation have?
Answer:
The radiation wavelength is 1.08 X 10⁻¹² m
Explanation:
Frequency is the ratio of speed of photon to its wavelength
F = c/λ
where;
c is the speed of the photon = 3 x 10⁸ m/s
λ is the wavelength of gamma ray = ?
F is the frequency of the gamma ray = 1/T
T is the period of radiation = 3.6x10⁻²¹ s
[tex]\frac{1}{T} = \frac{c}{\lambda}[/tex]
λ = T*C
λ = 3.6x10⁻²¹ * 3 x 10⁸
λ = 1.08 X 10⁻¹² m
Therefore, the radiation wavelength is 1.08 X 10⁻¹² m
A 52 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. The acceleration of gravity is 9.8 m/s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.
Answer:
6.1 m
Explanation:
m = Mass of person = 52 kg
h = Altitude
v = Velocity
Kinetic energy of the person on the ground
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}52\times 11^2\\ =3146\ J[/tex]
Kinetic energy of the person at the top
[tex]\dfrac{1}{2}52\times 1.2^2\\ =37.44\ J[/tex]
At the top the potential energy is given by
[tex]mgh=52\times 9.8h[/tex]
Balancing the energy of the system
[tex]3146=37.44+52\times 9.8h\\\Rightarrow h=\dfrac{3146-37.44}{52\times 9.8}\\\Rightarrow h=6.1\ m[/tex]
Her altitude is 6.1 m
Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/s2. How long after car A begins to accelerate does it take car A to overtake car B?A) 12.6 s B) 5.50 s C) 19.0 s D) 316 s E) Car A never overtakes car B.
Answer:
(c) 19.0s
Explanation:
Given Data
Car A speed v=22.0 m/s
Car B speed v=29.0 m/s
Car A distance S=300 m behind Car B
Car A acceleration a=2.40 m/s²
To find
Time required For Car A to take over Car B
Solution
We can represent Car A Coordinate by using equation of simple motion
[tex]X_{A} =vt+1/2at^{2}\\X_{A} =22t+(1/2)(2.40)t^{2}[/tex]
And Coordinates of car B equals
[tex]X_{B}=300+29t\\[/tex]
Car A is overtake car B when:
[tex]X_{A}=X_{B}\\ 22t+(1/2)(2.4)t^{2}=300+29t\\1.2t^{2}-7t-300=0\\ time=19.0s[/tex]
Option (C) 19.0s is correct one
By setting the distance traveled by car A equal to the distance traveled by car B plus the initial separation and solving the resulting quadratic equation, it is determined that car A will overtake car B in 20 seconds.
Explanation:To find out how long it takes for car A to overtake car B, we must calculate the respective positions of the cars as a function of time and find when they match. Car A is accelerating from an initial speed, while car B is moving at a constant speed.
The position of car A as a function of time can be found with the equation:
sA = vA0t + ½at2
Where:
The position of car B as a function of time (since car B is not accelerating) is simply:
sB = vBt
Where:
Car A will overtake car B when sA equals sB plus the initial 300 m separation. We can set up the equation and solve for t.
22t + ½(2.40)t2 = 29t + 300
Rearranging to solve for t gives a quadratic equation:
1.2t2 - 7t - 300 = 0
Using the quadratic formula or factoring, we solve for t. It yields t = 20 s (after discarding the negative solution which is not physically meaningful).
Therefore, it takes car A 20 seconds to overtake car B, which results in Option C being the correct answer.
An electroscope is charged by touching its top with positive glass rod. The electroscope leaves spread apart and the glass rod is removed. Then a negatively charged plastic rod is brought close to the top of the electroscope, but it does not touch. What happens to the leaves?
Answer: The leaves will spread.
Explanation: Since the top of the electroscope has been touched initially by a positive glass rod, the charge on it is positive ( this is charging by conduction).
By bringing a negative charged rod towards ( but not touching the top) the top of electroscope means we will be charging the electroscope by induction. Charging by induction implies that an opposite charge of the conductor (negative charged rod) we are using to charge will be formed on the conductor that we want to charge (already positive charged electroscope)
In this case, our plastic rod is negative and it is brought towards the top of the electroscope ( which is already positively charged), there will be an induced positive charge on the electroscope.
So we have positive charge on the electroscope by a positive rod ( charging by conduction) and a positive charge from a negative rod ( charging by induction) thus the leaves will spread meaning the charges are repelling based on the fact that opposite charges attract and like charges repel
In the presence of a negatively charged rod, the positively charged electroscope will have its positive charges attracted toward the negative charge. This re-distributes the charges within the electroscope causing the leaves to move closer together.
