Answer:
2 kg
Explanation:
Note: For the meter stick to be balanced,
Sum of clock wise moment must be equal to sum of anti clock wise moment
Wd = W'd' ................ Equation 1
Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.
make W' the subject of the equation
W' = Wd/d'............... Equation 2
Taking our moment about the support,
Given: W = mg = 1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm
Substitute into equation 2
W' = 9.8(50)/25
W' = 19.6 N.
But,
m = W'/g
m = 19.6/9.8
m = 2 kg.
To find the mass M suspended from the meter stick, we can use the principle of rotational equilibrium. By setting the torques on either side of the fulcrum equal to each other, we can calculate the mass M at the 75 cm mark to be 3 kg.
Explanation:To determine the mass M suspended from the meter stick at the 75 cm mark, we can use the principle of rotational equilibrium. Since the system is balanced, the torques on both sides of the meter stick must be equal.
The torque of the rock can be calculated using its weight, which is equal to its mass multiplied by the acceleration due to gravity. The torque of the mass M can be calculated by multiplying its mass by the distance from the fulcrum (75 cm) to its center of mass. By setting the torques equal to each other, we can solve for M.
The torque of the rock: TR = (1 kg)(9.8 m/s²)(0.75 m).
The torque of the mass M: TM = M(9.8 m/s²)(0.25 m).
Setting TR equal to TM and solving for M, we get M = (1 kg)(0.75 m)/(0.25 m) = 3 kg.
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How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice cube? = 6.02 kJ/mol 108 kJ 102 kJ 5.64 kJ 936 J none of the above
Answer:
How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice cube? = 6.02 kJ/mol
A. 108 kJ
B. 102 kJ
C. 5.64 kJ
D. 936 kJ
E. none of the above
5.64 kJ
Explanation:
The Heat of fusion is the heat energy required to dissolve a given mass of ice at melting point.
Step by Step CalculationThe heat energy required to dissolve ice can be calculated using the expression below;
Q = ΔH[tex]_{f}[/tex] x m ...............................................1
where Q is the heat energy required;
ΔH[tex]_{f}[/tex] is the heat of fusion for ice;
m is the mole
All the parameters above are provided in the question except m, so to get m we use the molar mass of water (also for ice) which is 18.01528 g/mol .
This means that 18.01528 g of ice is contained in one mole, therefore the mole for 16.87 g of ice is given as;
[tex]m = \frac{16.87g}{18.015g/mol}[/tex]
m = 0.9364 mole of ices
Now the parameters are complete, we are given;
ΔH[tex]_{f}[/tex] = 6.02 kJ/mol
m = 0.9364 mol
Q =?
Substituting into equation 1, we have
Q = 6.02 kJ/mol x 0.9364 mol
Q = 5.64 kJ
Therefore, the energy required to melt 16.87 g of ice is 5.64 kJ
The energy required to melt a 16.87 g ice cube can be calculated by first converting the mass to moles and then multiplying by the enthalpy of fusion. The calculated energy is 5.64 kJ.
Explanation:The amount of energy required to melt an ice cube can be calculated using the enthalpy of fusion, which is given as 6.02 kJ/mol for ice. In order to make the conversion, we need to convert the mass of the ice cube from grams to moles. Since the molecular weight of water is approximately 18.015 g/mol, the 16.87 g ice cube amounts to 0.937 mol of ice. We then multiply this amount by the enthalpy of fusion to obtain the required energy. Thus, Energy = (0.937 mol) * (6.02 kJ/mol) = 5.64 kJ. Therefore, it would take about 5.64 kJ of energy to melt the 16.87 g ice cube.
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A plastic pipe carries deionized water in a microelectronics clean room, and one end of it is capped. the water pressure is p0 = 60 psi, and the cap is attached to the end of the pipe by an adhesive. calculate the shear stress present in the adhesive.
Answer:
Where the height of the cap is less than or equal to the radius of the bore of the plastic pipe then the ⇵⇵shear stress is less than o equal to 60 psi.
Explanation:
Shear stress = F/A
where F = Applied force
A = Cross sectional area of the member experiencing the force
If the cap covers a section of the pipe of height, h then the area = 2πrh
Where the 60 psi pressure is acting on the pipe bore, we have P₀ = 60 psi = F₀/A₀ where A₀ = area of pipe bore ≈ πr². Therefore if F₀ = F then we have
F₀/A₀ = F₀/πr² and F/A = F₀/2πrh = where h greater than or equal to 0.5×r then the shear stress is less than or equal to 60 psi
To calculate the shear stress in the adhesive, use the formula for shear stress and the given pressure and area.
Explanation:To calculate the shear stress present in the adhesive, we can use the formula for shear stress:
Shear stress = Force / Area
In this case, the force exerted by the water pressure on the capped end of the pipe is equal to the pressure multiplied by the area of the capped end. Since the water pressure is given as 60 psi, we need to convert it to a consistent unit, such as Pascals. 1 psi is equal to approximately 6895 Pascals. The area of the capped end can be calculated using the formula for the area of a circle: Area = pi * radius^2.
