Answer:
Decrease the amount of carbon dioxide in the ecosystem.
Explanation:
becaus egthy
To increase the number of energy storage molecules that producers can make, Jaime should:
1. Increase the amount of sunlight in the ecosystem: Sunlight is a vital source of energy for photosynthesis, the process by which producers convert carbon dioxide and water into energy-rich molecules such as glucose. More sunlight can potentially increase the rate of photosynthesis, leading to more energy storage molecules being produced.
2. Decrease the amount of carbon dioxide in the ecosystem: While carbon dioxide is essential for photosynthesis, excessive levels can inhibit the process. By decreasing the amount of carbon dioxide in the ecosystem to an optimal level, Jaime can ensure that producers can efficiently utilize available resources to produce energy storage molecules.
So, both increasing sunlight and decreasing carbon dioxide can contribute to increasing the number of energy storage molecules that producers can make.
A ball of mass 1.00 kg is thrown at a door with velocity v1x = 12.0 m/s. The rectangular door has a mass of 30.0 kg and is 1.00 m wide. The ball strikes the door perpendicular to it, a horizontal distance 75.0 cm away from the hinges, and causes the door to swing open with an angular speed of 1.20 rad/s. What is the velocity v2x of the ball after striking the door? (The moment of inertia of a thin rectangular plate with mass M and dimensions a × b and the axis along the edge of length b is I = 1/3Ma2.)"
Answer:
The velocity of the ball after collision is [tex]v_{2x} = -4 m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m_b = 1.00\ kg[/tex]
The velocity of the ball is [tex]v_{1x}= 12.0 \ m/s[/tex]
The mass of the rectangular door is [tex]m_d = 30 \ kg[/tex]
The width of the door is [tex]a = 1.00 \ m[/tex]
The distance of impact from the hinge is [tex]L = 75 \ cm = \frac{75}{100} = 0.75 \ m[/tex]
The angular speed of the door is [tex]w = 1.20 \ rad/s[/tex]
So the moment of inertia of the door is given from the question as
[tex]I = \frac{1}{3} M a^2[/tex]
substituting values
[tex]I = \frac{1}{3} * 30 * (1)[/tex]
[tex]I = 10 \ kg \cdot m^2[/tex]
According to the law of angular momentum conservation
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_i = m_b * v_{1x} + Iw_i[/tex]
so [tex]w_i[/tex] is the initial angular speed of the door which is zero
So
[tex]L_i = m_b * v_{1x}[/tex]
[tex]L_f[/tex] is the final angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_f = I w + m_b v_{2x} * L[/tex]
So
[tex]m_b * v_{1x} = I w + m_b v_{2x} * L[/tex]
Substituting values
[tex]1 * 12 * 0.75 = 10* 1.2 * v_{2x} * 0.75[/tex]
[tex]v_{2x} = -4 m/s[/tex]
The negative sign show a reversal in the balls direction
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. The Moon has a mass of 7.36 × 1022 kg; the Earth has a mass of 5.98 × 1024 kg; and the Sun has a mass of 1.99 × 1030 kg. The distance between the Moon and the Earth is 3.84 × 108 m; the distance between the Earth and the Sun is 1.496 × 1011 m. Calculate the total gravitational potential energy for this arrangement.
Answer:
U = 5.37*10^33 J
Explanation:
The gravitational potential energy between two bodies is given by:
[tex]U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}[/tex]
G: Cavendish's constant = 6.67*10^-11 m^3/kg.s
For three bodies the total gravitational potential energy is:
[tex]U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}][/tex]
BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:
[tex]U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J[/tex]
hence, the total gravitational energy is 5.37*10^33 J
The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?
Answer:
The amount of electrons that flow in the given time is 3.0 C.
Explanation:
An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.
i.e I = [tex]\frac{Q}{t}[/tex] ...................(1)
It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.
In the given question, I = 1.5A, t = 2s, find Q.
From equation 1,
Q = I × t
= 1.5 × 2
= 3.0 Coulombs
The amount of electrons that flow in the given time is 3.0 C.
The force required to stretch a spring varies directly with the amount the spring is stretched. A spring stretches by 33 m when a 19 N weight is hung from it and the weight is at rest (at equilibrium). The 19 N weight is replaced by an unknown weight W so that the spring is stetched to a new equilibrium position, 17 m below the position if no weight were attached. The weight W is then displaced from equilibrium and released so that it oscillates.
