Answer:
[tex]6.60\cdot 10^5 Nm^2/C[/tex]
Explanation:
The electric flux through the rectangle is given by
[tex]\Phi = E A cos \theta[/tex]
where
E is the electric field strength
A is the area of the rectange
[tex]\theta[/tex] is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle
In this problem we have
E = 125 000 N/C
The area of the rectangle is
[tex]A=2.50 m \cdot 5.00 m=12.5 m^2[/tex]
and the angle is
[tex]\theta=65.0^{\circ}[/tex]
so, the electric flux is
[tex]\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C[/tex]
Juan and Anita each lift an identical stack of books onto a table, but Anita places the books on a table twice as high as Juan. Therefore, her actions involve twice as much__
Answer:
Work
Explanation:
Anita exerted the same amount of force but for twice the distance. So she did twice as much work.
Light passes through a single slit. If the width of the slit is reduced, what happens to the width of the central bright fringe? (a) The width of the central bright fringe does not change, because it depends only on the wavelength of the light and not on the width of the slit. (b) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes smaller. (c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. (d) The central bright fringe becomes narrower, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. (e) The central bright fringe becomes narrower, because the angle that locates the first dark fringe on either side of the central bright fringe becomes smalle
Answer:
(c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.
Explanation:
The formula that gives the angle of the first minimum of the diffraction pattern from a single-slit is
[tex]sin \theta = \frac{\lambda}{w}[/tex]
where
[tex]\lambda[/tex] is the wavelength of the light
w is the width of the slit
We see that the angle is inversely proportional to the width of the slit: therefore, if the width of the slit is reduced (so, w is decreased), the angle that locates the first minimum [tex]\theta[/tex] increases, and so the central bright fringe becomes wider.
Reducing the width of a single slit through which light passes leads to a wider central bright fringe. This occurs because the narrowed slit causes greater divergence of light rays, leading to a larger angle for locating the first dark fringe.
Explanation:
The phenomenon being discussed here pertains to physics, specifically light diffraction and interference. When light passes through a single slit, it exhibits a diffraction pattern with a central bright fringe flanked by smaller, dimmer fringes on either side. This pattern is influenced by the width of the slit and the wavelength of the light source.
For the scenario provided in the question, light passes through a single slit that is then narrowed. The correct answer is (c): The central bright fringe becomes wider because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.
This is explained by the fact that as the slit narrows, the rays of light diverge more upon exiting the slit, due to the wave nature of light. The larger spread of these rays leads to a larger angle for locating the first dark fringe, which subsequently results in a wider central bright fringe.
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In a laboratory, you determine that the density of a certain solid is 5.23×10−6kg/mm3. Convert this density into kilograms per cubic meter.Notice that the units you are trying to eliminate are now in the denominator. The same principle from the previous parts applies: Pick the conversion factor so that the units cancel. The only change is that now the units you wish to cancel must appear in the numerator of the conversion factor.
Answer:
[tex]5.23\cdot 10^3 kg/m^3[/tex]
Explanation:
The density of the solid is
[tex]d = 5.23\cdot 10^{-6}kg/mm^3[/tex]
we want to convert it into kg/m^3. We must note that:
[tex]1 m^3 = 1 m \cdot 1 m \cdot \1m =1000 mm\cdot 1000 mm \cdot 1000 mm=1\cdot 10^9 mm^3[/tex]
Therefore, the conversion can be done as follows:
[tex]d=5.23\cdot 10^{-6} \frac{kg}{mm^3} \cdot (1\cdot 10^9 \frac{mm^3}{m^3}) =5.23\cdot 10^3 kg/m^3[/tex]
In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:
Stage A: Friction with the atmosphere reduced the speed from 19300 km/h to 1600 km/h in 4.0 min.
Stage B: A parachute then opened to slow it down to 321 km/h in 94 s.
Stage C: Retro rockets then fired to reduce its speed to zero over a distance of 75 m.
Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant.