Once we have the force and area, we can substitute them into the formula for shear stress to find the value.
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A 92.6 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 735 (food) Calories. How high must the person climb? The acceleration due to gravity is 9.8 m/s 2 and 1 food Calorie is 103 calories. Answer in units of km.
Answer:
349 m
Explanation:
Parameters given:
Mass of climber, m = 92.6 kg
Amount of food calories = 735
1 food calorie = 103 calories
735 food calories = 75705 calories
1 joule is equal to 0.239 calories. Therefore, 75705 calories will be 316749.72 joules.
Hence, this is the amount of work the climber must do work off the food he ate.
Work done is given as:
W = Force * distance
W = m * g * h
h = W/(m * g)
h = 316749.72/(92.6 * 9.8)
h = 349 m
A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 27.2. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire.
a.) What is the current in the copper section? b.) What is the current in the silver section?
b.) What is the magnitude of E in the copper section?
c.) What is the magnitude of E in the silver section?
d.) What is the potential difference between the ends of the silver section of the wire?
Using physics concepts of Ohm's law, resistance calculation, and electric field magnitude calculation, we can find the current in copper and silver sections, the electric field magnitudes, and the potential difference in silver section. The potential difference across the silver section will be the same as that applied across the entire wire due to copper's negligible resistance.
Explanation:The student's question is related to electricity and magnetism, a branch of physics. In the scenario described, the resistance (R) of each wire section can be found using the formula R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Here, resistivity needs to be known from Table 27.2.
Once the resistance of each wire section is known, Ohm's law (V = IR) can be used to find the current (I) in each section. The same current will flow in both sections of the wire since it is a series circuit. The potential difference across each section can then be calculated using Ohm's law.
The magnitude of the electric field E in each section can be calculated from E = V/d, where V is the potential difference across the section and d is its length.
The potential difference across the silver section is the same as the potential difference applied across the entire wire because the copper wire has negligible resistance compared to silver.
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A 9-year-old has just fallen off the monkey bars. She has obvious deformity to her mid forearm. How should the radiographer care for her when taking x-rays?
Answer:
He will need to support the joint above and below the deformity, this will allow the xray to be taken without generating more damage to the child.
I hope you find this information useful and interesting! Good luck!
An air compressor takes normal air and pushes more and more of it in a rigid steel tank. The number of gas particles in the tank is increased. What happens to the temperature as more and more air is pushed into a tank which does not change size?
A. It increases
B. It stays the same
C. It decreases
Answer:
C. It decreases
Explanation:
The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualizing the particles of gas in the container moving with a greater energy when the temperature is increased.
A common example is cooking gas when refilled, there is a perceptible change in the temperature of the cylinder.
(a) Use the de Broglie relation λ = h/p to nd the wavelength of a raindrop with mass 1 mg and speed 1 cm/s. Is there a way to set up a raindrop-diraction experiment and detect the wave-like properties of rain?
Question:
a). Use the de Broglie relation λ=h/p to find the wavelength of a raindrop with mass m=1 mg and speed 1cm/s.
ii). Does it seem likely that the wave properties of a raindrop could be easily detected?
b). Find the wavelength of electrons with KE = 500 eV.
c). If a neutron has the same wavelength as blue light (λ=450 nm) what is it's KE?
ii). What if it's an electron?
Answer:
The answers to the question are
a). The wavelength of the raindrop is 6.626*10⁻²⁶ m
The properties of the rain drop will be hardly detected
b). The wavelength of the electrons is 5.491×10⁻¹¹ m
c). The KE of the neutron is 5.242510⁻²⁸ J
ii). For an electron it will increase to be KE (electron) = 9.6392639×10⁻²⁵ J
Explanation:
Using de Broglie relations, we have
p = h/λ and E = h·f also E = 1/2·m·v²
a). λ= h/p, E= p²/2·m, p = √(2·m·E), λ = h/√(2·m·E)
Where
λ=wavelength
E = energy
p = momentum
m = mass
The kinetic energy of the rain drop is [tex]\frac{1}{2}[/tex]×m×v² = 0.5×(1×10⁻⁶)(0.01)2
= 5× 10⁻¹¹ J
λ = h/√(2·m·E) = 6.626*10-34 Js/√(2×1×10⁻⁶×5× 10⁻¹¹)
= 6.626*10⁻²⁶ m
The properties of the rain drop will not be easily detected
b). The electron energy is equivalent to 500 eV ⇒500 eV × 1.6×10⁻¹⁹ J/eV
= 8×10⁻¹⁷ J
λ = h/√(2·m·E) = 6.626×10⁻³⁴ Js/√(2*×9.1×10⁻³¹×8×10⁻¹⁷)
= 5.491×10⁻¹¹ m
c). λ = h/√(2·m·E) then √(2·m·E) = h/ λ or E = (h/λ)²/(2·m)
= (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×1.674927471×10⁻²⁷)
E = 5.242510⁻²⁸ J
ii). For an electron, we have m = 9.10938356 × 10⁻³¹ kg
λ = (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×9.10938356×10⁻³¹) = 9.6392639×10⁻²⁵ J
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and rotates at 226 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.