Answer:
The period is [tex]T = 8.27 \ sec[/tex]
Explanation:
From the question we are told that
The extension of the spring is [tex]e_1 = 33 \ m[/tex]
The first weight applied is [tex]F_1 = 19 N[/tex]
The second weight applied is [tex]F_2 = W[/tex]
The second extension is [tex]e_2 = 17 \ m[/tex]
The spring constant of the spring is mathematically evaluated as
[tex]k = \frac{F_1}{e_1 }[/tex]
substituting values
[tex]k = \frac{19}{33 }[/tex]
[tex]k = 0.576[/tex]
We are told that
19 N extended the spring to 33 m
Then W N will extended it by 17 m
Therefore [tex]W = \frac{19 * 17}{33}[/tex]
[tex]W = 9.788 \ N[/tex]
Generally the period of the oscillation is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{M}{K} }[/tex]
where M is the mass of the W which is mathematically evaluated as
[tex]M = \frac{9.788}{9.8}[/tex]
[tex]M = 1.0 \ kg[/tex]
substituting values
[tex]T = 2 \pi \sqrt{\frac{1}{0.576} }[/tex]
[tex]T = 8.27 \ sec[/tex]
The force required to stretch a spring is directly proportional to the amount the spring is stretched. The force constant can be calculated using the equation F = kx. By substituting the given values into the equation, we can find the unknown weight W.
Explanation:The force required to stretch a spring is directly proportional to the amount the spring is stretched. This can be represented by the equation F = kx, where F is the force, k is the force constant, and x is the amount the spring is stretched. In this scenario, we are given that a 19 N weight stretches the spring by 33 m at equilibrium.
To find the force constant, we can use the equation:
k = F / x = 19 N / 33 m = 0.58 N/m
Next, we need to find the unknown weight W that will stretch the spring to a new equilibrium position 17 m below the position with no weight attached. Using the force constant we calculated, we can use the equation F = kx to find the force required:
F = kx = (0.58 N/m)(17 m) = 9.86 N
Therefore, the unknown weight W that will stretch the spring to the new equilibrium position is 9.86 N.
Which of the following statements is NOT true? When a light wave encounter a boundary where they must bounce off of a surface with a larger index of refraction, the phase of the wave flips by pi A monochromatic light source is one that consists of many colors The intensity of the center maximum is the largest in a single slit interference pattern The intensity of the center maximum is the largest in a double slit interference pattern
Answer: A monochromatic light source is one that consists of many colors
Explanation:
Monochromatic comes from "mono - one, chromatic - color", so a monochromatic wave has only one wavelength, this means that it has only one color (while it may have different shades or tones). So the correct option is the second one "A monochromatic light source is one that consists of many colors "
The other 3 statements are true.
The incorrect statement is that 'A monochromatic light source is one that consists of many colors', because a monochromatic light source indeed emits light of a single wavelength or color. The other claims regarding phase shift on reflection, and the location of highest intensity in single and double slit interference patterns are accurate.
Explanation:The statement that is NOT true among the given options is: 'A monochromatic light source is one that consists of many colors'. In fact, a monochromatic light source emits light of a single wavelength or color.
Further addressing the related concepts: When a light wave encounters a boundary where it must reflect off a surface with a larger index of refraction, the phase of the wave indeed flips by pi. This is an important concept in understanding wave reflection and refraction.
Regarding the intensity of the central maximum in both single and double slit interference patterns, it is accurate to say that the intensity of the center maximum is the most significant. This consequence is a direct outcome of the interference pattern produced within these experiments. In the central maximum, most wave paths align constructively, resulting in the highest intensity.
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What initially unknown quantity, together with the wavelength, is sufficient to calculate the stopping potential for 400 nmnm light? View Available Hint(s) Hint 1. Equation for stopping potentialopened hint Recall that the equation for the stopping potential is eV0=hf−ϕeV0=hf−ϕ.
Answer:
The initially known quantity, together with the wavelength, that is sufficient to calculate the stopping potential for electrons from the surface of a metal is called the WORK FUNCTION.
Explanation:
The stopping potential is defined as the potential that is required to stop electrons from being ejected from the surface of a metal when light with energy greater than the metal's work function/work potential is incident on the metal.
Given that light is known to be made up of photons, which carry energy in packets according to the frequencies of the light.
The photoelectric phenomenon explains that when light of a certain frequency that corresponds to an energy level that is higher than a metal's work function is incident on a metal, it will lead to electrons being ejected from the surface of the metal. The energy of the ejected electrons is then proportional to the difference between the energy level of the photons and the metal's work function.
