Answer:
Check the explanation
Explanation:
net angular momentum is conserved as there is no net external torque.
therefore. 0 = angular momentum of woman + angular momentum of wheel.
0 = 55x2x1.5 + 480xω
ω = -0.3056 rad/sec ,-ve sign shows that the direction is opposite to woman, ie counterclockwise.
work done by woman = total kinetic energy of the wheel and woman
=1/2 x 55 x 1.52 + 1/2 Iω2 = 1/2x55x1.52 + 1/2x540x0.30562= 41.8+73.35 = 115.14J
Final answer:
The turntable rotates clockwise with an angular speed of 0.188 rad/s, and the woman does 1.17 x 10^-2 J of work to set herself and the turntable into motion.
Explanation:
(a) The turntable rotates clockwise as viewed from above the system with an angular speed of 0.188 rad/s.
(b) The work done by the woman to set herself and the turntable into motion is 1.17 x 10^-2 J.
Angular Speed of Turntable
Determining the angular speed of the turntable involves considering the conservation of angular momentum before and after the woman starts moving.
Work Done by the Woman
As for the work done by the woman, it's important to note that work is related to the change in kinetic energy of the system. This calculation would consider the kinetic energies before and after the movement.
When you hear a noise, you usually know the direction from which it came even if you cannot see the source. This ability is partly because you have hearing in two ears. Imagine a noise from a source that is directly to your right. The sound reaches your right ear before it reaches your left ear. Your brain interprets this extra travel time (Δt) to your left ear and identifies the source as being directly to your right. In this simple model, the extra travel time is maximal for a source located directly to your right or left (Δt = Δtmax). A source directly behind or in front of you has equal travel time to each ear, so Δt = 0. Sources at other locations have intermediate extra travel times (0 ≤ Δt ≤ Δtmax). Assume a source is directly to your right.(a) Estimate the distance between a person's ears. (they gave us the answer of .2... apparently the program is messed up and we have to use .2(b) If the speed of sound in air at room temperature is vs = 338 m/s,find Δtmax. (Use your estimate.)
(c) Find Δtmax if instead you and the source are in seawater at the same temperature, where vs = 1534 m/s. (Use your estimate.)
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.
Understanding Sound Localization and Interaural Time Difference
When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.
(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).
(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:
Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.
(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:
Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.
This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in series. Initially, the switch is open and has been open for a very long time. At time t = 0 s, the switch is suddenly closed. How long after closing the switch will the potential difference across the inductor be 24 V?
Answer:
1.6 s
Explanation:
To find the time in which the potential difference of the inductor reaches 24V you use the following formula:
[tex]V_L=V_oe^{-\frac{Rt}{L}}[/tex]
V_o: initial voltage = 60V
R: resistance = 24-Ω
L: inductance = 42H
V_L: final voltage = 24 V
You first use properties of the logarithms to get time t, next, replace the values of the parameter:
[tex]\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s[/tex]
hence, after 1.6s the inductor will have a potential difference of 24V
To find the time when the potential difference across the inductor will be 24 V in an LR circuit, we need to use the formula for the time-dependent current in an RL circuit, and through mathematical manipulations, substitute into the equation for the induced emf in the inductor and solve for time.
Explanation:The time for the potential difference across the inductor to be 24V in an LR circuit can be calculated using the formula for the time-dependent current in an RL circuit, which accounts for how current evolves over time. According to Faraday's law, a changing current in an inductor generates an opposition voltage, its magnitude is determined by L (inductance of the inductor) times dI/dt (the rate of change of current).
In our case we have the equation for the current in an RL circuit when turned on: I(t) = V/R*(1 - e^(-R*t/L)) and the induced emf in the inductor V(L) = L*dI/dt. Our goal is to find the time when V(L) equals 24V. This requires substituting I(t) into the equation for V(L), setting V(L) equal to 24V, and solving for t.
This process involves the principles of RL circuits, the characteristic time constant, and Faraday's law.
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A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fuel before liftoff is 25.0 g, which includes fuel mass of 12.7 g. The engine fires for a total of 1.90 s. (Assume all the fuel is consumed.) (a) Calculate the average exhaust speed of the engine (in m/s). (b) This engine is situated in a rocket case of mass 63.0 g. What is the magnitude of the final velocity of the rocket (in m/s) if it were to be fired from rest in outer space with the same amount of fuel
Answer:
a) v = 786.93 m/s
b) v = 122.40 m/s
Explanation:
a) To find the average exhaust speed (v) of the engine we can use the following equation:
[tex] F = \frac{v\Delta m}{\Delta t} [/tex]
Where:
F: is the thrust by the engine = 5.26 N
Δm: is the mass of the fuel = 12.7 g
Δt: is the time of the burning of fuel = 1.90 s
[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]
b) To calculate the final velocity of the rocket we need to find the acceleration.