1- Find the rocket's acceleration (in m/s2) during stage A
2- Find the rocket's acceleration (in m/s2) during stage B
3-Find the rocket's acceleration (in m/s2) during stage C
Acceleration is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A) [tex]-20.5 m/s^2[/tex]
Let's convert the quantities into SI units first:
[tex]u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s[/tex]
[tex]v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s[/tex]
t = 4.0 min = 240 s
So the acceleration is
[tex]a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2[/tex]
B) [tex]-3.8 m/s^2[/tex]
As before, let's convert the quantities into SI units first:
[tex]u = 444.4 m/s[/tex]
[tex]v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s[/tex]
t = 94 s
So the acceleration is
[tex]a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2[/tex]
C) [tex]-53.0 m/s^2[/tex]
For this part we have to use a different formula:
[tex]v^2 - u^2 = 2ad[/tex]
where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find
[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2[/tex]
Final answer:
The rocket's acceleration was -20.49 m/s² during Stage A, -3.78 m/s² during Stage B, and -53.28 m/s² during Stage C, illustrating the varied methods employed to slow the vehicle during its descent onto Mars.
Explanation:
To calculate the rocket's acceleration during each stage of its descent onto Mars, we'll use the formula: a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the time taken for that change. It's essential to convert all units to meters and seconds for consistency.
Stage A:
Initial velocity = 19300 km/h = 5361.11 m/s
Final velocity = 1600 km/h = 444.44 m/s
Time = 4.0 min = 240 s
Acceleration = (444.44 m/s - 5361.11 m/s) / 240 s = -20.49 m/s²
Stage B:
Initial velocity = 1600 km/h = 444.44 m/s
Final velocity = 321 km/h = 89.17 m/s
Time = 94 s
Acceleration = (89.17 m/s - 444.44 m/s) / 94 s = -3.78 m/s²
Stage C:
This requires a different formula, a = v² / (2•d), where v is the final velocity (just before stage C, so 89.17 m/s), and d is the distance over which deceleration happens (75 m).
Acceleration = (89.17 m/s)² / (2• 75 m) = -53.28 m/s²
A small laser used as a pointer produces a beam of red light 5 mm in diameter, and has a power output of 5 milliwatts. What is the magnitude of the electric field in the laser beam?
Answer:
434.0 V/m
Explanation:
The power output of the laser is:
[tex]P=5 mW = 0.005 W[/tex]
while the radius of the beam is
[tex]r=\frac{5 mm}{2}=2.5 mm = 0.0025 m[/tex]
so the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2[/tex]
So the intensity of the laser beam is
[tex]I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2[/tex]
The intensity of a laser beam is related to the magnitude of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
Solving the formula for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m[/tex]
Which of he following explains the significance of the observations that some microbes can kill or inhibit the growth of other microbes?
A.) It led to the development of the first antibiotic
B.) It led to the development of the first antiviral
C.) It led to the development of the Theory of Evolution
D.) It led to the development of the Germ Theory of Disease
Some microbes can kill or inhibit the growth of other microbes led to the development of the first antibiotic.
Answer: Option A
Explanation:
Antibiotics are the substance that restricts the bacteria growth and replication. These are designed against microbes and it targets bacterial infections in or on a human body. Bacteria's comes under the microbe category that cause harm to the human body.
The substance that targets, restricts and kills microbial cells includes antibiotics, antiseptics and antiviral. The antibiotics that we use today can be produced in labs and they can be found in nature also. Hence, antibiotics are the one that kills or restricts the microbe's growth.
Consider two copper wires with the same cross-sectional area. Wire A is twice as long as wire B. How do the resistivities and resistances of the two wires compare? Check all that apply. Check all that apply. Wire A and wire B have the same resistance. Wire A has twice the resistance of wire B. Wire A and wire B have the same resistivity. Wire B has twice the resistivity of wire A. Wire B has twice the resistance of wire A. Wire A has twice the resistivity of wire B. SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Your answer indicates that you need to review resistivity and resistance. Provide Feedback Next Incorrect. Incorrect; Try Again; 4 attempts remaining. Feedback. Your answer indicates that you need to review resistivity and resistance. End of feedback. g
Answer:
Wire A has twice the resistance of wire B.
Wire A and wire B have the same resistivity.
Explanation:
- Resistivity is a property of a material, that tells how much is the material able to oppose to the flow of current through it. The value of the resistivity of a wire depends on the material only: this means that two wires made of the same material have same resistivity. Since both wire A and B here are made of copper, they have the same resistivity.
- Resistance of a wire instead is given by
[tex]R=\rho \frac{L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
Here, the two wires have same resistivity and same cross-sectional area, while wire A is twice as long as wire B (so, L for A is twice the value of L for B): therefore, the resistance of wire A will be twice that of wire B.
Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining hydrogen and oxygen gas to produce water and electrical energy. If a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?