Explanation:An electroscope is a device used to detect and measure electric charge. After it is charged with a positive charge (from the glass rod), the leaves spread apart due to like charges repelling each other. When a negatively charged plastic rod is brought near (but not touching) the electroscope, the positive charges in the electroscope are attracted to the negative charge of the rod. This causes the leaves of the electroscope to move closer together as most of the positive charges are attempting to move toward the top of the electroscope to be closer to the negative charge.
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Vector A has a magnitude of 5.0 m and points east, vector B has a magnitude of 2.0 m and points north, and vector C has a magnitude of 7.0 m and points west. The resultant vector A + B + C is given by
Answer:
The answer to your question is Vr = 2.83 m, to the Northwest
Explanation:
Data
Vector A = 5 m
Vector B = 2.0 m
Vector C = 7.0 m
Process
1.- Calculate the ∑Vx and ∑Vy
∑Vx = 5m - 7m = -2m Vectors substract because they are in opposite directions
∑Vy = 2m
2.- Calculate the resultant vector with the Pythagorean theorem
Vr² = Vx² + Vy²
Vr² = (-2)² + (2)²
Vr² = 4 + 4
Vr² = 8
Vr = 2.83 m
3.- Calculate the direction
tan Ф = 2/-2 = 1
tan⁻¹Ф = 45° to the Northwest
Final answer:
The resultant vector A + B + C has a magnitude of approximately 2.83 m and it points north-west, after subtracting the east-west components and adding the north-south component with no opposition.
Explanation:
To find the resultant vector A + B + C, we need to consider the directions and magnitudes of each vector. Vector A has a magnitude of 5.0 m and points east, vector B has a magnitude of 2.0 m and points north, and vector C has a magnitude of 7.0 m and points west. We can calculate the overall resultant vector by adding up the components in the east-west direction and the north-south direction separately.
Since east and west are opposite directions, we subtract the magnitudes of vectors A and C, which point in these directions:
East-West component: 5.0 m (east) - 7.0 m (west) = -2.0 m (west)Vector B points north and has no opposing southward vector, so its component remains unchanged:
North-South component: 2.0 m (north)Now, to find the resultant vector's magnitude, we can use the Pythagorean theorem:
Resultant magnitude = \\(\sqrt{(-2.0)^2 + 2.0^2} m\\) = \sqrt{4 + 4} m = \sqrt{8} m = 2.83 m (to two decimal places)
The resultant vector has a magnitude of approximately 2.83 m and it points north-west, considering that the east-west component is in the west direction while the north-south component points directly north.
If a bucket with water is hung to a spring balance and then a weight with the help of a thread is fully immersed inside the bucket such that it does not touch sides of the bucket, will the spring balance reading change or remain same?
Answer:
Explanation:
The weight fully immersed will displaced water of equal volume to itself. The weight of this water displaced = volume of water displaced × density of water × acceleration due to gravity. This increase in weight will lead the spring balance reading to change; increase because there is an increase in mass.
A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \vec{F} =(33 N)\hat{i} - (41 N)\hat{j} to the cart as it undergoes a displacement \vec{s} = (-9.4 m)\hat{i} - (3.1 m)\hat{j}.
Part A
How much work does the force you apply do on the grocery cart?
Express your answer using two significant figures.
W =
{\rm J}
Answer:
[tex]W=-183.1\ J[/tex]
Explanation:
Given:
force applied, [tex]\vec{F} =(33 N)\hat{i} - (41 N)\hat{j}[/tex]
displacement caused, [tex]\vec{s} = (-9.4 m)\hat{i} - (3.1 m)\hat{j}[/tex]
Work done by the force on the cart:
[tex]W=\vec F.\vec s[/tex]
[tex]W=[(33 N)\hat{i} - (41 N)\hat{j}].[(-9.4 m)\hat{i} - (3.1 m)\hat{j}][/tex]
[tex]W=-310.2+127.1[/tex]
[tex]W=-183.1\ J[/tex]
Negative work means that the force and displacement have an obtuse angle between them.
Answer:
-180 J
Explanation:
We are given that
Constant force=[tex]F=(33 N)\hat{i}-(41 N)\hat{j}[/tex]
Displacement=[tex]\vec{s}=(-9.4m)\hat{i}-(3.1m)\hat{j}[/tex]
We have to find the work done .