The acceleration (a) can be calculated as follows:
[tex] a = \frac{F}{m} [/tex]
In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:
[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]
Now, the acceleration is:
[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]
Finally, the final velocity of the rocket can be calculated using the following kinematic equation:
[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]
I hope it helps you!
Final answer:
The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.
Explanation:
To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.
Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s
The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.
Δ(momentum) = Δm × v_ex
Substituting the impulse and solving for v_ex, we get:
v_ex = Impulse / Δm
v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s
Part B: Final Velocity of the Rocket in Space
The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:
V_final = v_ex × ln(m_initial / m_final)
Where:
m_initial = mass of the rocket with fuel = 25.0 g + 63.0 g = 88.0 g = 0.088 kgm_final = mass of the rocket without fuel = 25.0 g - 12.7 g + 63.0 g = 75.3 g = 0.0753 kgCalculating the final velocity:
V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.1 cm ( 0.581 m) and the flow speed of the petroleum is 10.1 m/s. At the refinery, the petroleum flows at 5.85 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
volume flow rate:________m^3/sdiameter________cm
Answer:
volume flow rate:2.68m³/s
diameter :58.27cm
Explanation:
flow rate = Q = πr² v = amount per second that flows through the pipe
Given:
pipe's diameter= 0.581
r= 0.581/2=>0.2905m
speed 'v'= 10.1 m/s
Q= (3.142)(0.2905)²(10.1)
so
volume flow rate= 2.68m³/s
->if no oil has been added or subtracted or compressed then Q is the same everywhere
therefore,
Q= πr²v
2.68 = π(d/2)²(5.85)
d= (2.68x4) /(3.142 x 5.85)
d= 0.5827m =>58.27cm
Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. (Use t as the independent variable, y as the dependent variable, R as the room temperature, and k as a proportionality constant.)
Answer:
[tex]\frac{dy}{dt}=k(y_t-R)[/tex]
Explanation:
According to Newton’s law of cooling, the rate of loss of heat from a body and the difference in the temperature of the body and its surroundings are proportional to each other.
[tex]\frac{dy}{dt}=k(y_t-R)[/tex]
Here, [tex]y_t[/tex] represents temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.
The equation that represents Newton's Law of Cooling for this particular situation is dy/dt = k(yt-R)
What is Newton’s law?
According to Newton’s law of cooling refer, the rate of loss of heat from a body and also the difference in the temperature of the body and also its surroundings are proportional to each other.
dy/dt is = k(yt-R)
Therefore, yt conveys temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.
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The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 kg/m3. (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius = 0.280 m)? (b) For comparison, determine the weight of a jetliner whose mass is 2.44 × 105 kg.
Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
The force exerted by the water on the sea underwater vehicle's window at the depth of the Mariana Trench is approximately 2.73 x 10⁷ N. For comparison, the weight of a jetliner with a mass of 2.44 x 10⁵ kg is 2.39 x 10⁶ N. Thus, the force exerted on the underwater vehicle at the Mariana's depth is an order of magnitude greater than the weight of the jetliner.
Explanation:To calculate the force that the water exerts on the observation window of an underwater vehicle, you first need to find the pressure at the depth of the Mariana Trench. The pressure is given by the formula P = ρgh, where P is pressure, ρ is density, g is acceleration due to gravity, and h is height (in this case, depth).
Substituting the given values, the pressure at a depth of 11,000 m is about 1.11 x 10⁸ Pa. The force exerted on the window can be found using F = PA, where A is the area of the window. With a radius of 0.280 m, the area of the window (A = πr²) is 0.246 m². Thus, the force applied by the water on the window is F = (1.11 x 10⁸ Pa) (0.246 m²) = 2.73 x 10⁷ N approx.
For the second part of the question, the weight (force) of the jetliner is given by the equation W = m × g. Using the given mass (2.44 x 10⁵ kg) and the acceleration due to gravity (9.8 m/s²), the weight of the jetliner is 2.44 x 10⁵ kg × 9.8 m/s² = 2.39 x 10⁶ N, which is an order of magnitude less than the force exerted on the underwater vehicle at the depth of the Mariana Trench.