A.
0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts
Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.
10,000 W / 15.8 A = V
632.9, or 633.
The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.
What is electric power?Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.
The given data in the problem is;
V is the voltage = ? Volt (V)
Electric current (I)= 15.8 amps (A)
P is the power =10.0 kilowatts =10⁴ watt
The formula for the power is given as;
[tex]\rm P= V I \\\\\ 10^4= V \times 15.8 \\\\ V=632.9 \ V[/tex]
The voltage of the electricity will be 63.29 V.
Hence, option B is correct.
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A 170 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and has vx = -28 cm/s. Determine the following: (a) the period (b) the angular frequency (c) the amplitude (d) the phase constant (e) the maximum speed (f) the maximum acceleration (g) the total energy (h) the position at t = 0.4 s
(a) 0.5 s
In a simple harmonic motion, the period is equal to the reciprocal of the frequency:
[tex]T=\frac{1}{f}[/tex]
where f is the frequency
For this simple harmonic oscillator, the frequency is
f = 2.0 Hz
So the period is
[tex]T=\frac{1}{2.0 Hz}=0.5 s[/tex]
(b) 12.56 rad/s
The angular frequency is given by
[tex]\omega = 2 \pi f[/tex]
where
f is the frequency
In this problem,
f = 2.0 Hz
So the angular frequency is
[tex]\omega = 2 \pi (2.0 Hz)=12.56 rad/s[/tex]
(d) 0.419 rad
The displacement of the system can be written as
[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)
where A is the amplitude and [tex]\phi[/tex] is the phase constant.
The velocity is the derivative of the displacement:
[tex]v(t) = x'(t) = -A\omega sin(\omega t+\phi)[/tex] (2)
Here we know that
at t=0, x(5)=5.0 cm and v(t)=-28 cm/s. So we can rewrite the ratio (2)/(1) as
[tex]\frac{v(t)}{x(t)}=\frac{-28 cm/s}{5.0 cm}=\frac{-\omega A sin(\phi)}{A cos(\phi)}=-\omega tan \phi[/tex]
And re-arranging the equation we can find the phase constant:
[tex]\phi = tan^{-1} (\frac{v}{\omega x})=tan^{-1} (\frac{-28 cm/s}{(5.0 cm)(12.56 rad/s)})=0.419 rad[/tex]
(c) 5.47 cm
The displacement of the system can be written as
[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)
at t=0, x=5.0 cm, so using the values we found for [tex]\omega, \phi[/tex] we can now solve the equation to find A, the amplitude:
[tex]A=\frac{x}{cos(\omega t+\phi)}=\frac{5.0 cm}{cos(0.419 rad)}=5.47 cm[/tex]
(e) 68.7 cm/s
The maximum speed in a simple harmonic system is given by
[tex]v=\omega A[/tex]
where in this case we have
[tex]\omega=12.56 rad/s[/tex]
[tex]A=5.47 cm[/tex]
Substituting the numbers into the formula, we find
[tex]v=(12.56 rad/s)(5.47 cm)=68.7 cm/s[/tex]
(f) 862.9 cm/s^2
The maximum acceleration in a simple harmonic system is given by
[tex]a=\omega^2 A[/tex]
where in this case we have
[tex]\omega=12.56 rad/s[/tex]
[tex]A=5.47 cm[/tex]
Substituting the numbers into the formula, we find
[tex]a=(12.56 rad/s)^2(5.47 cm)=862.9 cm/s^2[/tex]
(g) 0.04 J
The total energy of the system is equal to the kinetic energy when the speed of the system is maximum: this occurs at x=0 (equilibrium position), where the elastic potential energy is zero, and all the energy is just kinetic energy:
[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]
where we have
m = 170 g = 0.170 kg is the mass
[tex]v_{max}=68.7 cm/s = 0.687 m/s[/tex] is the maximum speed
Substituting into the equation,
[tex]E=K=\frac{1}{2}(0.170 kg)(0.687 m/s)^2=0.04 J[/tex]
(h) 3.65 cm
The position of the system is given by
[tex]x(t) = A cos (\omega t+\phi)[/tex]
where we have
[tex]\omega=12.56 rad/s[/tex] is the angular frequency
[tex]A=5.47 cm[/tex] is the amplitude
[tex]\phi = 0.419 rad[/tex] is the phase constant
Substituting t=0.4 s, we find the position at this time:
[tex]x = (5.47 cm) cos ((12.56 rad/s)(0.4 s)+0.419 rad)=3.65 cm[/tex]
The SHM the oscillating and periodic motion of an object. a) T = 0.5 s. b) ω = 12.56 rad/seg. c) A = 5.476 cm. d) Φ = 0.4186 rad. e) Vmax = 68.778 cm/s. f) a = 863 m/s². g) Ek = 0.04012 J. h) X = 4.08 cm
What is simple harmonic motion, SHM?It is a rectilinear movement performed by a oscillating and periodic movil.