Basically, it is the excess energy after overcoming the work function that rejects the electrons.
So, to prevent this excess energy from ejecting electrons from a metal's surface, an energy thay matches this excess must be in place to stop electrons from coming out. This energy/potential required to stop the ejection of electrons, is called the stopping potential.
The stopping potential is given as
eV₀ = hf - ϕ
The stopping potential (eV₀) them depends on the hf and the ϕ.
hf is the energy of the photons, where h is Planck's constant and f is the photons' frequency which is further given as
f = (c/λ)
c = speed of light (speed of the photons)
λ = wavelength of the photons.
The other quantity, ϕ, is the metal's work function; the amount of energy needed to be overcome by the photons before ejection of electrons is possible. It is the minimum energy that the light photoms must possess to even stand a chance of being able to eject electrons from a metal's surface.
So, the stopping potential is the difference between the energy of the photons (obtained using the photons' frequency, wavelength and/or speed) and the metal's work function.
Hope this Helps!!!!
[100 POINTS] Which of the following do not make their own energy through nuclear fusion? Select all that apply.
A. giant star
B. protostar
C. dwarf star
D. black hole
E. neutron star
F. main sequence star
Answer:
The correct answer is :
A . Giant star
B . proto star
F .main sequence star
Explanation:
hope this helps
TRY
An electric oven of resistance 2002
drawn a current of 10A. it works 3
hours a daily. Calculate the weekly
bill when energy is $0.48 per kwh
Hint: P=1?R
Answer:
$2018.016
Explanation:
Applying,
P = I²R.............. Equation 1
Where P = power, drawn by the oven, I = current drawn by the oven, R = resistance of the oven.
Given: I = 10 A, R = 2002 Ω
Substitute this values into equation 1
P = 10²(2002)
P = 100(2002)
P = 200200 W.
P = 200.2 kW
If the electric oven works 3 hours daily,
Then the total time it works in a week = 3×7 = 21 hours.
E(kWh) = P(kW)×t(h)
E = 200.2×21
E = 4204.2 kWh.
If 1 kWh of energy cost $0.48,
Then, 4204.2 kWh = $4204.2(0.48) = $2018.016
constant torque 2.5 kW motor drives a riveting machine. The mass of the moving parts including the flywheel is 125 kg at a radius of gyration of 700 mm. One riveting operation absorbs 1000 J of energy and takes one second. Speed of the flywheel is 240 rpm before riveting. (a) Determine the reduction in speed after one riveting operation. (b) Modify the mass moment of inertia to the flywheel if the reduction in speed is to be limited to 35 rpm.
Answer:
speed after one riveting operation = 230.5 rpm
mass moment of inertia = 9.90 kg
Explanation:
given data
torque energy = 2.5 kW = 2500 W
mass = 125 kg
radius = 700 mm
energy = 1000 J
Speed of the flywheel N = 240 rpm
solution
we know here Speed of the flywheel N so here angular speed ω
ω = [tex]\frac{2\pi N}{60}[/tex] .....................1
ω = [tex]\frac{2\pi 240}{60}[/tex]
ω = 25.13 rad/s
so here
change in energy is
ΔE = E1 - E2 ..............2
ΔE = 2500 - 1000
ΔE = 1500 J/sec
and
I = mr² .........3
I = 125 × 0.7²
I = 61.25 kg-m²
and ΔE is express as here
ΔE = 0.5 × I × (ω² - ω1² ) ........4
put here value and we get
1500 = 0.5 × 61.25 × (25.13² - ω1² )
ω1 = 24.13 rad /s
and
reduction in speed after one riveting operation will be
N = [tex]\frac{24.13\times 60 }{2\pi }[/tex]
speed after one riveting operation = 230.5 rpm
and
for 35 rpm
ω1 = [tex]\frac{2\pi 35}{60}[/tex]
ω1 = 3.65
so ΔE will be here
ΔE = 0.5 × mr² × (ω² - ω1² ) ....................5
put here value and we get m
1500 = 0.5 × m (0.7)² × (25.13² - 3.65² )
solve it we get
mass moment of inertia = 9.90 kg
Problem 9) If you had light of wavelength 700nm when emitted, what would the wavelength be for an observer who is a red shift z = 0.05 from that source?
Answer:
The answer is 735 Nanometer
Explanation:
Solution to this problem
Given that:
Solution)
The Redshift is categorized by the difference related between wavelength(emitted) and wavelength (observer).
Now,
The wavelength can be computed as :
1+z=\lambda (observer)\div \lambda (emitted)
hence,
1 + 0.05 = λ(observer) / 700 nm
Gives
= 1.05*700 = 735 nm
Therefore, The wavelength from that source for an observer would be 735 nanometer
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Complete Question
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 [tex]\Omega[/tex]. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300[tex]\Omega[/tex] . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Answer:
a
The additional resistance is [tex]R_z = 4.4 \Omega[/tex]
b
The rate at which internal energy increase at the supply is [tex]Z_1 = 32 W[/tex]
c
The rate at which internal energy increase in the battery is [tex]Z_1 = 32 W[/tex]
d
The rate at which internal energy increase in the added series resistance is [tex]Z_3 = 70.4 W[/tex]
e
the increase rate of the chemically energy in the battery is [tex]C = 48 W[/tex]
Explanation:
From the question we are told that
The open circuit voltage is [tex]V = 40.0V[/tex]
The internal resistance is [tex]R = 2 \Omega[/tex]
The emf of each battery is [tex]e = 6.00 V[/tex]
The internal resistance of the battery is [tex]r = 0.300V[/tex]
The charging current is [tex]I = 4.00 \ A[/tex]
Let assume the the additional resistance to to added to the circuit is [tex]R_z[/tex]
So this implies that
The total resistance in the circuit is
[tex]R_T = R + 2r +R_z[/tex]
Substituting values
[tex]R_T = 2.6 +R_z[/tex]
And the difference in potential in the circuit is
[tex]E = V -2e[/tex]
=> [tex]E = 40 - (2 * 6)[/tex]
[tex]E = 28 V[/tex]
Now according to ohm's law
[tex]I = \frac{E}{R_T}[/tex]
Substituting values
[tex]4 = \frac{28}{R_z + 2.6}[/tex]
Making [tex]R_z[/tex] the subject of the formula
So [tex]R_z = \frac{28 - 10.4}{4}[/tex]
[tex]R_z = 4.4 \Omega[/tex]
The increase rate of internal energy at the supply is mathematically represented as
[tex]Z_1 = I^2 R[/tex]
Substituting values
[tex]Z_1 = 4^2 * 2[/tex]
[tex]Z_1 = 32 W[/tex]
The increase rate of internal energy at the batteries is mathematically represented as
[tex]Z_2 = I^2 r[/tex]
Substituting values
[tex]Z_2 = 4^2 * 2 * 0.3[/tex]
[tex]Z_2 = 9.6 \ W[/tex]
The increase rate of internal energy at the added series resistance is mathematically represented as
[tex]Z_3 = I^2 R_z[/tex]
Substituting values
[tex]Z_3 = 4^2 * 4.4[/tex]
[tex]Z_3 = 70.4 W[/tex]
Generally the increase rate of the chemically energy in the battery is mathematically represented as
[tex]C = 2 * e * I[/tex]
Substituting values
[tex]C = 2 * 6 * 4[/tex]
[tex]C = 48 W[/tex]
Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The secondary coil of this transformer is being replaced so that its output can be 730 kV for more efficient cross-country transmission on upgraded transmission lines.
Randomized Variables
Vi = 315 kV
V2 = 730 kV
(a) What is the ratio of turns in the new secondary to the number of turns in the old secondary?
(b) What is the ratio of new current output to the old current output for the same power input to the transformer?
Answer:
2.320.43Explanation:
12.0 kv primary voltage
315 kv secondary voltage ( converted voltage ) V1 or Vo
v2 (Vn)= 730 kv new secondary voltage
a) Ratio of turns in 730 kv to turns in 315 kv
[tex]\frac{Vn}{Vo} = \frac{Nn}{No}[/tex] = [tex]\frac{730}{315}[/tex] therefore the ratio of turns = 2.317 ≈ 2.32
B) ratio of the new current output to the old current output for the same power input to the transformer
since the power input is the same
[tex]\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }[/tex] equation 1
Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current
therefore equation 1 becomes
[tex]\frac{In}{Io} = \frac{Vo}{Vn}[/tex] = 315 / 730 = 0.43
Final answer:
The ratio of turns in the new secondary compared to the old secondary is 2.32, and the ratio of the new current output to the old current output for the same power input to the transformer is approximately 0.432.
Explanation:
Transformer Coil Turn Ratio and Current Output Ratio
In a transformer, the ratio of the voltage across the secondary coil to the voltage across the primary coil is equal to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. For the same power input to a transformer, the current output will be inversely proportional to the voltage output. Thus, when the voltage output is increased, the current output decreases proportionally.
(a) Ratio of Turns in the New Secondary to the Old Secondary
To find the ratio of turns in the new secondary compared to the old secondary, we use the ratio of the new voltage to the old voltage:
New ratio of turns (N2) / Old ratio of turns (N1) = V2 / V1 = 730 kV / 315 kV = 2.32.
(b) Ratio of the New Current Output to the Old Current Output
For the same power input, P = VI, where P is the power, V is the voltage and I is the current. Keeping the power constant and increasing the voltage output results in a decrease in current. Therefore, the ratio of the new current to the old current is inversely proportional to the ratio of the new voltage to the old voltage:
New current (I2) / Old current (I1) = V1 / V2 = 315 kV / 730 kV = 0.432.
Which of the following statements ARE correct?
Group of answer choices:
A) If net torque is zero, the object will be moving at constant velocity or remain static.
B) If the net torque is zero, the object will not rotate or rotate at constant angular velocity.
C) If net torque is zero, the angular momentum is conserved.
D) If net force is zero, the object will be rotate at constant angle velocity or not rotate at all.
Answer:
Option A, C and D are correct statements.
Explanation:
Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter.
The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
1. It will be in rotational equilibrium and the angular velocity of the body remains constant.
2. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity.
3. Angular momentum will be conserved.
If the forces on an object are balanced, the net force is zero.
If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.
By considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
The given problem is based on the concept and fundamentals of Torque. Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter. The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
It will be in rotational equilibrium and the angular velocity of the body remains constant. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity. Angular momentum will be conserved. If the forces on an object are balanced, the net force is zero. If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.Thus, we can conclude that by considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
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The switch is moved to position A at t = 0. a) What is the time constant in position A b ) How long does it take the capacitor voltage to reach 9V? Assume the initial voltage of the capacitor is 0 V. c) What is the time constant when Switch is moved to position B? d) How long after moving the switch to position B does it take for the current to reach -6 ma (miliamp)?
Answer:
b
Explanation:
Choose the appropriate word choices below that correctly complete the blanks in the following three sentences: At the beginning, the object is at the origin and moving to the left. In the first half of its motion, the object moves to the left with ________ velocity. Then, the motion of the object changes. In the second half of its motion, the object________ due to a ________ acceleration that is ________ in magnitude.
Answer:
Constant
Accelerates
Constant
Positive
A hollow metal ball hangs on a long massless string. The device is driven by a small side-to-side oscillator with a constant very small amplitude and an adjustable frequency. The driver begins to oscillate at a very low frequency. The driver's frequency is then made to increase slowly and steadily. The frequency of the driver continues to increase regardless of what happens to the ball, eventually reaching very high frequencies. Predict what will happen to the ball over time as the driver frequency slowly increases.
Answer:
If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.
Explanation:
As the frequency of driven force increases, the amplitude of the oscillation of the ball increases. This type of oscillation is referred to as forced oscillation.
At a particular higher frequency, the amplitude of oscillation becomes maximum. This frequency is referred to as the resonant frequency.
If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.
An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.
Decrease the mass of the car slightly.
Increase the mass of the car slightly.
Decrease the launch speed of the car slightly.
Increase the launch speed of the car slightly.
Answer:
we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically
Explanation:
Kinetic energy is
K = ½ m v²
the speed of the expensive we can find it r
v² = v₀² + 2 a x
we can find acceleration with Newton's second law
F = m a
a = F / m
F= cte
substitute in the velocity equation
v² = v₀² + 2 F/m x
let's substitute in the kinetic energize equation
K = ½ m (v₀² + 2 F/m x)
K = ½ m v₀² + f x
we see that the kinetic energy depends on two tomines
in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy
As the launch speed increases the initial energy increases quadratically
we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically
When a conducting plate swings through a magnetic field, which is true?
a. Energy is transferred from mechanical energy to thermal energy.
b. Energy is transferred from mechanical energy to the magnetic field.
c. Energy is transferred from the magnetic field to thermal energy.
d. Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy.
e. Energy is transferred from the magnetic field to mechanical energy.
Answer:
Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )
Explanation:
when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.
An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.
Answer:
Energy is transferred from mechanical energy to thermal energy. (A)
Explanation:
As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.
A loud speaker has an input of 79 dB. If the volume of the sound is turned up so that the output intensity is 100 times greater, what is the new sound intensity level?
Answer:
99 dB
Explanation:
To find the new sound intensity level you calculate first the initial intensity by using the following formula:
[tex]\beta=10log(\frac{I}{I_o})\\\\10^{\frac{\beta}{10}}=10^{log(\frac{I}{I_o})}\\\\10^{\frac{\beta}{10}}=\frac{I}{I_o}\\\\I=I_o10^{\frac{\beta}{10}}[/tex]
where β is the sound level of 79dB and Io is the hearing threshold of 10^-12 W/m^2. By replacing you obtain:
[tex]I=(10^{-12}W/m^2)10^{\frac{79}{10}}=7.94*10^{-5}W/m^2[/tex]
The new sound intensity level is given by:
[tex]\beta'=10log(\frac{100I}{I_o})=10log(\frac{100(7.94*10^{-5}W/m^2)}{10^{-12}W/m^2})\\\\\beta'=99\ dB[/tex]
hence, the answer is 99 dB
A what is a negatively charged Atoms
Answer:
Negatively charged or positively charged atom is generally termed as ANION/CATION. Short Explanation: If an atom loses electrons or gains protons, it will have a net positive charge and is called a Cation. If an atom gains electrons or loses protons, it will have a net negative charge and is called an Anion.
Explanation:
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor Ks of 1.25 is appropriate. (a) Find Fc, Fi, (F1)a, and F2, assuming operation at the maximum tension limit. (b) Find Ha, nfs, and belt length.
Answer:
a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,
b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm
Explanation:
Given that:
γ= 9.5 kN/m³ = 9500N/m3
b = 6 inches = 0.1524 m
t = 0.0013 mm
d = 2 inches = 0.0508 m
n = 1750 rpm
[tex]H_{nom}=2hp=1491.4W[/tex]
L = 9 ft = 2.7432 m
Ks = 1.25
g = 9.81 m/s²
a)
[tex]w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m[/tex]
[tex]V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s[/tex]
[tex]F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N[/tex]
[tex](F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N[/tex]
[tex]T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm[/tex]
[tex]F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N[/tex]
[tex]F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N[/tex]
b)
[tex]H_a=1491*1.25=1863.75W[/tex]
[tex]n_f_s=\frac{H_a}{H_{nom}K_S }=1[/tex]
dip = [tex]\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm[/tex]
A) The values of Fc, Fi, ( F1)a and F2
Fc = 4.15 N Fi = 435.65 N( F1 )a = 640 N F2 = 239.6 NB ) The values of Ha, nfs and belt length
Ha = 1863.75 Nnfs = 1belt length = 11.8 mmGiven data :
Angular velocity ratio of large pulley = 0.5
width of polyamide ( b ) = 6 inches = 0.1524 m
pulley width ( d ) = 2 inches = 0.0508 m
center to center distance ( L ) = 9 ft = 2.7432 m
angular speed of small pulley ( n ) = 1750 rev/min
power of small pulley ( Hnom ) = 2 hp = 1491.4 Watts
Ks = 1.25
γ = 9500 N/m³
g = 9.81 m/s²
t = 0.0013 mm
A) Determine the values of Fc, Fi, F1a and F2
w = γ * b * t
= 9500 * 0.1524 * 0.0013 = 1.88 N/m
V = [tex]\frac{\pi dn}{60}[/tex] = ( π * 0.0508 * 1750 ) / 60 = 4.65 m/s
T = 10.17 Nm ( calculated value )
i) Fc = [tex]\frac{wV^2}{g}[/tex] ---- ( 1 )
Insert values into equation ( 1 )
Fc = ( 1.88 * 4.65² ) / 9.81
= 4.15 N
ii) Fi = [tex]\frac{(F1)a+F2 }{2}[/tex] - Fc ----- ( 2 )
where ( F1 )a = 0.1524 * 6000 * 0.7 * 1 = 640 N
F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex] = 640 - [tex]\frac{2 *10.17}{0.0508}[/tex] = 239.6 N
back to equation ( 2 )
Fi = [ ( 640 + 239.6 ) / 2 ] - Fc
= 435.65 N
iii) ( F1 )a = b * Fa * Cp * Cv
= 0.1524 * 6000 * 0.7 * 1 = 640 N
iv) F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex]
= 640 - [tex]\frac{2 *10.17}{0.0508}[/tex] = 239.6 N
B) Find the values of Ha, nfs and belt length
i) Ha = 1491 * 1.25
= 1863.75 W
ii) nfs = [tex]\frac{Ha}{Hnom*Ks } = 1[/tex]
iii) belt length ( dip ) = [tex]\frac{L^2w}{8fi}[/tex]
= ( 2.7432 * 1.88 ) / 435.65 = 11.8 mm
Hence we can conclude that the values of the variables in the question is as listed above
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A conducting rod of mass M with negligible electrical resistance slides on a pair of frictionless, horizontal, parallel, conducting rails separated by a distance L. The two rails are connected by an electrical resistance of R. A uniform magnetic field B is directed vertically upward in the entire region. a. Looking from above, is the conventional current flowing cw or cow? b. Determine the current through the resistor. c. Find the force necessary to keep the rod moving at a constant speed.