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Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the seven seconds that I’m watching it (so from t = 0 to t = 7). (a) What is the velocity function, v(t), for the motion of the squirrel? (b) What is the acceleration function, a(t), for the motion of the squirrel? (c) At the four second mark, is the squirrel moving up the tree or down the tree? Justify your answer.
Answer:
Explanation:
Given
Position of squirrel is given by
[tex]s(t)=t^3-12t^2+36t[/tex]
Velocity is given by
[tex]v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}[/tex]
[tex]v(t)=3t^2-12\times 2t+36[/tex]
[tex]v(t)=3t^2-24t+36[/tex]
(b) acceleration is given by
[tex]a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}[/tex]
[tex]a(t)=6t-24[/tex]
(c)at [tex]s(3)=3^3-12(3)^2+36(3)[/tex]
[tex]s(3)=27\ m[/tex]
at [tex]s(4)=4^3-12(4)^2+36(4)[/tex]
[tex]s(4)=16\ m[/tex]
at [tex]t=3\ s[/tex] Position is [tex]27\ m[/tex] and at [tex]t=4\ s[/tex] position is [tex]16\ m[/tex]
therefore squirrel is moving down
The light pattern shown is a result from a beam of light being passed through a single slit. The pattern is created from constructive and destructive interference. This phenomenon is due to
A) the wave nature of light.
B) the particle nature of light.
C) the resonance effect of light.
D) the photoelectric effect of light.
Answer:
A
Explanation:
When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.0 km/s and the secondary, or S, wave has a speed of about 4.5 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.2 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far (in terms of m) is the seismograph from the earthquake?
Answer:[tex]d=7.94\times 10^5\ m[/tex]
Explanation:
Given
Speed of Primary wave [tex]v_1=8\ km/s[/tex]
Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]
difference in timing of two waves are [tex]77.2\ s[/tex]
Suppose both travel a distance of d km then
[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]
[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]
Subtract (ii) from (i)
[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]
[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]
[tex]d[0.0972]=77.2[/tex]
[tex]d=794.23\ km[/tex]
[tex]d=7.94\times 10^5\ m[/tex]
Final answer:
To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.
Explanation:
When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.
Let the distance be d, then:
Time for P-wave to travel d: d / 8.0 km/sTime for S-wave to travel d: d / 4.5 km/sThe difference in travel time is 77.2 s, so: d / 4.5 km/s - d / 8.0 km/s = 77.2 sTo find the distance d, we solve the equation:
8.0×d - 4.5×d = 77.2 s × (8.0 km/s × 4.5 km/s)3.5×d = 77.2 s × 36 km²/s²d ≈ 618.94 kmNow, to convert kilometers to meters:
d ≈ 618.94 km × 1,000 m/kmd ≈ 618,940 mTherefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.
The density of mercury is 13.546 g/cm3 . Calculate the pressure exerted by a column of mercury 76 cm high. Give your answer in Pascals and lbf/in2 . 2. The density of water is 62.43 lbm/ft3 . Calculate the pressure exerted by a column of water 25 ft high. Give your answer in Pascals and lbf/in2 3. What is the power required to pump 10 kg/s water from a height of 5 meters to a height of 30 meters? Report the power in Watts and hp. 4. Water is being pumped by the application of pressure at point 1, see below, up to a height of 50 m at a rate of 1 kg/s. At the top, point 2 the pressure is nearly one atmosphere (105 N/m2 )
Answer:
The required solution is 100890 Pa and 14.3lb/in²
Explanation:
See attached image
Answer:
1. p = 14.63 lb/in² or 100890.608 Pa
2. p = 74676 Pa or 10.83 lb/in²
3. P = 2450 W or 3.28 hp
4. [tex]p_{1}[/tex] = 490105 N/m²
Explanation:
1. Let's begin by listing out the given parameters:
density of mercury = 13.546 g/cm³ = 13546 kg/m³,
height of column = 76 cm = 0.76 m, acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 13546 * 9.8 * 0.76
p = 100890.608 Pa
To get the answer in lb/in², divide by 6895
p = 100890.608 ÷ 6895 = 14.632
p = 14.63 lb/in²
2. Let's list out the parameters given:
density of water = 62.43 lbm/ft³ = 62.43 * 16.018 = 1000kg/m³,
height of column = 25 ft = 25 ÷ 3.281 = 7.62 m,
acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 1000 * 9.8 * 7.62
p = 74676 Pa
To convert from Pa to lb/in², divide by 6895
p = 74676 ÷ 6895
p = 10.83 lb/in²
3. Let's list out the parameters given:
mass flow rate (ṁ) = 10 kg/s, [tex]h_{1}[/tex] = 5 m, [tex]h_{2}[/tex] = 30 m, Δh = 30 - 5 = 25 m, g = 9.8 m/s²
Using Power = Energy (Potential Energy) ÷ Time
Energy (Potential Energy) = m g h
Power = mgΔh ÷ t; m÷ t = ṁ
Substitute ṁ into the equation
P = ṁ g h = 10 * 9.8 * 25
P = 2450 W
To convert from W to hp, divide by 746
P = 2450 ÷ 746 = 3.284
P = 3.28 hp
4. Let's list out the parameters given:
height (Δh) = 50 m, ṁ = 1 kg/s, g = 9.8 m/s²,
p2 = 105N/m², ρ = 1000 kg/m³
Using Bernoulli's Equation,
p1 + ½ρ([tex]v_{1}[/tex])² + ρgh1 = p2 + ½ρ([tex]v_{2}[/tex])² + ρgh2
Assuming steady state flow; [tex]v_{2}[/tex] = [tex]v_{1}[/tex] ⇒ [tex]v_{2}[/tex] - [tex]v_{1}[/tex] = 0
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ½ρ([tex]v_{2}[/tex] - [tex]v_{1}[/tex])² + ρg([tex]h_{2}[/tex] - [tex]h_{1}[/tex])
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ρgΔh
[tex]p_{1}[/tex] - 105 = 1000 * 9.8 * 50
[tex]p_{1}[/tex] = 490000 + 105 = 490105
[tex]p_{1}[/tex] = 490105 N/m²
The number of protons in the nucleus of an atom determines the species of the atom, i.e., the element to which the atom belongs. An atom has the same number of protons and neutrons. But the electron number cannot be used instead because
A. electrons are not within the nucleus
B. electrons are negatively charged
C. electrons can be removed from or added to an atom
D.electrons are lighter than protons
The number of protons in the nucleus of an atom determines the element that the atom belongs to.
Explanation:The subject of this question is Chemistry. The number of protons in the nucleus of an atom determines the species or element to which the atom belongs. This is because each element has a unique number of protons, known as the atomic number. The other options are not correct because:
Electrons are not within the nucleus: Electrons are found in electron shells outside the nucleus of an atom.Electrons are negatively charged: While electrons are negatively charged, this does not affect their use in determining the species of an atom.Electrons can be removed from or added to an atom: While electrons can be added or removed from an atom, their number does not define the species of the atom.Electrons are lighter than protons: Although electrons are lighter than protons, this is not the reason why their number cannot be used to determine the species of an atom.Learn more about protons in the nucleus here:https://brainly.com/question/30654172
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A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for gas heating and cooking in a period of 1 month. Given that 1.00 therm is equal to 29.329.3 kWh, how many milligrams of mass would need to be converted directly to energy each month to meet the energy needs for the home?
Answer:
Explanation:
Given that,.
A house hold power consumption is
475 KWh
Gas used is
135 thermal gas for month
Given that, 1 thermal = 29.3 KWh
Then,
135 thermal = 135 × 29.3 = 3955.5 KWh
So, total power used is
P = 475 + 3955.5
P =4430.5 KWh
Since 1 hr = 3600 seconds
So, the energy consumed for 1hr is
1KW = 1000W
P = energy / time
Energy = Power × time
E = 4430.5 KWhr × 1000W / KW × 3600s / hr
E = 1.595 × 10^10 J
So, using Albert Einstein relativity equation
E = mc²
m = E / c²
c is speed of light = 3 × 10^8 m/s
m = 1.595 × 10^10 / (3 × 10^8)²
m = 1.77 × 10^-7 kg
Then,
1 kg = 10^6 mg
m = 1.77 × 10^-7 kg × 10^6 mg / kg
m = 0.177mg
m ≈ 0.18 mg
Total monthly energy consumption of the house in kilowatt-hours (kWh) is calculated. Then, with the help of the equation E = mc², where c is the speed of light, the equivalent mass of energy in kilograms is found by converting energy to Joules and solving for mass.
Explanation:The household uses 475475 kWh of electrical energy and 135135 therms for gas heating and cooking monthly. Considering that 1.00 therm is equal to 29.329.3 kWh, total energy consumption for the month in kWh can be calculated as (475475 + (135135 × 29.3)) kWh.