It is periodic because it repeats in a certain intervale of time.
The time -in seconds- it takes to the movil to make a complete oscillation is called Period, T.
The frequency, f, refers to the number of complete oscillations (cycles) completed per second. f = 1/T (Hz).
When considering a simple harmonic motion, we need to know that there will be always a restoring force (F), that tends to take the movil to the original position when it moves a distance named amplitud.
There are some significant components to consider in this motion,
The position of the movil, X (m) ⇒ initial, equilibrium, and in other positionsThe acceleration, a (m/s²)Velocity, V (m/s)The amplitud, A (m)Frequency, f (Hz)The initial phase, Φ (rad)Period, T (s)Available data:
m = 170 gf = 2 Hzt = 0 → x(t) = 5cm v(t) = -28 cm/s(a) the period, T
T = 1/f
T = 1/2
T = 0.5 s
(b) the angular frequency, ω
ω = 2πf
ω = (2π)2Hz
ω = 12.56 rad/seg
(c) the amplitude, A
It can be cleared from the following displacement formula,
X = A sen (ω t + Φ)
First, we need to get Φ.
Φ = arctan (v/ωx) = tan⁻¹(v/ωx)
Φ = tan⁻¹(-28/12.56x5)
Φ = tan⁻¹(-28/62.8)
Φ = tan⁻¹(0.445)
Φ = 0.4186 rad
So now we can calculate A
X = A sen (ω t + Φ)
5 = A sen (12.56 x 0 + 0.4186)
5 = A sen (0.4186)
A = 5/cos(0.4186)
A = 5/0.913
A = 5.476 cm
(d) the phase constant, Φ
Φ = arctan (v/ωx) = tan⁻¹(v/ωx)
We did this in the previous step.
Φ = 0.4186 rad
(e) the maximum speed, Vmax
Vmax = A ω
Vmax = 5.476 x 12.56
Vmax = 68.778 cm/s
(f) the maximum acceleration
a = ω ² A
a = 12.56² x 5.476
a = 157.75 x 5.476
a = 863 m/s²
(g) the total energy
Total energy = kinetic energy + Potential energy = Ek + Ep
At x=0 ⇒ v=max, Ep = 0 ⇒ Et = Ek + Ep = Ek
Ek = 1/2 mv²
Ek = 1/2 (0.170 kg x 0.687²m/s)
Ek = 0.04012 J.
(h) the position at t = 0.4 s
X = A sen (ω t + Φ)
X = 5.476 cm sen (12.56 rad/seg x 0.4 + 0.4186 rad)
X = 5.476 sen (5.024 + 0.4186)
X = 5.476 sen (5.4426)
X = 4.08 cm
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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 106 J/kg . Answer in units of J. 006 (part 2 of 3) 10.0 points Find the amount of heat added to the water to accomplish this process. Answer in units of J.
1. 408.4 J
The work done by a gas is given by:
[tex]W=p\Delta V[/tex]
where
p is the gas pressure
[tex]\Delta V[/tex] is the change in volume of the gas
In this problem,
[tex]p=1.01\cdot 10^5 Pa[/tex] (atmospheric pressure)
[tex]\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3[/tex] is the change in volume
So, the work done is
[tex]W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J[/tex]
2. 10170 J
The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:
[tex]Q = m \lambda_v[/tex]
where
m is the mass of the water
[tex]\lambda_v = 2.26\cdot 10^6 J/kg[/tex] is the specific latent heat of vaporization
The initial volume of water is
[tex]V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3[/tex]
and the water density is
[tex]\rho = 1000 kg/m^3[/tex]
So the water mass is
[tex]m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg[/tex]
So, the amount of heat added to the water is
[tex]Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J[/tex]
Most of the funding for research comes from the federal government or ? And is provided to Principal Investigators (PIs) through the organizations for which they work.