We know that
Work done=[tex]F\cdot s[/tex]
Using the formula
Work done=[tex](33i-41j)\cdot (-9.4i-3.1j)[/tex]
Work done =[tex]33i\cdot (-9.4)i+41j\cdot 3.1 j[/tex]
By using rule [tex]i\cdot i=j\cdot j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=i\cdot k=k\cdot j=j\cdot i=0[/tex]
Work done=[tex]-310.2+127.1[/tex]
Work done=-183.1 J
We have to write answer in two significant figures.
When units digit 3 is less than 5 then digits on left side of 3 remains same and digits on right side of 3 and 3 will be replace by zero
Work done=-180 J
Hence, the work done =-180 J
In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of the tank through a nozzle that is directed straight upward. The opening of the nozzle is 2.86 m below the surface of the water. (a) Find the speed at which the water leaves the nozzle. (b) Ignoring air resistance and viscous effects, determine the height (relative to the submerged end of nozzle) to which the water rises.
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
What scenarios best describes how the hawaiian islands formed in the pacific ocean?
Answer:
Magma generated from a hot spot burned through the overlying plate to create volcanoes.
Explanation:
The Earth’s outer crust is made up of a series of tectonic plates that move over the surface of the planet. In areas where the plates come together, volcanoes will form in most cases. Volcanoes could also form in the middle of a plate, where magma rises upward until it erupts on the seafloor which is called a hot spot.
Hawaiian Islands were formed by such a hot spot occurring in the middle of the Pacific Plate. While the hot spot itself is fixed, the plate is moving. As the plate moved over the hot spot, the string of islands that make up the Hawaiian Island were formed.
Marcus used a toaster oven in the morning.He notices that when he plug it in and turn it on the coils inside begin to glow red what transformation are taking place
Answer:Conversion of electric energy to Heat energy
Explanation:Energy is a quantitative energy measured in JOULES or KILOJOULES which must be transferred to a material for a job to be done. It has also been described as the capacity to do work.
In electric toasters the ELECTRIC ENERGY FROM THE SOURCE IS TRANSFERRED INTO THE TOASTER TO BE CONVERTED TO HEAT ENERGY NEEDED TO TOAST FOODS. Other electrical appliances which converts electric energy to Heat energy includes ELECTRIC BOILERS, ELECTRIC COOKERS etc.
if a material has a half-life of 24 hours, how long do you have to wait until the amount of radioisotope is 1/4 its original amount?
Answer:
48 hours
Explanation:
Using the formula,
R/R' = 2ᵃ/ᵇ..................... Equation 1
Where R = Original amount, R' = Radioactive remain, a = Total time, b = half life.
Given: b = 24 hours,
Let: R = X, then R' = X/4.
Substitute into equation 1
X/(X/4) = 2ᵃ/²⁴
4 = 2ᵃ/²⁴
2² = 2ᵃ/²⁴
Equating the base and solving for a
2 = a/24
a = 24×2
a = 48 hours.
Hence the time = 48 hours
The half-life of a material is the time it takes for half of the atoms in a sample to decay. If a material has a half-life of 24 hours, then after 48 hours (i.e., two half-lives), the amount of the material will be 1/4 of its original amount.
Explanation:The half-life concept applied in this question falls under the subject of Physics, specifically nuclear physics. The half-life of a radioactive material is the time it takes for half of the atoms in a sample to decay. If a material has a half-life of 24 hours, then after 24 hours, half of the original material would remain.
Given this, if we want the amount of the material to be only 1/4 of its original amount, we simply wait for another half-life. Remember, each half-life reduces the original amount by half. So after the first 24 hours (the first half-life), half of the material would have decayed. If you wait another 24 hours (another half-life), half of what remained after the first half-life would decay again, leaving you with 1/4 of the original amount.
So, you would have to wait 48 hours for the amount of radioisotope to be 1/4 of its original amount assuming the half-life is 24 hours.
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A shoreline runs north-south, and a boat is due east of the shoreline. The bearings of the boat from two points on the shore are 110° and 100°. Assume the two points are 550 ft apart. How far is the boat from the shore?
Answer:
2930.90 ft
Explanation:
*Attached are two rough sketches I made to represent the problem.
In diagram 2, the bearings are represented relative to the boat's position.
To find x, the distance between the boat and point having bearing 110° to the boat, we can use sine rule:
(sin 10°) / 550 = (sin 100) / x
=> x = (550 * sin 100°) / sin 10°
x = 3119 ft
Having found this, we can now find the distance between the host and the shore, as represented in diagram 1.