Answer:
a. Rotational inertia: 5.21kgm²
b. Magnitude of it's angular momentum: 123.32kgm²/s
Explanation:
Length of the rod = 3.46m
Weight of the rod = 12.8 N
Angular velocity of the rod= 226 rev/min
a. Rotational Inertia (I) about its axis
The formula for rotational inertia =
I = (1/12×m×L²) + m × ( L ÷ 2)²
Where L = length of the rod
m = mass of the rod
Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.
Acceleration due to gravity = 9.81m/s²
Mass of the rod = 12.8N/ 9.81m/s²
Mass of the rod = 1.305kg
Rotational Inertia =
(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²
Rotational Inertia = 1.3019115 + 3.9057345
Rotational Inertia = 5.207646kgm²
Approximately = 5.21kgm²
b. The magnitude of the rod's angular momentum about the rotational axis is calculated as
Rotational Inertia about its axis × angular speed of the rod.
Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60
= (226×2π) /60
= 23.67 rad/s
Rotational Inertia = 5.21kgm²
The magnitude of the rod's angular momentum about the rotational axis
= 5.21kgm²× 23.67 rad/s
= 123.3207kgm²/s
Approximately = 123.32kgm²/s
The energy expenditure value of traveling by car is 3.6 mj/passenger-kilometer. The value for traveling by train is 1.1 mj/passenger-kilometer. What would be the best way to increase the efficiency of traveling by car?
Answer:
Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.
Explanation:
Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.
Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:
[tex]\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s[/tex]
In a particular television picture tube, the measured beam current is 23.3 µA . How many electrons strike the tube screen every 28 s ? The fundamental charge is 1.602 × 10−19 C. Answer in units of electrons.
Answer:
Explanation:
Given that
Beam current (i)=23.3µA
And the time to strike(t)=28s
Also, a fundamental charge e=1.602×10^-19C
Then, the charge quantity is given as,
q=it
Then, q=23.3×10^-6×28
q=6.524×10^-4C
Also, the number of electron N is given as
q=Ne
Therefore, N=q/e
So, N=6.524×10-4/1.602×10^-19
N=4.072×10^15
There are 4.072×10^15 electrons strike the tube screen every 28 s.
When The fundamental charge is 1.602 × 10−19 C So, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
Calculate of Units of Electrons
Given that information as per the question
Beam current (i) is =23.3µA
And also the time to strike(t)=28s
Also, a fundamental charge e=[tex]1.602\times10\wedge -19C[/tex]
Then, the charge quantity is given as,
q is =it
Then, q=[tex]23.3\times 10\wedge-6×28[/tex]
q is =[tex]6.524\times10\wedge -4C[/tex]
Also, When the number of electron N is given as
Now, q=Ne
Therefore, N is =q/e
So, N is = [tex]6.524\times10-4/1.602\times10\wedge-19[/tex]
N is = [tex]4.072\times 10\wedge 15[/tex]
Therefore, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
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Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the horizontal (see below). He exerts a force of 535 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
Answer:
W = 3.4x0³ J.
Explanation:
The work done by the man is given by the following equation:
[tex] W = F_{t}\cdot d [/tex] (1)
where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.
We need to find first the total force Ft, which is:
[tex] Ft = Fm + W [/tex]
where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.
[tex] F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N [/tex]
Hence, the work is:
[tex] W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J [/tex]
Therefore, the work done by the man is 3.4x10³ J.
I hope it helps you!
This image shows a stream of positively charged particles being directed at gold foil. The positively charged particles are called "alpha particles” and each one is like a nucleus without any electrons.
What is the best explanation for why a particle is striking point X?
Alpha particles are mostly empty space, so they move in random directions, and one of these alpha particles can strike point X.
The gold atoms contain negative electrons, so when alpha particles strike the gold they keep moving but in various directions.
If an alpha particle hits the gold foil, a gold nucleus splits and a particle from it flies out at point X.
When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
Answer:
D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
Explanation:
D)When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
What happens to alpha particles that pass close to the nucleus of a gold atom?The gold nucleus and alpha particle are both definitely charged therefore there is a repulsive pressure between the (gold) nucleus and the alpha particle. This causes the alpha particle to be deflected through a massive angle.
Maximum alpha debris surpassed instantly through the gold foil, which implied that atoms are ordinarily composed of open space. a few alpha particles had been deflected barely, suggesting interactions with different definitely charged particles in the atom.
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A comet is approaching Earth at a known velocity. Indicate the proper sequence that describes the wavelength of light you measure as it first approaches, then passes, and finally recedes from Earth.
Answer:
Explanation:
When it was approaching, the wavelength is blue-shifted and the rest wavelength when it was receding the wavelength is redshifted.
Any planet is an extremely faint light source compared to its parent star. Comets are considered as the most primitive objects in the Solar System.
For a body falling freely from rest (disregarding air resistance), the distance the body falls varies directly as the square of the time. If an object is dropped from the top of a tower 490 ft high and hits the ground in 7 sec, how far did it fall in the first 5 sec?
Explanation:
Below is an attachment containing the solution.