Einstein’s famous equation E = mc² relating energy (E) and mass (m) allows us to calculate the equivalent mass of this energy.
Here, c is the speed of light. We convert the energy into Joules (1 kWh = 3.6 × 10⁶ J) and then solve for m to find out the mass in kilograms that would need to be converted into energy each month. The question seems to have a typo in the values, you might need to correct them to calculate the accurate amount.
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Buffy is rolling along in her 10.4 kg wagon at 3.4 m/s (in the positive direction) when she jumps off the back. She continues to move forward at 1.7 m/s relative to the ground. This causes her wagon to go speeding forward at 8.04 m/s relative to the ground. How much does Buffy weigh
Answer:
Buffy weigh's 28.39 kg
Explanation:
Given;
mass of Buffy and wagon, M = 10.4 kg
final velocity of Buffy - wagon system, v = 3.4 m/s
Buffy's velocity relative to the ground, u₁ = 1.7 m/s
Wagon's velocity relative to the ground, u₂ = 8.04 m/s
Buffy's velocity = 1.7 - 3.4 = - 1.7 m/s
Wagon's velocity = 8.04 - 3.4 = 4.64 m/s
Apply the principle of conservation of linear momentum;
1.7 x m = 4.64 x 10.4
where;
m is mass of wagon
1.7 m = 48.256
m = 48.256 / 1.7
m = 28.39 kg
Therefore, Buffy weigh's 28.39 kg
A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:
(a) Find the change in potential energy of the capacitor.
(b) Does the potential energy increase or decrease? Explain
Answer:
Explanation:
capacitance = 20 x 10⁻⁶ F .
potential V = 12 V
charge = CV
= 20 x 10⁻⁶ x 12
Q = 240 x 10⁻⁶ C
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶
= 1440 x 10⁻⁶ J
b )
In this case charge will remain the same but capacity will be increased 4 times
new capacity C = 4 x 20 x 10⁻⁶
= 80 x 10⁻⁶
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶
= 360 x 10⁻⁶ J .
potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J
If a light bulb is missing or broken in a parallel circuit, will the other light bulb right ?
The theory of nuclear astrophysics is that all the heavy elements like uranium are formed in the interior of massive stars. These stars eventually explode, releasing these elements into space. If we assume that at the time of the explosion there were equal amount of U-235 and U-238, how long ago did the star(s) explode that released the elements that formed our Earth? The present U-235/U-238 ratio is 0.00700. [The half-lives of U-235 and U-238 are 0.700 × 109 yr and 4.47 × 109 yr.]
Answer:
t = 5.94x10⁹ years.
Explanation:
The time of the explosion can be calculated using the decay equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]
Where:
N(t): is the quantity of the element at the present time
N(0): is the quantity of the element at the time of explosion
λ: is the decay constant
t: is the time
Knowing that the present U-235/U-238 ratio is 0.00700 and that at the time of the explosion there were equal amount of U-235 and U-238, we have:
[tex]\frac{N_{U-235}}{N_{U-238}} = \frac{N_{U-235_{0}}e^{-\lambda_{U-235} t}}{N_{U-238_{0}}e^{-\lambda_{U-238} t}}[/tex] (1)
The decay constant is equal to:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]
For the U-235 we have:
[tex] \lambda_{U-235} = \frac{ln(2)}{0.700 \cdot 10^{9} y} = 9.90 \cdot 10^{-10} y^{-1} [/tex]
For the U-238 we have:
[tex] \lambda_{U-238} = \frac{ln(2)}{4.47 \cdot 10^{9} y} = 1.55 \cdot 10^{-10} y^{-1} [/tex]
By introducing the values of [tex]\lambda_{U-235}[/tex] and [tex]\lambda_{U-238}[/tex] into equation (1) we have:
[tex]0.00700 = \frac{e^{-9.90 \cdot 10^{-10} t}}{e^{-1.55 \cdot 10^{-10} t}}[/tex]
[tex]0.00700 = e^{(-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t}[/tex]
[tex]ln(0.00700) = (-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t[/tex]
[tex]t = \frac{ln(0.00700)}{-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}} = 5.94 \cdot 10^{9} y[/tex]
Therefore, the star exploded 5.94x10⁹ years ago.
I hope it helps you!
The explosion of the star that released the uranium that formed Earth happened approximately 6 billion years ago, determined using the half-lives of U-235 and U-238 and the current U-235/U-238 ratio.