Using trigonometric function of SOHCAHTOA, we have that:
cos 20° = y / 3119
=> y = 3119 * cos 20°
y = 2930.90 ft
Hence, the distance between the boat and the shore is 2930.90 ft
A 4.0-kg mass is moving with speed 2.0 m/s. A 1.0-kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?
Answer
given,
mass of object 1 = 4 Kg
Speed of object 1 = 2 m/s
mass of object 2 = 1 Kg
speed of object 2 = 4 m/s
KE of the object 1 = [tex]\dfrac{1}{2}MV^2[/tex]
= [tex]\dfrac{1}{2}\times 4 \times 2^2[/tex]
= 8 J
KE of the object 2 = [tex]\dfrac{1}{2}MV^2[/tex]
= [tex]\dfrac{1}{2}\times 1 \times 4^2[/tex]
= 8 J
Kinetic energy of both the object is same hence,Work done by both the object will also be same.
It is given that braking force is same in both cases.
So, distance travel by Both the object will be same.
Both of them cover the same amount of distance.
Given that;
4 kg mass is moving with speed 2.0 m/s
1.0 kg mass is moving with speed 4.0 m/s.
Both objects are brought to a halt by the same steady braking force.
Their momentum will be proportionate to their weight in the opposite direction = 1:4 ratio
Distance before coming to rest
v² - u² =2as
So,
s1/s2 = 16/16
Both of them cover the same amount of distance.
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Merry-go-rounds are a common ride in park play-grounds. The ride is a horizontal disk that rotates about a vertical axis at their center. A rider sits at the outer edge of the disk and holds onto a metal bar while someone pushes on the ride to make it rotate. Suppose that a typical time for one rotation is 6.0 s and the diameter of the ride is 16 ft.
A) For this typical time, what is the speed of the rider in m/s?
B) What is the rider's radial acceleration, in m/s?
C) What is the rider's radial acceleration if the time for one rotation is halved?
The speed of the rider is 0.81 m/s, the radial acceleration is 0.338 m/s², and if the time for one rotation is halved, the speed becomes 0.405 m/s and the radial acceleration becomes 0.085 m/s².
Explanation:A) To calculate the speed of the rider in m/s, we can use the formula:
Speed = Distance / Time
The distance traveled by the rider in one rotation is equal to the circumference of the ride, which is given as the diameter multiplied by π (pi).
Therefore, the distance = 16 ft × π
To convert this distance to meters, we multiply by the conversion factor 0.3048 m = 1 ft.
So, the distance in meters = 16 ft × 0.3048 m/ft × π
Given that the time for one rotation is 6.0 s, we can now calculate the speed:
Speed = (16 ft × 0.3048 m/ft × π) / 6.0 s
Simplifying this equation gives us:
Speed ≈ 0.81 m/s
B) The radial acceleration of the rider can be calculated using the formula:
Radial Acceleration = (Speed)² / Radius
Given that the radius of the ride is half the diameter, which is 8 ft, we can substitute the values into the formula:
Radial Acceleration = (0.81 m/s)² / (8 ft × 0.3048 m/ft)
Simplifying this equation gives us:
Radial Acceleration ≈ 0.338 m/s²
C) If the time for one rotation is halved, the speed of the rider will also be halved because speed is distance divided by time. Therefore, the new speed would be 0.81 m/s / 2 = 0.405 m/s.
The radial acceleration can then be calculated using this new speed and the same formula as in part B:
Radial Acceleration = (0.405 m/s)² / (8 ft × 0.3048 m/ft)
Simplifying this equation gives us:
Radial Acceleration ≈ 0.085 m/s²
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If you fire a bullet through a board, it will slow down inside and emerge at a speed that is less than the speed at which it entered. Does light, then, similarly slow down when it passes through glass and also emerge at a lower speed? Defend your answer.
Answer:
RUn he got gun
Explanation:
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.60 105 Pa. Assuming that the top and bottom surfaces of the cap each have an area of 3.70 10-4 m2, obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.
Answer:
F tread = 21.8N
Explanation:
In order to find the force that the screw thread exert on the cap, use equation 11.3 taking into consideration that the cap is in equilibrium
Making the vertical net force equal zero .