The distance covered by a freely falling object in a certain time is calculated by the formula d = 1/2 * g * t^2 where d is the distance, g is acceleration due to gravity and t is time. When plugging in the values g = 32 ft/sec^2 and t = 5 sec, we find that the object would have fallen 400 ft in the first 5 seconds.
Explanation:In physics, the distance that a free-falling object covers is given by the formula d = 1/2 * g * t^2, where g is the acceleration due to gravity. In typical physics problems, g is approximated as 32 feet/second^2 on Earth. If we plug t = 5 sec into the equation, we get:
d = 1/2 * 32 ft/sec^2 * (5 sec)^2 = 400 ft.
Therefore, the object would have fallen 400 ft in the first 5 seconds of free fall.
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You're having a hard time pushing a refrigerator across the kitchen floor. The force of your own push is 993 N. The force of friction opposing your own push is 973 N. If the refrigerator has a mass of 355 kg, what is the refrigerator's acceleration
Answer:
The acceleration of the refrigerator is [tex]a= 0.056 ms^{-2}[/tex]
Explanation:
The expression of the equation of the net force acting on the refrigerator is as follows;
F-f= ma
Here, F is the applied force, f is the force of friction, m is the mass and a is the acceleration.
It is given in the problem that you're having a hard time pushing a refrigerator having mass 355 kg across the kitchen floor. The force of your own push is 993 N. The force of friction opposing your own push is 973 N.
Put F= 993, f= 973 N and m = 355 kg in the above expression of the equation to calculate the acceleration of the refrigerator.
993 - 973 = (355)a
20 = 355 a
[tex]a= 0.056 ms^{-2}[/tex]
Therefore, the acceleration of the refrigerator is [tex]a= 0.056 ms^{-2}[/tex].
The refrigerator's acceleration is 0.0563 m/s², calculated by subtracting the force of friction from the pushing force and then dividing by the refrigerator's mass.
Explanation:To find the acceleration of the refrigerator, we first need to calculate the net force acting on it. The net force is the difference between the force of the push and the force of friction.
Step 1: Calculate the Net Force
Net Force = Force of the push - Force of friction
Net Force = 993 N - 973 N
Net Force = 20 N
Step 2: Apply Newton's Second Law of Motion
According to Newton's Second Law of Motion, the acceleration (a) can be calculated using the formula:
a = Net Force / Mass
Substituting the known values:
a = 20 N / 355 kg
a ≈ 0.0563 m/s²
The refrigerator's acceleration when being pushed across the kitchen floor is approximately 0.0563 meters per second squared.
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A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.16 m/s. Suppose the collision is elastic.
Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar system. have the Sun at their exact center. are highly inclined to the ecliptic. are almost circular, with low eccentricities.
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
[tex]T^{2} = a^{3}[/tex]
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
Which of the following represents energy in its most disordered form? Group of answer choices Chemical-bond energy Electromagnetic (light) energy Heat energy Potential energy The kinetic energy of a moving object
Answer: Heat Energy
Explanation:
Heat is energy in its most disordered form. heat energy is the random jostling of molecules and is therefore not organized. As cells perform the chemical reactions that generate order within, some energy is inevitably lost in the form of heat. Because the cell is not an isolated system, the heat energy produced by the cell is quickly dispersed into the cell's surroundings where it increases the intensity of the thermal motions of nearby molecules. This increases the entropy of the cell's environment and keeps the cell from violating the second law of thermodynamics.
A generator uses a coil that has 100 turns and a 0.50-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made
Answer:
38 m
Explanation:
Number of turns=N=100
Magnetic field=B=0.50 T
Frequency of the generator=f=60 Hz
Rms value of emf=[tex]E_{rms}=120 V[/tex]
We have to find the length of the wire from which the coil is made.
Peak value of emf=[tex]E_0=E_{rms}\sqrt 2=120\times \sqrt 2=169.7 V[/tex]
Length of wire=[tex]4\sqrt{\frac{NE_0}{2\pi fB}}[/tex]
Substitute the values
Length of wire=[tex]4\times \sqrt{\frac{169.7\times 100}{0.50\times 2\pi\times 60}}[/tex]
Length of wire=38 m
Hence, the length of wire from which the coil is made=38 m
To determine the length of wire in a generator coil with 100 turns in a 0.50-T field and 120 V rms at 60 Hz, one can use the formula for rms emf of a generator and solve for the area of one turn to find the length per turn and multiply by the number of turns.
Explanation:The question is asking to determine the length of wire used to make a coil in a generator. The generator has 100 turns of wire, operates with a 0.50-T magnetic field, and has an rms value of the emf of 120 V with a 60.0 Hz frequency. Assuming the turns are squares, we can use the formula for the rms value of the emf (Erms) for a generator, which is Erms = NABωrms, where N is the number of turns, A is the area of the turn, B is the magnetic field, and ωrms is the rms angular velocity. The rms angular velocity ωrms is related to the frequency (f) by the equation ωrms = 2πf/√2.
To find the side length (L) of the square turns, we rearrange the formula to solve for A and then take the square root. Once we have L, we multiply by 4 to get the perimeter of one turn and then by 100 to find the total length of wire needed for all turns.