Answer:
Explanation:
Given
mass of rod is M
Distance between rails is L
If the direction of magnetic field is directed upwards
If rod is moving right then,
Area enclosed is increasing therefore a Clockwise current will induce
and if rod is moving left then enclosed area is decreasing therefore an anticlockwise current will setup in the circuit
(b)Induced EMF will be
[tex]E=BvL[/tex]
where v=velocity of rod
and current is [tex]I=\frac{V}R}[/tex]
[tex]I=\frac{BvL}{R}[/tex]
(c)Force necessary to move rod with constant velocity v is
[tex]F_{net}=F-BIL=0[/tex]
[tex]F=BIL[/tex]
[tex]F=B\times \frac{BvL}{R}\times L[/tex]
[tex]F=\frac{B^2L^2v}{R}[/tex]
2. As a pendulum swings, its energy is constantly converted between kinetic
energy and potential energy. At which point in its swing does a pendulum have
the greatest potential energy and the least kinetic energy? *
O
A. at point F
O
B. at point G
O
C. at point H
O
D. at point J
answer fast
Answer:
at point F
Explanation:
To know the point in which the pendulum has the greatest potential energy you can assume that the zero reference of the gravitational energy (it is mandatory to define it) is at the bottom of the pendulum.
Then, when the pendulum reaches it maximum height in its motion the gravitational potential energy is
U = mgh
m: mass of the pendulum
g: gravitational constant
The greatest value is obtained when the pendulum reaches y=h
Furthermore, at this point the pendulum stops to come back in ts motion and then the speed is zero, and so, the kinetic energy (K=1/mv^2=0).
A) answer, at point F
Answer:
at point F
Explanation:
A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the slit. Find the widths (lengths, not angles)of(a) the central maximum,(b) the first maximum beyond the central maximum, and(c) the second maximum beyond the central maximum.
Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
[tex]m\lambda=asin\theta[/tex]
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
[tex]y=\frac{m\lambda D}{a}[/tex]
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
[tex]w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm[/tex]
b) the width of first maximum is y2-y1:
[tex]w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm[/tex]
c) and for the second maximum:
[tex]w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm[/tex]
A distant object will form what type of image on the retina of an emmetropic eye?
If 1.0 joule of work is required to move a charge of 1.0 coulomb between two points is an electric field the potential difference between these two point is?
Answer:1V
Explanation: simply put it
V=W/q
=1/1
=1V
The potential difference between these two point is 1V which can be determined from the equation of potential difference that involves work done and charge.
What is Potential difference?Potential difference is the work done per unit charge. A potential difference of 1 V means that 1 joule of work is done per coulomb of charge. Potential difference in a circuit is measured using a voltmeter which is placed in parallel with the component of interest in the circuit.
Given:
Work = 1.0 Joule
Charge = 1.0 coulomb
V=W/q
V=1/1
V=1 V
Thus, the potential difference is one Volts.
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You throw a small toy ball that is covered with suction cups at the center of your glass patio door. When it hits it sticks to the glass, and because the door was not latched, it causes the door to swing open with an angular velocity of 0.19 rad/s. If the ball has a mass of 120 g and the patio door can be treated as a uniform box that is 2.3 m high, 1.0 m wide, and 0.030 m thick with a mass of 7.5 kg what speed did you throw the ball at
Answer:
Explanation:
moment of inertia of the door M = 1 /3 m ( l² + b² + d² )
= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )
= 1 / 3 x 7.5 x 6.2909
I = 15.7272.
moment of inertia of the door + small toy
15.7272 + .12 x .5²
= 15.7272 + .03
= 15.7572
Applying conservation of angular momentum law
mvr = M ω
m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.
.12 x v x .5 = 15.7572 x .19
v = 49.89 m /s .
What needs to be emphasized in this step when reading?
"Turn the two-bottle assembly over, and place the assembly on a table with the filled bottle on top. Watch the water slowly drip down into the lower bottle.... Now, rapidly rotate the bottles in a horizontal circle a few times. Observe the formation of a funnel-shaped vortex..."
a.
rapidly rotate the bottles in a horizontal circle
c.
both A and B
b.
Watch the water
d.
neither A nor B
Answer:
I would say A since it's important part of the instructions.
Explanation:
Answer:
A
Explanation:
In 1923, the United States Army (there was no United States Air Force at that time) set a record for in-flight refueling of airplanes. Using two refueling planes, an Airco DH-4B biplane was able to remain in flight for 37 h . During the flight, the refueling planes were able to air-transfer a total of 687 gallons of fuel to the plane in 9 refueling transfers. Assume that the refueling nozzle had a diameter of 1.45 in and each refueling took 2.45 min to perform. Calculate the velocity of the fuel through the nozzle. Assume that the fuel filled the entire cross-sectional area of the nozzle.
Answer:
Explanation:
Total time of transfer oil fuel = 2.45 x 9 = 22.05 minutes
= 22.05 x 60
= 1323 s
Total volume of ful transferred = 687 gallon
= .0037854 x 687
= 2.6 m³
radius of pipe = .5 x 1.45 inch
= .5 x 1.45 x 2.54 x 10⁻² m
r = .018415 m
cross sectional area
= π r²
a = 3.14 x .018415²
= 10.648 x 10⁻⁴ m²
If v be the velocity
volume of fuel coming out a x v x t , a is cross sectional area , v is velocity and t is time .
10.648 x 10⁻⁴ x v x 1323 = 2.6
v = 1.845 m / s
The velocity of the fuel through the refueling nozzle was calculated to be approximately 1.845 meters per second, using the area of the nozzle and the volume of fuel transferred over the total duration of the refuelings.
Explanation:To calculate the velocity of the fuel through the nozzle during the in-flight refueling mentioned, we will need to use the formula for the volume flow rate, which is Q = A × v, where Q is the volume flow rate, A is the cross-sectional area of the nozzle, and v is the velocity of the fuel.
The cross-sectional area A of the nozzle can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the nozzle. Given the diameter of the nozzle is 1.45 inches, the radius r is half of that, which is 0.725 inches. However, we should convert this to meters to use SI units: 0.725 inches × 0.0254 meters/inch = 0.018415 meters. So, A = π × (0.018415 meters)^2 ≈ 1.066 × 10^{-3} m^2.
To find the total volume of fuel transferred, we need to convert 687 gallons to cubic meters. There are 3.78541 liters in a gallon and 1,000 liters in a cubic meter, so the volume V in cubic meters is 687 gallons × 3.78541 liters/gallon × 1 m^3/1000 liters = 2.600 cubic meters.
Given that each refueling took 2.45 minutes and there were 9 refuelings, the total time t taken for the fuel transfer in seconds is 9 transfers × 2.45 minutes/transfer × 60 seconds/minute = 1321.5 seconds.
Now we can calculate the volume flow rate Q = V / t, which equals 2.600 m^3 / 1321.5 s = 1.967 × 10^{-3} m^3/s.
Finally, we can solve for the velocity v using Q = A × v, so v = Q / A which is (1.967 × 10^{-3} m^3/s) / (1.066 × 10^{-3} m^2) = 1.845 m/s.
Thus, the velocity of the fuel through the nozzle was approximately 1.845 meters per second.
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Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous light. If an exoplanet existed with the same mass (5.972 x 10^24 kg) and twice the radius of earth (1.27 x 10^7 m) and half the distance (7.5 x 10^10 m) from Betelgeuse what would the intensity of the light at the surface of that "earth" look like? a) What is the intensity of Betelgeuse at the "earth’s" surface?
Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:
[tex]Intensity \ I = \frac{P}{4 \pi r^2}[/tex]
[tex]I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}[/tex]
I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
Answer:
1.97*10^14 W/m^2
Explanation:
To find the intensity of the light emitted by Betelgeuse you taken into account that the light from Betelgeuse expands spherically in space.
You use the following formula:
[tex]I=\frac{P}{A}=\frac{P}{4\pi r^2}[/tex] (1)
I: intensity of light
r: distance from Betelgeuse to the exoplanet = (1/2)*7.5*10^10m
P: power = 100,000*3.486*10^31 W
By replacing the values of the parameters in (1) you obtain:[tex]I=\frac{100000(3.486*10^{31}W}{4\pi (0.5*7.5*10^{10}m)^2}=1.97*10^{14}\frac{W}{m^2}[/tex]