Sum Fy= - F tread+ Inside -F outside=0
F tread = F inside- F out side = P inside A- P out side A =
(P inside- P outside) A.=
((160000pa)-(101000pa))* 0.00037
21.8N
Using Newton's Version of Kepler's Third Law II The Sun orbits the center of the Milky Way Galaxy every 230 million years at a distance of 28,000 light-years. Use these facts to determine the mass of the galaxy. (As we'll discuss in Chapter Dark Matter, Dark Energy, and the Fate of the Universe, this calculation actually tells us only the mass of the galaxy within the Sun's orbit.) M= solar billion years
Final answer:
To find the mass of the Milky Way galaxy, we apply a version of Kepler's Third Law using the orbital period and radius of the Sun's orbit. We convert units to meters and seconds, calculate the Sun's orbital velocity, and use this alongside the gravitational constant to estimate the galaxy's mass within the Sun's orbit.
Explanation:
To determine the mass of the Milky Way galaxy using the information given, we can refer to a version of Kepler's Third Law tailored for galactic scales, which allows us to estimate the mass of the galaxy based on the orbital period and radius of an orbiting object—in this case, our Sun. The version of the law used in galactic dynamics is:
M = (v^2 x R) / G
where M is the mass of the galaxy within the Sun's orbit, v is the orbital speed of the Sun, R is the radius of the Sun's orbit, and G is the gravitational constant.
Calculation Steps:
First, we need to convert the orbital period from million years to seconds, as follows: 230 million years x (365.25 days/year) x (24 hours/day) x (3600 seconds/hour).
Next, convert the radius of the Sun's orbit from light-years to meters using the fact that one light-year is approximately 9.461 x 10^15 meters.
Now, we can calculate the orbital velocity of the Sun using the circumference of its orbit (2 x π x R) and the orbital period found in step 1 to obtain v = (2 x π x R) / period.
Finally, apply Kepler's Third Law to find the mass M using the velocity v from step 3, the radius R from step 2, and the known value of the gravitational constant G.
Performing these calculations would result in an estimate for the Milky Way's mass within the Sun's orbit.
Important Note
It is critical to understand that these calculations only provide the mass within the Sun's orbit. There is additional mass outside the Sun's orbit, much of which is thought to be dark matter, that is not accounted for in this simple model.
At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.40 m/s^2. At the same instant, a truck, traveling with a constant speed of 15.5 m/s, overtakes and passes the automobile.
How far beyond its starting point does the automobile overtake the truck?
How fast is the automobile traveling when it overtakes the truck?
By setting the distance equations for both the accelerating automobile and the constant-speed truck equal to each other, we find the automobile overtakes the truck 151.666 meters beyond its starting point, traveling at 31.008 m/s.
Explanation:To solve the problem, we must consider the equations of motion for both the automobile and the truck. The key to solving this problem is to set their distances equal to find the point where the automobile overtakes the truck, because this equal distance indicates the same position for both vehicles at the same time.
For the automobile, initially at rest and accelerating from the traffic light, we use the equation of motion s = ut + (1/2)at2, where s is the distance, u is the initial velocity (0 m/s in this case), a is the acceleration (2.40 m/s2), and t is the time. Substituting the known values, we get s = 0*t + (1/2)*2.40*t2 = 1.2t2.
For the truck, traveling at a constant speed, the distance covered is simply s = vt, with v representing velocity (15.5 m/s). So, the distance equation for the truck is s = 15.5t.
To find when the automobile overtakes the truck, we set their distance equations equal: 1.2t2 = 15.5t. Solving for t, we get t = 12.92 seconds. Substituting t back into either vehicle's distance equation gives us the distance at which they are equal, resulting in 151.666 meters.
To find the speed of the automobile at the moment of overtaking, we use the formula for final velocity in terms of acceleration and time: v = u + at. Substituting the given values, we get v = 0 + 2.40*12.92, resulting in a speed of 31.008 m/s.
You watch distant Sally Homemaker driving nails into a front porch at a regular rate of 1 stroke per second. You hear the sound of the blows exactly synchronized with the blows you see. And then you hear one more blow after you see the hammering stop. Explain how you calculate that Sally is 340 m away from you.?
Answer:
The velocity of sound of an echo is given as:
v = 2d/t, where d is the distance the sound source and the reflecting surface.
The time take for the stroke to be heard is 2s, because the rate is one stoke per second(one stroke in 1s). It means it will be heard after 2s, after the reflection of the sound wave.
v is speed of sound in air(Value 240m/s)
Therefore, d = vt/2 = (340 x 2)/2
d = 340m.
Sally is 340m away.
Explanation:
The above question is an application of echo.
Echo is a sound heard after the reflection of sound wave.
The distance covered will be twice the distance between the source of the sound and the reflecting surface. This is because the sound will travel a certain distance to a reflecting surface and travels back equal distance to the source after getting reflected.
A lunar eclipse can only happen during a(1) new moon.(2) solstice.(3) first quarter moon.(4) full moon.(5) perihelion passage of the Sun.
Answer:
(4) full moon.
Explanation:
Lunar eclipse can only occur on a full moon night when the sun the earth and the moon are very much in a straight line.
During this period the the light of the sun that incidents on the moon is blocked by the earth and so we have the phases of the moon due to the relative motion of the three bodies which partially enables the light of the sun to reach the moon.
The moon appears orange-red during this time because the light that reaches the moon is after the refraction through the earth's atmosphere from which the other wavelengths have been absorbed by the earth's atmosphere.
Lunar eclipse can only occur at night and hence it can only be observed from about half of the earth.
A lunar eclipse occurs when the full moon moves into Earth's shadow, which can only happen when the Sun, Earth, and Moon are nearly aligned. This event is more common and widely visible than a solar eclipse, which requires a new moon and occurs when the Moon blocks the Sun.
A lunar eclipse occurs when the Moon enters the shadow of Earth. For a lunar eclipse to happen, the Sun, Earth, and Moon must be nearly in a straight line. The Moon must be in its full moon phase, as this is the only time when the alignment allows Earth's shadow to fall on the entire face of the Moon that is visible from Earth. This alignment does not occur at every full moon due to the inclination of the Moon's orbit. However, when it does, the shadow of Earth can cover about four full moons, given the length of Earth's shadow is about 1.4 million kilometers, and the distance to the Moon is roughly 384,000 kilometers.
It's important to differentiate between a lunar and a solar eclipse; the latter occurs when the Moon passes in front of the Sun, blocking it from view, and this can only happen during a new moon. The solar eclipse also requires the celestial bodies to be in the same plane called the ecliptic. Unlike a solar eclipse, a lunar eclipse is visible to all on the night side of Earth, making it an event observed more frequently from any given place on Earth.
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to compress the gas by a factor of 11.0, starting from its initial volume?
Final answer:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work required to compress the gas by a factor of 11.0 is -11,000 J.
Explanation:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work done on a gas during compression is given by the equation:
Work = Pressure ×Change in Volume
In this case, we know that the work required to compress the gas to half its initial volume is 550 J. Let's assume the initial volume of the gas is V. So the work done to compress the gas by a factor of 11.0 can be calculated as:
Work = Pressure × Change in Volume
550 J = Pressure ×(V/2 - V)
550 J = Pressure ×(-V/2)
Solving for Pressure, we get:
Pressure = -1100 J/V
Therefore, the work required to compress the gas by a factor of 11.0 is -1100 J/V× (11V - V) = -11,000 J (negative sign indicates work is done on the gas).
1. A woman driving at the 45 mi/hour speed limit on the entrance ramp to the highway accelerates at a constant rate and reaches the highway speed limit of 65 mi/hour in 6.00 s. What distance does the car travel during that acceleration? (Make the simplifying assumption that she is traveling in a straight line and be careful with your units)
Answer:
s = 147.54 m
Explanation:
given,
initial velocity,u = 45 mi/h
1 mph = 0.44704 m/s
45 mph = 45 x 0.44704 = 20.12 m/s
final velocity, v = 65 mi/h
v = 65 x 0.44704 = 29.06 m/s
time, t = 6 s
acceleration, [tex]a = \dfrac{v-u}{t}[/tex]
[tex]a = \dfrac{29.06-20.12}{6}[/tex]
a = 1.49 m/s²
distance travel by the car
using equation of motion
v² = u² + 2 a s
29.06² = 20.12² + 2 x 1.49 x s
2.98 s = 439.6692
s = 147.54 m
distance traveled by the car is equal to 147.54 m
Brownian motion is due to:
a. The random movement of pollen granules suspended in water.
b. The random fluctuation of the energy content of the environment. Thermal noise.
c. The random fluctuation of the energy content of the environment and thermal noise.
Answer:
option B
Explanation:
Brownian motion is the random movement of the microscopic size particle suspended in a liquid or gas.
Brownian motion is the result of the collision of the fast-moving particles.
This phenomenon was described by Robert Brown in the year 1927 i.e. it is named Brownian motion.
Brownian motion is due to the random fluctuation of the energy in the environment which leads to the zig-zag movement of the Particle.
Hence, the correct answer is option B.