The question regarding the age of the elements from a star explosion can be addressed using the concept of radioactive decay and isotope half-lives, specifically of Uranium-235 (U-235) and Uranium-238 (U-238). Using the current U-235/U-238 ratio of 0.00700 and knowing the half-lives of U-235 (0.700 × 109 years) and U-238 (4.47 × 109 years), we can calculate that the star exploded approximately 6 billion years ago based on the change in the U-235/U-238 ratio from an assumed equal initial amount. This estimate is consistent with the age of the solar system and the time it would take for such materials to coalesce into a planetary body like Earth.
Which pair shows the law of reflection?
Answer:
The answer is A and C .
Explanation:
Reflection of object is reflected through the Normal (mirror) .
*incident angle = refracted angle
Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+β .
To evaluate the final kinetic energy of a supply spacecraft under the given tractor beam force, you would need to integrate the force over the displacement. Direct calculation requires specific values for α, β, and displacement x, which are not provided.
Explanation:The question involves calculating the final kinetic energy of a supply spacecraft under a specific tractor beam force, F(x)=αx3+β. This calculation would typically require the integration of the force over the displacement, as kinetic energy can be evaluated through the work-energy principle where work done by a force in moving an object is equal to the change in kinetic energy of the object. Here, without specific values or further context provided for α, β, or the displacement, x, a direct calculation cannot be made. However, in physics, especially in the study of mechanics and spacecraft dynamics, understanding how forces affect motion and energy forms the basis for analyzing and optimizing space missions. Evaluating the final kinetic energy would typically involve integrating the force function over the spacecraft's path, considering initial conditions, and any external forces or resistances.
A wedge-shaped air film is made by placing a small slip of paper between the edges of two thin plates of glass 12.5 cm long. Light of wavelength 600 nm in air is incident normally on the glass plates. If interference fringes with a spacing of 0.200 mm are observed along the plate, how thick is the paper? This form of interferometry is a very practical way of measuring small thicknesses.
Answer:
The thickness of the paper is [tex]t = 188\mu m[/tex]
Explanation:
From the question we are told that
The length of the wedge-shaped air film [tex]L = 12.5 cm = \frac{12.5}{100} = 0.125m[/tex]
The wavelength of the light is [tex]\lambda = 600nm = 600* 10^{-9}m[/tex]
The spacing of the interference fringe is [tex]D = 0.200mm = \frac{0.200}{1000} = 0.2*10^{-3} m[/tex]
For destructive interference the thickness is mathematically represented as
[tex]t =\frac{\lambda * L}{2 * D }[/tex]
Substituting values
[tex]t = \frac{600 * 10^{-9} * 0.125 }{2 * 0.2 *10^{-3}}[/tex]
[tex]t = 188\mu m[/tex]
A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed
Answer:
Explanation:
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
The wok done by force will be converted into rotational kinetic energy
F x d = 1/2 I ω²
F is force applied , d is displacement , I is moment of inertia of disc and ω
is angular velocity of disc
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
The angular speed should be 17.18 rad / s
Calculation of the angular speed:Since
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
Now the work done by force should be converted into the rotational kinetic energy
F x d = 1/2 I ω²
here,
F is the force applied,
d is displacement,
I is moment of inertia of disc
and ω is angular velocity of disc
So,
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
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A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a tuning fork over the opening. The sound waves that travel down into the tube will reflect from the surface of the water and come back to the tuning fork. This is like a half open pipe with a fixed end condition on the water side and an open end condition on the openThe condition for resonance is if the depth of the air column is an odd integer multiple of lamda/4. The speed of sound is 345 m/s.A) If a tuning fork was used for this experiment, what is the depth of the air column that will satisfy the resonance condition for the fundamental mode?B) What is the depth of L3the air column for the 3rd harmonic resonance?C) Is there a depth that will result in a 2nd harmonic resonance? Explain
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:
[tex]f_n=\frac{nv_s}{4L}[/tex]
n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:
[tex]f_1=\frac{v_s}{4L}[/tex]
B) For the 3rd harmonic you have:
[tex]f_3=\frac{3v_s}{4L}[/tex]
C) For the 2nd harmonic:
[tex]f_2=\frac{2v_s}{4L}[/tex]
One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.15 m, the speed of the 3.12 x 10-3 kg particle is 131 m/s. Find the initial separation between the particles.
Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Explanation:
Given
mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]
mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]
Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]
Now final speed of first particle is [tex]v_1=131\ m/s[/tex]
Final separation between particles is [tex]r=0.15\ m[/tex]
As there is no external force therefore linear momentum is conserved
[tex]0+0=m_1v_1+m_2v_2[/tex]
[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]
[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]
[tex]v_2=-57.56\ m/s[/tex]
Conserving total energy
Initial Kinetic energy +Initial Potential energy=Final Kinetic energy +Final Potential energy
[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]
[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]
[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.
Conservation of Linear MomentumGiven that the masses of the two particles are;
[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]
Also, the charges of both the particles are equal.
[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]
The final separation between the particles is;
[tex]r_f = 0.15\,m[/tex]
Also, the final speed of the first particle is;
[tex](v_1)_{final} = 131\,m/s[/tex]
There is no external force applied here; so by the law of conservation of linear momentum, we have;
[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]
But initially, the particles are at rest, so the initial velocities are zero.
[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]
[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]
Conservation of EnergyNow, applying the law of conservation of total energy, we get;
[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]
But initially, the particles are at rest; so they have no initial kinetic energy.
They have electrostatic potential alone initially.
[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]
Substituting the known values, we get;
[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]
[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]
[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]
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Interference in 2D. Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.025 mm apart? (b) What would be the separation between maxima in intensity on a screen located 1 m from the slits? (c) What would the angular separation be if the slits were 2.5 mm apart? (d) What would the separation between maxima be on a screen 25 mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful given the size of your pupil. The spacing between cones on the retina is about 10 µm. (Of course, you would never want to look directly into a laser beam like this....) (e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by π? Would the separation change? 3. Combined two source interference and
Answer:
Explanation:
λ
given λ = 700 nm
a) for first maxima, d*sin(θ) = λ
sin(theta) = λ/d
= 700*10^-9/(0.025*10^-3)
= 0.028
theta θ = sin^-1(0.028)
= 1.60 degrees
b) given R = 1 m,
delta_y = λ*R/d
= 700*10^-9*1/(0.025*10^-3)
= 0.028 m or 2.8 cm
c) for first maxima, d*sin(θ) = λ
sin(θ) = λ/d
= 700*10^-9/(2.5*10^-3)
= 0.00028
theta = sin^-1(0.00028)
= 0.0160 degrees
d) R = 25 mm = 0.025 m
δ_y = λ*R/d
= 700*10^-9*0.025/(2.5*10^-3)
= 7*10^-6 m or 7 micro m
e) No. But the position of maxima and minima will be shifted.
How much heat is required, in joules, to complete the of 15.0g of water’s phase change from liquid to steam?
A. 150.7 J
B. 33,900 J
C. 62.7 J
D. 5,010 J
Answer: 62.7
Explanation:
The specific heat capacity of water is 4.186 J you multiply 4.186 and 15.0 and you will get 62.7
A solid copper sphere hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it vibrates with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. The density of water, copper, and steel is 1000kg/m³, 8960kg/m³, and 8050kg/m³ respectively
Answer:
291.509 Hz
Explanation:
Fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform.
It is a vital concept in musical instruments and many aspects .
See the attached file for the solution to the given problem.
En una competición de tiro con arco, la diana de 80 cm de diámetro se encuentra a 50 m de distancia y a 1,5 metros del suelo. En uno de los tiros la flecha sale a 230 km hora con una de 3,5 grados desde una altura de 1,60 metros,despreciando el rozamiento con el aire.¿ Impactará la flecha en la diana? En caso afirmativo¿ con qué velocidad y en qué dirección?
Answer:
a .
15.68
m/s
(b).
21.72
m
Explanation:
To determine if the arrow will hit the target, we can analyze the projectile motion. By calculating the time of flight and horizontal displacement, we can determine if the arrow will hit the target. The velocity and direction of the arrow at impact can also be determined using the horizontal and vertical components of velocity.
To determine if the arrow will hit the target, we need to analyze the projectile motion of the arrow. First, we need to calculate the time of flight using the equation t = 2 * (v * sin(theta))/g, where v is the initial velocity of the arrow and theta is the launch angle. Next, we can calculate the horizontal displacement using the equation x = v * cos(theta) * t, where x is the distance traveled horizontally. Comparing the calculated horizontal displacement with the distance to the target, we can determine if the arrow will hit the target.
Given the initial velocity of the arrow (230 km/hr), launch angle (3.5 degrees), and distance to the target (50 m), we can convert the initial velocity to m/s (230 km/hr * (1000 m/km) * (1 hr/3600 s)) and use the formulas to calculate time of flight and horizontal displacement. If the horizontal displacement is less than or equal to the distance to the target, the arrow will hit the target.