The quantity of heat Q that changes the temperature L1Tof a mass mof a substance isgiven by Q = cmt:T, where c is the specific heat capacity of the substance. Forexample,forH20,c=1caljg'C",Andfora change of phase, the quantity of heat Q that changes the phase of a mass m is Q = ml., where L is the heat of fusion or heat of vaporization of the substance. For example, for H20, the heat offusion is 80 cal/g (or 80 kcaljkg) and the heat of vaporization is 540 cal/g (or 540 kcaljkg). Use these relationships to determine the number of calories to change (a) 1 kg ofO°C ice to O°C ice water, (b) 1 kg ofO°C ice water to 1 kg of 100°C boiling water, (c) 1 kg of 100°C boiling water to 1 kg of 100°C steam, and (d) 1 kg ofO°C ice to 1 kg of 100°C steam.
Answer:
a) Q = 80,000 cal
b) Q = 100,000 cal
c) Q = 540,000 cal
d) Q = 720,000 cal
Explanation:
a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion
[tex]Q_{l} = ML_{f}[/tex] = 1 * 80
[tex]Q_{l}[/tex] = 80 kCal = 80,000 cal
b) 1 kg of O°C ice water to 1 kg of 100°C boiling water
Specific heat capacity, c = 1000cal/kg.C
[tex]Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal[/tex]
c) 1 kg of 100°C boiling water to 1 kg of 100°C steam
Latent heat of vaporization is needed for this conversion
[tex]Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal[/tex]
d) 1 kg of O°C ice to 1 kg of 100°C steam.
Q = [tex]Q_{L} + Q_{c} + Q_{v}[/tex]
Q = 80,000 + 100,000 + 540,000
Q = 720,000 cal
Final answer:
To change 1 kg of ice at 0°C to water at 0°C, 80 kcal of energy is required. To change 1 kg of water at 0°C to boiling water at 100°C, 100 cal of energy is required. To change 1 kg of boiling water at 100°C to steam, 540 kcal of energy is required.
Explanation:
The quantity of heat required to change the phase of a substance is given by the equation Q = mL, where m is the mass of the substance and L is the heat of fusion or heat of vaporization. For example, to change 1 kg of ice at 0°C to water at 0°C, the energy required is Q = (1 kg)(80 kcal/kg) = 80 kcal. To change 1 kg of water at 0°C to boiling water at 100°C, the energy required is Q = (1 kg)(1 cal/g °C)(100°C) = 100 cal. To change 1 kg of boiling water at 100°C to steam, the energy required is Q = (1 kg)(540 kcal/kg) = 540 kcal.
A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 300 N applied to its edge causes the wheel to have an angular acceleration of 0.876 rad/s2.
(a)
What is the moment of inertia of the wheel (in kg · m2)?
_____ kg · m2
(b)What is the mass (in kg) of the wheel?
_________ kg
(c)The wheel starts from rest and the tangential force remains constant over a time period of 6.00 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
________ rad/s
Answer:
a) [tex]I = 113.014\,kg\cdot m^{2}[/tex], b) [tex]m = 2075.556\,kg[/tex]
Explanation:
a) The turntable has the following physical model by using Newton's laws:
[tex]F \cdot R = I \cdot \alpha[/tex]
The moment of inertia is:
[tex]I = \frac{F\cdot R}{\alpha}[/tex]
[tex]I = \frac{(300\,N)\cdot(0.33\,m)}{0.876\,\frac{rad}{s^{2}} }[/tex]
[tex]I = 113.014\,kg\cdot m^{2}[/tex]
b) The moment of inertia for a solid cylinder:
[tex]I = \frac{1}{2}\cdot m \cdot R^{2}[/tex]
The mass of the turntable is:
[tex]m = \frac{2 \cdot I}{R^{2}}[/tex]
[tex]m = \frac{(2)\cdot (113.014\,kg\cdot m^{2})}{(0.33\,m)^{2}}[/tex]
[tex]m = 2075.556\,kg[/tex]
The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.35 rad/s2. It accelerates for 28.3 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 58.9 s after it begins rotating.
Answer:
Total angle through which the wheel has turned 58.9s after it begins rotating is 1709.52 rad
Explanation:
The image attached would offer a better explanation
Answer:
The answer to the question is;
The total angle through which the wheel has turned 58.9 s after it begins rotating is approximately 1709.67 rad.