The federal government and industry are the main sources of funding for research, providing funding to Principal Investigators (PIs) through their organizations. The U.S. economy has increasingly relied on industry-funded research, but the government still plays a significant role in funding research.
Explanation:The federal government is one of the main sources of funding for research, along with industry. The government provides funding to Principal Investigators (PIs) through the organizations they work for. Over time, the U.S. economy has relied more heavily on industry-funded research and development (R&D). However, the government still plays a significant role in funding research, especially in areas where private firms are not as active.
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Most funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through their organizations.
Explanation:
Most of the funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through the organizations for which they work. The federal government, through agencies such as the National Institutes of Health (NIH) and the National Science Foundation (NSF), provides grants for research in various fields. Industry-funded research, on the other hand, is supported by companies and private entities who invest in research and development projects.
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The probable question can be: Complete the following sentence. Most of the funding for research comes from the federal government or ______ and is provided to Principal Investigators (PIs) through the organizations for which they work.
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Which scientist invented a model of the atom that most closely resembles the modern electron cloud model
Answer:
Niels Bohr
Explanation:
Niels Bohr was scientist whose model showed electrons location around a nucleus and all in an orbit.The orbit had varying energies according to their sizes.This model further indicated that when an electron is lost or gained, radiation is absorbed or emitted.The modern atomic model was prepared from quantum mechanics which shows how matter differ in atomic and subatomic levels.In this model, the chances of finding an electron is higher where the cloud is more dense.Bohrs model resembles the electron cloud model is that they both talk about position of electrons in an atom, and the energy associated with size of the orbit occupied by the electron.
____ is undeniably accepted by scientists all over the world as the primary language of science.
Answer:
Latin
Explanation:
In order for the scientists to have a common and official name for a particular thing that can be understood by every scientist in the world, a single language has been established for the purpose. The language chosen is the Latin language. The official scientific names are given in this language, so it is a necessity for the scientists to know and understand this language. The terms that are commonly used are regional, and they come in many different languages, which is why this language has been chosen. Occasionally, the ancient Greek language is used as well, though much less than the Latin.
Answer: English
Explanation: I would say the Answer is English but I’m not Sure!
can someone help me?!!!!!
The answer is:
The first option, the walker traveled 360m more than the actual distance between the start and the end points.
Why?
Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).
So, calculating we have:
Traveler:
[tex]Distance=NorthCoveredDistance+EastCoveredDistance[/tex]
[tex]Distance=4*180m+3*180m=720m+540m=1260m[/tex]
Actual distance between the start and the end point (displacement):
[tex]ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m[/tex]
Now, to calculate how much farter did the traveler walk, we need to use the following equation:
[tex]DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m[/tex]
Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.
Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.
Have a nice day!
Part A Which statement is true about nuclear fusion? Which statement is true about nuclear fusion? Nuclear fusion is not considered a good method for the production of electricity because the by-products of nuclear fusion are even more toxic and long-lived than those from nuclear fission. The reaction responsible for the hydrogen bomb is four hydrogen nuclei combining to form a helium nucleus. Nuclear fusion reactions are difficult to accomplish because they only occur at extremely high temperatures. Several research groups around the world have been able to use fusion reactions on a small scale to produce a net output of energy (more energy is produced than was needed to get the fusion reaction to occur).
Answer:
Nuclear fusion reactions are difficult to accomplish because they only occur at extremely high temperatures
Explanation:
Nuclear fusion is a nuclear process that occurs when nuclei of light elements fuse together forming a nucleus of a heavier element. In this process, the total mass of the final nucleus is smaller than the sum of the masses of the initial nuclei, so part of the mass has been converted into energy according to the equation
[tex]E=\Delta mc^2[/tex]
where
[tex]\Delta m[/tex] is the mass change
c is the speed of light
Due to the huge value of c ([tex]c=3.0\cdot 10^8 m/s[/tex]), we see that even for a small amount of mass change, the energy released in this process is huge.
In fact, nuclear fusion is the process that occurs in the core of the stars, that allow stars to continuously produce huge amount of energy. However, nuclear fusion can only occur at very high temperatures (million of Kelvins), because the nuclei need to have enough kinetic energy in order to overcome the electrostatic repulsion that occurs between them (nuclei are made of protons, so they naturally repel each other). This is the reason why it is currently difficult to accomplish nuclear fusion reactions on Earth, because it is not easy to create environments with such extremely high temperatures.
Several research groups around the world have been able to use fusion reactions on a small scale to produce a net output of energy (more energy is produced than was needed to get the fusion reaction to occur) is true about nuclear fusion. Correct option is d.
a. This statement is not entirely true. Nuclear fusion is considered a potentially promising method for the production of electricity because it has the potential to generate vast amounts of clean energy with abundant fuel (e.g., isotopes of hydrogen). However, one of the challenges of nuclear fusion is dealing with the by-products, which can include high-energy neutrons that can activate surrounding materials, making them radioactive. The radioactivity from fusion by-products may not be as long-lived or as toxic as the by-products of nuclear fission, but it still presents technical challenges that need to be addressed for practical fusion power plants.
b. This statement is not accurate. The reaction responsible for the hydrogen bomb (thermonuclear bomb) is a two-stage process. The first stage involves the fission of uranium or plutonium, which produces the high temperatures and pressures needed to initiate the fusion reaction. The second stage is the fusion of hydrogen isotopes (deuterium and tritium) to form helium and release additional energy.
c. This statement is true. Nuclear fusion reactions require extremely high temperatures and pressures to overcome the electrostatic repulsion between positively charged atomic nuclei. The most promising approach for achieving these conditions is by using magnetic confinement or inertial confinement, which are methods that require sophisticated technology and substantial energy input.
d. This statement is true. Although nuclear fusion is not yet fully realized as a viable large-scale energy source, there have been significant advances in fusion research, and several research groups and experimental reactors have achieved net energy gain, where more energy is produced from the fusion reaction than is used to sustain it. However, it's important to note that the current challenges lie in sustaining and scaling these reactions to achieve continuous, controlled, and economically feasible fusion energy production.
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Complete question is:
Which statement is true about nuclear fusion?
a. Nuclear fusion is not considered a good method for the production of electricity because the by-products of nuclear fusion are even more toxic and long-lived than those from nuclear fission.
b. The reaction responsible for the hydrogen bomb is four hydrogen nuclei combining to form a helium nucleus.
c. Nuclear fusion reactions are difficult to accomplish because they only occur at extremely high temperatures.
d. Several research groups around the world have been able to use fusion reactions on a small scale to produce a net output of energy (more energy is produced than was needed to get the fusion reaction to occur).
X-rays with an energy of 265 keV undergo Compton scattering from a target. If the scattered rays are deflected at 41.0° relative to the direction of the incident rays, find each of the following. (a) the Compton shift at this angle _________nm (b) the energy of the scattered x-ray __________keV (c) the kinetic energy of the recoiling electron ___________keV
The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:
[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex] (1)
Where:
[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.
[tex]\theta=41\°[/tex] the angle between incident phhoton and the scatered photon.
We are told the scattered X-rays (photons) are deflected at [tex]41\°[/tex]:
[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))[/tex] (2)
[tex]\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m[/tex] (3)
But we are asked to express this in [tex]nm[/tex], so:
[tex]\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}[/tex]
[tex]\Delta \lambda=0.000595nm[/tex] (4)
(b) the energy of the scattered x-rayThe initial energy [tex]E_{o}=265keV=265(10)^{3}eV[/tex] of the photon is given by:
[tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex] (5)
From this equation (5) we can find the value of [tex]\lambda_{o}[/tex]:
[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex] (6)
[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{265(10)^{3}eV}[/tex]
[tex]\lambda_{o}=4.682(10)^{-12}m[/tex] (7)
Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:
[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]
Then:
[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex] (8)
[tex]\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m[/tex]
[tex]\lambda'=5.277(10)^{-12}m[/tex] (9)
Knowing the wavelength of the scattered photon [tex]\lambda'[/tex] , we can find its energy [tex]E'[/tex] :
[tex]E'=\frac{h.c}{\lambda'}[/tex] (10)
[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{5.277(10)^{-12}m}[/tex]
[tex]E'=235.121keV[/tex] (11) This is the energy of the scattered photon
(c) Kinetic energy of the recoiling electron
If we want to know the kinetic energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon in the wavelength shift, which is:
[tex]K_{e}=E_{o}-E'[/tex] (12)
[tex]K_{e}=265keV-235.121keV[/tex]
Finally we obtain the kinetic energy of the recoiling electron:
[tex]E_{e}=29.878keV[/tex]
Answer:
The first one:
the energy of the scattered x-ray
The answer for last on:
Kinetic energy of the recoiling electron