To find the velocity and direction of the arrow at impact, we can use the equations v_x = v * cos(theta) and v_y = v * sin(theta), where v_x is the horizontal component of velocity and v_y is the vertical component of velocity. Since the angle of impact is not specified, the direction of the arrow at impact will depend on the particular situation.
In a double‑slit interference experiment, the wavelength is λ = 452 nm λ=452 nm , the slit separation is d = 0.190 mm d=0.190 mm , and the screen is D = 49.0 cm D=49.0 cm away from the slits. What is the linear distance Δ x Δx between the eighth order maximum and the third order maximum on the screen?
Answer:
Δx = 5.82mm
Explanation:
To find the distance between the eight maximum and the third one you use the following formula:
[tex]x_m=\frac{m\lambda D}{d}[/tex] (1)
λ: wavelength = 452*10^-9 m
m: order of the fringes
D: distance to the scree = 0.49m
d: distance between slits = 0.190*10^-3 m
you use for m=8 and m=3, then you calculate x8 - x3:
[tex]x_8=\frac{8(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=9.32*10^{-3}m\\\\x_3=\frac{3(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=3.49*10^{-3}m\\\\\Delta x_{8,3}=5.82*10^{-3}m=5.82mm[/tex]
hence, the distance between these fringes is 5.82mm
A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial velocity of 3.33 m/s. The car then runs up the frictionless slope, gaining 0.108 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance
Answer:
2.994 m/s
Explanation:
m = Mass of the car = 100 g
[tex]v_1[/tex] = Initial velocity = 3.33 m/s
h = Height = 0.108 m
g = Acceleration due to gravity = 9.81 m/s²
We know that energy in the system is conserved so
[tex]\dfrac{1}{2}mv_1^2=mgh+\dfrac{1}{2}mv_f^2\\\Rightarrow v_f=\sqrt{2\left({\dfrac{1}{2}v_1^2-gh}\right)}\\\Rightarrow v_f=\sqrt{2\left(\dfrac{1}{2}3.33^2-9.81\times 0.108\right)}\\\Rightarrow v_f=2.994\ m/s[/tex]
The final velocity of the car is 2.994 m/s
The final velocity of the car is approximately 6.05 m/s.
Explanation:To find the final velocity of the car, we can use the principles of conservation of energy. In this case, the initial kinetic energy of the car is equal to its final potential energy. The initial kinetic energy can be calculated using the formula K.E. = 1/2 mv^2, where m is the mass of the car (100 g = 0.1 kg) and v is the initial velocity (3.33 m/s).
The final potential energy can be calculated using the formula P.E. = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the change in altitude (0.108 m).
Setting the initial kinetic energy equal to the final potential energy, we can solve for the final velocity of the car:
1/2 mv^2 = mgh
Substituting the given values, we get:
1/2 (0.1 kg)(v^2) = (0.1 kg)(9.8 m/s^2)(0.108 m)
Simplifying and solving for v, we find that the final velocity of the car is approximately 6.05 m/s.
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[03.02]
If you were going to describe the relationship between current, voltage, and power, you could say: (1 point)
Group of answer choices
"If there is a decrease in power, there must have been an increase in either current or voltage."
"If you decrease your voltage, and all other factors remain the same, you will have an increase in power."
"In order to gain more power you would need to increase either current or voltage."
"The only way to increase your power is to change your voltage and keep your current the same."
Answer:
in other to gain more power you would need to increase either current or voltage
Explanation:
In other to gain more power, you would need to increase either current or voltage,because power is directly proportional to voltage when current is kept constant.also power is directly proportional to current when voltage is kept constant
A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9930c. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?
Answer:
Her current biological age = 24.02 years
Explanation:
From time dilation equation, we know that;
t = t_o * [√(1-(v²/c²))]•L_o
Where;
t = dilated time
t_o = stationary time
v = the speed of the moving object
c = the speed of light in a vacuum
First, let's convert the rest time (t_o) from light years to years.
Thus;
t_o = [c/0.993c] * [42.2]
c will cancel out and we now have;
t_o = 42.5 years
Since t = t_o * [√(1-(v²/c²))]
Thus; t = 42.5 * [√(1-(0.993²c²/c²))]
t = 42.5/[√(1 - (0.993²))]
t = 42.5 * 0.1181
t = 5.02 years
Since the astronaut was 19 years old when the probe left the earth, thus;
Her current biological age now the probe has reached Capella, will be;
19 + 5.02 = 24.02 years