Explanation:
To solve the question we note the equation for the motion of the flywheel as
ω₂ = ω₁ + α·t
Where:
ω₁ = Initial angular velocity = 0 rad/s as the body is initially at rest
ω₂ = Final angular velocity
α = angular acceleration = 1.35 rad/s²
t = Time = 28.3 s
Plugging in the values, we find ω₂
ω₂ = 0 + 1.35 rad/s²× 28.3 s = 38.205 rad/s
Since the acceleration is constant, only the mean velocity is required to determine the angle traveled during the first 28.3 seconds thus
Average velocity
ω[tex]_{average}[/tex]= [tex]\frac{Final .Velocity +Initial . Velocity}{2} = \frac{\omega_2+\omega_1}{2} = \frac{38.205 rad/s+0 rad/s}{2}[/tex]
= 19.1025 rad/s
The total angle traveled in 28.3 s is ω[tex]_{average}[/tex] × time
= 19.1025 rad/s × 28.3 s = 540.60075 rad
After this the remaining time left is
58.9 s - 28.3 s = 30.6 s
Since the flywheel is moving at a constant velocity of 38.205 rad/s during the last 30.6 s we have
Angle traveled in 30.6 s at an angular velocity of 38.205 rad/s is given by
Angle traveled = Time × Angular velocity = 30.6 s × 38.205 rad/s
= 1169.073 rad
Therefore, the total angle traveled by the flywheel in 58.9 s is given by
540.60075 rad + 1169.073 rad = 1709.67375 rad ≈1709.67 rad.
the total angle traveled by the flywheel in 58.9 s ≈ 1709.67 rad.
A car of mass m = 1030 kg is traveling down a θ = 13-degree incline. When the car's speed is v0 = 14 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
a. Write an expression for the magnitude of the force of kinetic friction
b. Write an expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline.
c. Calculate the distance the car travels down the hill 1 in meters until it comes to a stop at the end
Answer:
a. [tex]F_f = \mu mg cos\theta[/tex]
b. h = Lsinθ
c. 22.78 m
Explanation:
a. The kinetic friction is the product of kinetic coefficient and normal force N, which is the gravity force in the direction normal to the incline
[tex]F_f = \mu N = \mu mg cos\theta[/tex]
b. As the car travels a distance L down the incline of θ degrees, vertically speaking it would have traveled a distance of:
h = Lsinθ
As we can treat L and h in a right triangle where L is the hypotenuse and h is a side length in opposite of incline angle θ
c. Let g = 9.81 m/s2. the acceleration caused by kinetic friction according to Newton's 2nd law is
[tex]a = F_f/m = \mu g cos\theta = 0.45*9.81*cos13^o = 4.3 m/s^2[/tex]
We can use the following equation of motion to find out the distance traveled by the car:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v = 0 m/s is the final velocity of the car when it stops, [tex]v_0[/tex] = 14m/s is the initial velocity of the car when it starts braking, a = -4.3 m/s2 is the deceleration of the car, and [tex]\Delta s[/tex] is the distance traveled, which we care looking for:
[tex]0^2 - 14^2 = 2(-4.3)\Delta s[/tex]
[tex]\Delta s = 14^2 / (2*4.3) = 22.78 m[/tex]
a. An expression for the magnitude of the force of kinetic friction is [tex]\(f_{\text{friction}} = \mu_k \cdot N\).[/tex]
b. An expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline is: [tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. The car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
The detailed explanation is as follows:
a. The magnitude of the force of kinetic friction can be calculated using the formula:
[tex]\(f_{\text{friction}} = \mu_k \cdot N\),[/tex]
Where:
[tex]\(f_{\text{friction}}\)[/tex] is the force of kinetic friction,
[tex]\(\mu_k\)[/tex] is the coefficient of kinetic friction (given as 0.45),
[tex]\(N\)[/tex] is the normal force.
The normal force can be calculated using the equation:
[tex]\(N = m \cdot g \cdot \cos(\theta)\),[/tex]
Where:
[tex]\(m\)[/tex] is the mass of the car (1030 kg),
[tex]\(g\)[/tex] is the acceleration due to gravity (approximately 9.81 m/s²),
[tex]\(\theta\)[/tex] is the angle of the incline (13 degrees converted to radians).
b. The magnitude of the change in the car's height [tex](\(h\))[/tex] along the y-direction can be found using trigonometry. When the car travels a distance [tex]\(L\)[/tex] down the incline, the vertical displacement [tex](\(h\))[/tex] can be calculated as:
[tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. To calculate the distance the car travels down the hill until it comes to a stop, you can use the work-energy theorem. The work done by the force of kinetic friction will be equal to the initial kinetic energy of the car. The work-energy theorem is given as:
[tex]\(W = \Delta KE\),[/tex]
Where:
[tex]\(W\)[/tex] is the work done by friction (negative, as it opposes motion),
[tex]\(\Delta KE\)[/tex] is the change in kinetic energy.
The initial kinetic energy is:
[tex]\(KE_0 = \frac{1}{2} m v_0^2\).[/tex]
The final kinetic energy is zero because the car comes to a stop.
So, the work done by friction is:
[tex]\(W = -\frac{1}{2} m v_0^2\).[/tex]
Now, you can use the work-energy theorem to find the distance \(L\) down the incline:
[tex]\(W = -\frac{1}{2} m v_0^2 = \Delta KE = KE_f - KE_0\),[/tex]
Where [tex]\(KE_f = 0\)[/tex] (final kinetic energy).
Solve for [tex]\(L\):[/tex]
[tex]\(-\frac{1}{2} m v_0^2 = -\mu_k m g L \cos(\theta)\).[/tex]
Now, solve for [tex]\(L\):[/tex]
[tex]\[L = \frac{v_0^2}{2 \mu_k g \cos(\theta)}.\][/tex]
Substitute the known values:
[tex]\[L = \frac{(14 m/s)^2}{2 \cdot 0.45 \cdot 9.81 m/s^2 \cdot \cos(13^\circ)} \approx 94.69 \, \text{meters}.\][/tex]
So, the car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
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Two charges A and B are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.15 m from the spot, while charge B is 0.48 m from it. Find the ratio qB/qA of the charges.
Answer:
qa/qb = 0.3125
Explanation:
Let the distance of the point from first charge (qa) be ra.
Likewise, let the distance of the point from the second charge (qb) be ra
Now, from the question, ra=0.15m
While rb = 0.48m
Normally, we know that:
The electric potential due to a point charge, q, at a point located at a distance, r, away from it is given by the equation;
V = q/(4π(ϵo)r)
We know that 1/(4π(ϵo)) cam be said to ne K.
Therefore, V = Kq/r
Where K = 9 × 10^(9) V.m/C
Now, since from the question, the electric potential at the point is the same due to each of the charges, their electric potential will be the same, thus;
Va = Vb
So, (Kqa) / ra = (Kqb) / rb
This gives us; qa / ra = qb / rb
So rearranging, we get;
qa/qb = ra/rb = 0.15/0.48 = 0.3125
The ratio qB/qA of the charges is 3.2.
The electric potential owing to a point charge Q at a distance r from the charge is given as:
V = kQ / r
where k is the electrostatic constant.
Since the potentials due to charges A and B are equal at the given spot, we can write:
VA = VB
Substituting the values:
kQA / 0.15 = kBQB / 0.48
Simplifying:
qB/qA = 0.48 / 0.15 = 3.2
A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis of the hole is a distance b from the axis of the cylinder, where a 6 b 6 R (Fig. P22.58). The solid material of the cylinder has a uniform volume charge density r. Find the magnitude and direction of the electric field E S inside the hole, and show that E S is uniform over the entire hole
The electric field inside the hollowed insulator can be calculated considering the missing charge in the cylindrical hole. This electric field is uniform and its magnitude is E = ρ.b/ϵ0; where ρ is the uniform charge density and ϵ0 is the permittivity constant. The direction of the electric field is from the axis of the cylinder toward the hole for positive ρ and opposite for negative ρ.
Explanation:The question refers to the electric field inside a hollowed cylindrical insulator. This is a physics problem related to the study of electrostatics. The concept of electric field refers to the influence exerted in the space around a charged object, which can result in forces on other charged objects.
We know the electric field inside a uniformly charged insulator is zero because charges move to the surface in an insulator. In this case, though, the insulator is not uniform; it has a cylindrical hole. So, we must calculate the contribution from the missing charge in the hole.
The electric field generated by the missing volume, dV, with charge density ρ is given by Coulomb's law, where the electric field corresponds to the charge divided by the square of the distance: E = k.Q/R^2.
Because the volume dV, at radius r from the axis of the cylinder has a charge equal to the volume of the cylindrical disk multiplied by ρ, we have dQ = ρ.dV and thus E = k.ρ.dV/R^2. Integrating over the total volume of the cylindrical hole, we find that the electric field is uniform and its magnitude, E, is given by E = ρ.b/ϵ0, where ϵ0 is the permittivity constant.
The direction of the E is from the axis of the cylinder towards the hole if ρ is positive, because positive charges repel, and vice versa if ρ is negative.
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To find the electric field inside the hole of the insulating cylinder, we use the superposition principle and Gauss's law, resulting in a uniform electric field due to the subtracted cylindrical charge distribution.
The question is asking for the magnitude and direction of the electric field inside a cylindrical hole bored inside another charged cylindrical insulator. To determine the electric field (ES) inside the hole, we apply the principle of superposition, which involves considering the charge distribution that would have been present had there been no hole and subtracting the charge distribution of a cylinder of radius a with charge density -ρ, placed such that it would carve out the hole.
Using Gauss's Law, the electric field due to the entire charged cylinder without the hole would be zero inside the cylinder. Therefore, the electric field inside the hole is solely due to the cylindrical charge distribution we subtracted. Since the cylindrical charge distribution is symmetric, the electric field ES within the hole will also be uniform and directed radially outward from the axis of the imaginary cylinder we subtracted. The magnitude of ES can be calculated using Gauss's Law and considering the symmetry of the problem.
A proton (????p = +????, mp = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron (???????? = +????, m???? = 2.0 u) and an alpha particle (???????? = +2????, m???? = 4.0 u) are accelerated from rest through the same potential difference ????, and then enter the same region of uniform magnetic field ????⃗⃗ , moving perpendicularly to the direction of the magnetic field.
a. What is the ratio of the proton’s kinetic energy Kp to the alpha particle’s kinetic energy K?????
b. What is the ratio of the deuteron’s kinetic energy K???? to the alpha particle’s kinetic energy K?????
c. If the radius of the proton’s circular orbit ????p = 10 cm, what is the radius of the deuteron’s orbit ?????????
d. What is the radius of the alpha particle’s orbit ?????????
Answer:
a. 1/2 b. 1/2 c, 20 cm d. 40 cm
Explanation:
Here is the complete question
A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = +2, = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.
A) What is the ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy ?
B) What is the ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic energy ?
C) If the radius of the proton’s circular orbit = 10 cm, what is the radius of the deuteron’s orbit ?
D) What is the radius of the alpha particle’s orbit ?
Solution
a. For both particles, kinetic energy = electric potential energy
For proton K.E= K₁ = 1/2m₁v₁² = +eV , for alpha particle K.E = K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
b. For both particles, kinetic energy = electric potential energy
For deuteron K₁ = 1/2m₁v₁² = +eV , for alpha particle K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,
mv²/r = Bev
r₁ = mv/Be
Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁
So, radius of deuteron's circular orbit equals
r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 × 10 cm = 20 cm
d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.
So, radius of alpha particle's circular orbit equals
r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 × 10 cm = 40 cm
_____ has the potential to superimpose digital data over real photos so that GPS maps can be combined with real pictures of stores and streets to help people locate their position.
Augmented reality has the potential to superimpose digital data over real photos so that GPS maps can be combined with real pictures of stores and streets to help people locate their position.
Explanation:
An engaging perception of an original globe atmosphere, where by computer-generated perceptual knowledge the transformation of real-world entities take place and also by multiple sensory modalities, involving somatosensory, visual, auditory, haptic and olfactory forms, thus known as augmented reality.
AR app uses GPS and camera from a smartphone to deploy an augmented reality-enabled GPS navigation system. As in the web, AR tool termed as Real View Navigation is accessible to all Android and iOS clients. Google is brought its first virtual reality walking directions, now recognized as Live Experience on Google Maps.
What is the angular velocity of the second hand on a clock? (Hint: It takes the second hand 6060 seconds to rotate 2π2π radians. Divide the number of radians by the number of seconds so that your answer has units of radians/second.)
Explanation:
Below is an attachment containing the solution.
6. Two blocks are released from rest at the same height. Block A slides down a steeper ramp than Block B. Both ramps are frictionless. The blocks reach the same final height indicated by the lower dashed line. Block B weighs more than Block A.
a. Is the work done by the gravitational force on Block A (greater/less than/equal to) the work done by the gravitational force on Block B? Explain your reasoning.
b. Is the speed of Block A (greater/less than/equal to) the speed of Block B? Explain your reasoning.
c. Is the momentum of Block A (greater/less than/equal to) the momentum of Block B? Explain your reasoning.
Answer:
a. the work done by the gravitational force on Block A is less than the work done by the gravitational force on Block B.
b. the speed of Block A is equal to the speed of Block B.
c. the momentum of Block A is less than the momentum of Block B.
Explanation:
a. The work done by the gravitational force is equal to:
w = m*g*h
where m is mass, g is the standard gravitational acceleration and h is height. Given that both blocks are released from rest at the same height, then, the bigger the mass, the bigger the work done.
b. With ramps frictionless, the final speed of the blocs is:
v = √(2*g*h)
which is independent of the mass of the blocks.
c. The momentum is calculated as follows:
momentum = m*v
Given that both bocks has the same speed, then, the bigger the mass, the bigger the momentum.
The work done by gravitational force on blocks A and B is equal as the work is independent of the path. Both blocks have the same speed when they reach the final height due to the conversion of potential energy into kinetic energy. However, the momentum of Block B is greater due to its larger mass.
Explanation:This question is about the principles of work, energy and momentum in physics. Let's address each part of it:
Work done by gravitational force: The work done by the gravitational force on both blocks A and B is equal. This is because work done by gravity depends only on the change in height, which is the same for both blocks, and the weight of the block. So, even though Block B is heavier as it weighs more than Block A, the work done is the same because the force of gravity has to move the blocks the same vertical distance.Speed of the blocks: The speed of both blocks A and B will be equal when they reach the final height. This is because, in the absence of friction and air resistance, the blocks will convert all of their gravitational potential energy at the start into kinetic energy at the end, irrespective of their weights or the steepness of the ramps.Momentum of the blocks: The momentum of Block B is greater than Block A. This is because momentum depends on both the mass and the velocity of an object. Even though both blocks have the same speed at the bottom, Block B is heavier, therefore, it will have more momentum.Learn more about Physics principles here:https://brainly.com/question/15872863
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Help please!
Which setup could not be used to induce an electric current?
Spinning wire coils within an array of fixed permanent magnets
Spinning electromagnets within an array of fixed wire coils
Spinning permanent magnets within an array of fixed permanent magnets
Spinning wire coils within an array of fixed electromagnets
Answer:
Definitely Spinning permanent magnets within an array of fixed permanent magnets
Explanation:
Any relative motion between magnets (be they permanent or electromagnetic) and a coil of wire will induce an electric current in the coil.
What will not induce an electric current is the relative motion between the two coils of wire (because there is no change in magnetic field), or the relative motion between two magnets (there are no coils of wire to induce the current into).
Therefore, spinning permanent magnets within an array of fixed permanent magnets does not induce an electric current.
Answer:
Spinning permanent magnets within an array of fixed permanent magnets
Explanation: