Answer:
All of the following are true EXCEPT, the angle of refraction is 30 degrees
Explanation:
It is given that,
The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. We know that when a ray of white light moves through the air and strikes the surface of water in a beaker, refraction occurs. Let r is the angle of refraction. It can be calculated using Snell's law as :
[tex]n=\dfrac{sin\ i}{sin\ r}[/tex]
[tex]sin\ r=\dfrac{sin\ i}{n}[/tex]
[tex]sin\ r=\dfrac{sin(30)}{1.33}[/tex]
[tex]r=22.08^{\circ}[/tex]
So, the angle of refraction is 22.08 degrees.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since, it is incident at an angle of 30 degrees, so the angle of reflection will be 30 degrees. Also, total internal reflection will result, depending on the critical angle.
So, the wrong statement is the angle of refraction is 30 degrees. It is equal to 22.08 degrees.
If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle experiences the greater acceleration? Explain your answers.
Answer:
Small sports car.
Explanation:
Lets take
mass of the small car = m
mass of the truck = M
As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.
Therefore
F = m a
a=Acceleration of the car
[tex]a=\dfrac{F}{m}[/tex]
F= M a'
a'=Acceleration of the massive truck
[tex]a'=\dfrac{F}{M}[/tex]
Here given that M > m that is why a > a'
Therefore car will experiences more acceleration.
Suppose that you have a reflection diffraction grating with n= 125 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. PART A
Two visible lines in the sodium spectrum have wavelengths 498\rm nm and 569 \rm nm. What is the angular separation \Delta \theta of the first maxima of these spectral linesgenerated by this diffraction grating?
PART B
How wide does this grating need to be to allow you to resolvethe two lines 589.00 and 589.59 nanometers, which are a well knownpair of lines for sodium, in the second order (m=2)?
Answer:
[tex]0.54^{\circ}[/tex]
3.99322032 mm
Explanation:
n = Lines per mm = 125
Seperation between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]
[tex]\lambda_1[/tex] = 498 nm
[tex]\lambda_2[/tex] = 569 nm
We have the expression
[tex]dsin\theta_1=m\lambda_1[/tex]
For first maximum m = 1
[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]
[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]
Angular separation is given by
[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]
Angular separation is [tex]0.54^{\circ}[/tex]
Now
[tex]\lambda_1[/tex] = 589 nm
[tex]\lambda_2[/tex] = 589.59 nm
[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]
We have the relation
[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]
Width is given by
[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]
The width is 3.99322032 mm
The angular separation \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°
The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm
Calculations and Parameters:n = Lines per mm
= 125
The Separation between slits is given by:
d= 1/n
d= 1/125
= 0.008mm.
Where
line 1 = 498nm
line 2 = 569nm
The first maximum m= 1 will be:
θ1= 3.57°
θ2= 4.08°
The angular separation would be:
θ2- θ1= 0.54°.
Now, to find the width is:
w= N/n
= 499.15254/125
= 3.99322032 mm.
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If a force of 163 newtons is applied to the end of the jack handle the end moves down 15 cm. If lifting part of the jac raises the car bumper 1.75 cm what force is the jack exerting on the bumper?
The force exerted on the bumper is 1397 N
Explanation:
We can solve this problem by using the equilibrium of the torques: in fact, the torque exerted on one side of the jack must be equal to the torque exerted on the other side of the jack.
Therefore, we can write:
[tex]F_h d_h = F_b d_b[/tex]
where
[tex]F_h = 163 N[/tex] is the force applied to the end of the jack handle
[tex]d_h = 15 cm[/tex] is the distance between the force applied on the handle and the pivot
[tex]F_b[/tex] is the force exerted by the jack on the car bumper
[tex]d_b = 1.75 cm[/tex] is the distance between this force and the car bumper
And solving for [tex]F_b[/tex], we find:
[tex]F_b = \frac{F_h d_h}{d_b}=\frac{(163)(15)}{1.75}=1397 N[/tex]
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The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.
Answer:
27.5 days
0.92 month
Explanation:
[tex]r[/tex] = radius of the orbit of moon around the earth = [tex]3.85\times10^{8} m[/tex]
[tex]M[/tex] = Mass of earth = [tex]5.98\times10^{24} m[/tex]
[tex]T[/tex] = Time period of moon's motion
According to Kepler's third law, Time period is related to radius of orbit as
[tex]T^{2} = \frac{4\pi ^{2} r^{3} }{GM}[/tex]
inserting the values, we get
[tex]T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec[/tex]
we know that
1 day = 24 hours = 24 x 3600 sec = 86400 s
[tex]T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days[/tex]
1 month = 30 days
[tex]T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month[/tex]
Final answer:
The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.
Explanation:
To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.
We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:
Convert the given distance to meters: 3.85 x 10^8 m.Calculate the period using Kepler's third law equation:Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.
The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provides 0.09-N thrust at maximum power.By how much does the engine cause the craft's speed to change in 1 week of running at maximum power? Describe any assumptions you made.
Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,
[tex]T= m_s a[/tex]
[tex]a=\dfrac{T}{m_s}[/tex]
[tex]a =\dfrac{0.09}{435}[/tex]
a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
A 0.47 kg mass is attached to a spring with a spring constant of 130 N/m so that the mass is allowed to move on a horizontal friction-less surface. The mass is released from rest when the spring is compressed 0.12 m. A) Find the force on the mass at the instant the spring is released. B) Find the acceleration of the mass at the instant the spring is released.
Answer:
F= 15.6 N, a= 33.2 m/s^2
Explanation:
mass= m = 0.47 kg
spring constant= k = 130N/m
spring compression = x = 0.12 m
a).
force on the mass= F = k*x
F = 130 * 0.12 N
F= 15.6 N
b).
Acceleration of mass= a=?
F= ma
a=F / m
a= 15.6/ 0.47 m/s^2
a= 33.2 m/s^2
Final answer:
The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.
Explanation:
Understanding Spring Force and Acceleration
To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.
To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.
Your car has stalled and you need to push it. You notice as the car gets going that you need less and less force to keep it going. Suppose that for the first 15 m your force decreased at a constant rate from 210 N to 45 N. How much work did you do on the car?
Answer:
675 Joules
Explanation:
Considering that work can be calculated with the following formula:
W = Fx D
Where:
W = work
F = force
D = distance
We can directly use this formula in case the applied force remains constant
In case the force does not remain constant, we can calculate the work as a change, this is :
ΔW = ΔFxΔD
For this scenario, we have:
W₁ = F₁xD₁
F₁ = 210 N, D = 0 m
W₁ = 210Nx0m = 0 Joules
W₂ = F₂xD₂
F₂ = 45 N , D₂ = 15 m
W₂ = 45Nx15m = 675 Joules
Finally: ΔW = Total work performed when moving the car 15 m
ΔW = W₂ - W₁ = 675 Joules - 0 Joules = 675 Joules
Final answer:
The amount of work done on the car while pushing it is 1575 Joules.
Explanation:
Work is defined as the product of force and displacement. In this case, the force required to keep the car moving decreases as the car gets going. Work can be calculated using the formula:
Work = Force × Distance
Given that the force decreased from 210 N to 45 N over a distance of 15 m, we can calculate the work done as follows:
Work = (210 N + 45 N) / 2 × 15 m = 1575 J
Therefore, the amount of work done on the car while pushing it is 1575 Joules.
A flat uniform circular disk has a mass of 3.97 kg and a radius of 85.7 cm. It is suspended in a horizontal plane by a vertical wire attached to its center. If the disk is rotated 2.42 rad about the wire, a torque of 0.0688 N·m is required to maintain that orientation. Calculate
(a) the rotational inertia of the disk about the wire,
(b) the torsion constant, and
(c) the angular frequency of this torsion pendulum when it is set oscillating.
Answer:
1.457881265 kgm²
0.02842 Nm/rad
0.13962 rad/s
Explanation:
M = Mass = 3.97 kg
R = Radius = 85.7 cm
[tex]\tau[/tex] = Torque = 0.0688 Nm
[tex]\theta[/tex] = Angle of rotation = 2.42 rad
Moment of inertia about the center of the disk is given by
[tex]I=\dfrac{1}{2}MR^2\\\Rightarrow I=\dfrac{1}{2}\times 3.97\times 0.857^2\\\Rightarrow I=1.457881265\ kgm^2[/tex]
The rotational inertia of the disk about the wire is 1.457881265 kgm²
Torque is given by
[tex]\tau=\kappa \theta\\\Rightarrow \kappa=\dfrac{\tau}{\theta}\\\Rightarrow \kappa=\dfrac{0.0688}{2.42}\\\Rightarrow \kappa=0.02842\ Nm/rad[/tex]
The torsion constant is 0.02842 Nm/rad
Time period is given by
[tex]T=2\pi\sqrt{\dfrac{I}{\kappa}}[/tex]
Angular frequency is given by
[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\sqrt{\dfrac{\kappa}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{0.02842}{1.457881265}}\\\Rightarrow \omega=0.13962\ rad/s[/tex]
The angular frequency of this torsion pendulum when it is set oscillating is 0.13962 rad/s
Which technique is best for manual in-line stabilization of a person floating faceup on the surface?
Answer:
vise grip
Explanation:
Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea
MLS provides a means of stability to the cervical column for a patient in trauma. During this technique, a patient is restricted from moving his or her cervical collar. The vise grip can be used for a patient with neck injury. The technique is used to roll a patient to face up to prevent further injuries.
Answer: the technique which is best for manual in-line stabilization of a person floating faceup on the surface is vice grip.
Explanation:
Vice grip is a rescue technique used to prevent further injury to a guest who is suspected of having suffered a spinal injury as the typical posture is floating faceup on the surface.
Which statement is true regarding radioactive particles subjected to an electric field?
Answer:There are three types of radiation
Alpha, Beta and Gamma radiation
Explanation: In an electric field produced by two parallel charged plates alpha particle would be deflected toward a - plate following a parabolic path, beta rays toward a +plate following a parabolic path and gamma radiation either - or + source.
What type of wave is shown above?
A. transverse wave
B. longitudinal wave
C. surface wave
D. electromagnetic wave
Answer:
its a A. Transverse wave
Identify the method of thermal energy transfer.
Conduction, convection, radiation
A:
B:
C:
radiation
convection
conduction
trust me dawg
The method of thermal energy transfer in this three cases:
A: Radiation
B: Convection
C: Conduction
What are the methods of thermal energy transfer?Heat transfer is defined by thermodynamic systems as "The transfer of heat over the system boundary caused by a temperature difference between the system and its surroundings."
There are several ways for heat to go from one place to another. Conduction, convection, and radiation are some of the several ways that heat is transferred.
Conduction is the process of energy being transferred from one medium particle to another while the particles are in close proximity to one another.
The flow of fluid molecules from higher temperature regions to lower temperature regions is referred to as convection.
Radiant heat is the name for thermal radiations. Emission of electromagnetic waves results in the production of thermal radiation. These waves remove the energy from the body that is releasing them.
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A ball of mass 24 g is attached to a cord of length 0.463 m and rotates in a vertical circle. What is the minimum speed the ball must have at the top of the circle so the cord does not become slack? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:2.13 m/s
Explanation:
Given
mass of ball [tex]m=24 gm[/tex]
Length of cord [tex]L=0.463 m[/tex]
acceleration due to gravity [tex]g=9.8 m/s^2[/tex]
minimum velocity after which cord will slack
[tex]\frac{mv^2}{r}=mg[/tex]
[tex]v=\sqrt{rg}[/tex]
[tex]v=\sqrt{0.463\times 9.8}[/tex]
[tex]v=2.13 m/s[/tex]
Please !!! I really need help !!! How do I understand these ?!!!!
Answer
The answer for the first one I think is false.
The second one would be true i think. I hope i got it right and have a wonderful day
Answer:
True
False
Explanation:
From 0 to E, the train moves a distance of 55 m.
From F to J, the train moves a distance of 59 m.
The total distance is 55 + 59 = 114 m.
The displacement is the difference between the final position and initial position. Here, the distance between J and 0 is -4 m.
What is the general aerobic activity recommendation for an exercise frequency that is at least five days per week?
Explanation:
Minimum of aerobic activity 150 minutes, or a mix of moderates and intensive exercise 75 minutes of vigorous aerobic activity a week. We can extend this practice throughout the week with the instructions. Mild aerobic workouts might include practices like quick strolling or swimming, while activity like running can include strong aerobic activity.
A space station in the form of a large wheel, 283 m in diameter, rotates to provide an "artif icial gravity" of 9.5 m/s 2 for people located at the outer rim. What is the frequency of the rotational motion for the wheel to produce this effect? Answer in units of rev/min.
Answer:
The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
Explanation:
Given that,
Acceleration = 9.5 m/s²
Diameter = 283 m
We need to calculate the frequency of the rotational motion for the wheel to produce this effect
Using formula of rotational frequency
[tex]a= r\omega^2[/tex]
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
Where, r = radius
a = acceleration
[tex]\omega[/tex] = rotational frequency
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{9.5\times2}{283}}[/tex]
[tex]\omega=0.259\ rad/s[/tex]
The frequency in rev/min
[tex]\omega=0.259\times\dfrac{60}{2\pi}[/tex]
[tex]\omega=2.473\ rev/min[/tex]
Hence, The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
N2 is non-reactive, however, if you can get it to react with H2, it does so in a 1:3 mole ratio (N2:H2) and releases 92.0 kJ of thermal energy. That means which if these statements are true?
A. This reaction is exothermic.
B. This reaction has a negative enthalpy of reaction.
C. All of the above are true.
D. This reaction has a negative enthalpy of reaction.
Answer:
C. all above is true.
Explanation:
Energy releasing reactions are exothermic. Total energy of products ( [tex] N_2 H_2[/tex] ) is less than the total energy of reactants ( [tex] H_2 + N_2 [/tex] ) gives negative enthalpy change.
hint: prefix exo- means "outside, external".
A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent?
The mass of the mountain is 3.002 × 10^16 kg and its fraction of Earth's mass is approximately 4.92 × 10^-8. These results are unreasonable due to the large mass of the mountain compared to Earth and the assumptions made in the question.
To calculate the mass of the mountain, we can use Newton's Law of Universal Gravitation. The gravitational force exerted by the mountain is equal to 2.00% of the person's weight. Since weight is equal to mass multiplied by acceleration due to gravity, we can set up the equation:
0.02mg = GMm / d^2
where G is the gravitational constant, M is the mass of the mountain, m is the mass of the person, and d is the distance between them. Since the person's weight is equal to mg, we can rewrite the equation as:
0.02mg = (GMm / d^2)
Dividing both sides by mg gives us:
0.02 = (GM / d^2)
Now we can solve for the mass of the mountain (M):
M = (0.02d^2 / G)
Substituting the given values (d = 10.0 km, G = 6.673 × 10^-11 Nm²/kg²),
M = (0.02 × 10000^2) / (6.673 × 10^-11)
M = 3.002 × 10^16 kg
The mass of the mountain is 3.002 × 10^16 kg.
Comparing the mass of the mountain with that of Earth, we can use the equation:
Mountain's Mass / Earth's Mass = 3.002 × 10^16 / (6 × 10^24)
Mountain's Mass / Earth's Mass = 4.92 × 10^-8
The mass of the mountain is approximately 4.92 × 10^-8 of Earth's mass.
These results are unreasonable because the mass of the mountain and its fraction of Earth's mass are too large. It is unlikely for a mountain to have such a massive mass compared to the entire Earth.
The premises that are unreasonable or inconsistent in this scenario are the assumption that the gravitational force exerted by the mountain is 2.00% of the person's weight and the assumption that the distance between the mountain and the person is 10.0 km.
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The calculated mass of the mountain is 2.937 × 10¹⁷ kg, which is 4.91 × 10⁻⁸ of Earth's mass. This result is unreasonable because the mass and fraction are too large. The assumption about the gravitational force exerted by the mountain is inconsistent.
Let's start by calculating the mass of the mountain, given the gravitational force it exerts.
(a)
Given data:
Distance between person and mountain, r = 10.0 km = 10,000 mGravitational force exerted by mountain on person, F = 2.00% of the person's weightThe weight of the person is W = mg, where m is the mass of the person and g is the gravitational acceleration (9.8 m/s²).
The gravitational force F is given by Newton's law of gravitation: F = G * (m_p * M) / r², where:
G = gravitational constant (6.674 × 10-11 N·m²/kg²)m_p = mass of the personM = mass of the mountainGiven that F = 0.02 * mg, we can equate:
0.02 * mg = G * (m * M) / (10,000 m)²
Cancel out m and solve for M:
M = (0.02 * g * (10,000 m)²) / G
M = 2.937 × 10¹⁷ kg
(b)
Mass of Earth ME = 5.972 × 10²⁴ kg
Fraction of Earth's mass = M / ME
Fraction = 2.937 × 10¹⁷ kg / 5.972 × 10²⁴ kg = 4.91 × 10⁻⁸
(c)
The mass of the mountain and its fraction of the Earth's mass are excessively large. Such a massive mountain would be geologically and physically improbable.
(d)
The gravitational force assumed to be exerted by the mountain is too large. In reality, a mountain would exert a much smaller percentage of gravitational force on a person.
A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastward. Carrying a bad penny, he walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high
(a) In unit-vector notation, what is the displacement of the penny from start to its landing point?
(b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
[tex]\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m[/tex]
Magnitude of his displacement = 2729.47 m
Answer:
Explanation:
d1 = 1300 m east
d2 = 2400 m north
d3 = 640 m downward
(a)
The displacement of penny is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}[/tex]
(b) For the return journey of man, the displacement is given by
[tex]\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}[/tex]
The magnitude of the displacement is given by
[tex]d=\sqrt{1300^{2}+2400^{2}}=2729.47 m[/tex]
a bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go
Answer:
20 metres
Explanation:
Speed = distance ÷ time
[tex]s = \frac{d}{t} [/tex]
If we substitute the values:
[tex]4 = \frac{d}{5} [/tex]
[tex]20 = d[/tex]
Answer: 20m
Explanation:
Speed = Distance/time taken
Speed = 4.00m/s
Time taken = 5.00s
Distance = D = ?
We insert the values in the formula
4.00m/s = D/5.00s
Multiply through by 5
D = 20m
Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?
Answer:
a)
6.33 x 10⁸ N/C
Direction : Towards negative charge.
b)
1.11125 x 10²⁰ m/s²
Direction : Towards positive charge.
Explanation:
a)
[tex]Q_{1}[/tex] = magnitude of negative charge = 25 x 10⁻⁶ C
[tex]Q_{2}[/tex] = magnitude of positive charge = 50 x 10⁻⁶ C
[tex]r_{1}[/tex] = distance of negative charge from point P = 0.02 m
[tex]r_{2}[/tex] = distance of positive charge from point P = 0.08 m
Magnitude of electric field at P due to negative charge is given as
[tex]E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C[/tex]
Magnitude of electric field at P due to positive charge is given as
[tex]E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C[/tex]
Net electric field at P is given as
[tex]E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C[/tex]
Direction:
Towards the negative charge.
b)
[tex]m[/tex] = mass of the electron placed at P = 9.31 x 10⁻³¹ C
[tex]Q_{1}[/tex] = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
Acceleration of the electron due to the electric field at P is given as
[tex]a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}[/tex]
Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.
Answer:
Explanation:
qA = - 25 x 10^-6 C
qB = 50 x 10^-6 C
AP = 2 cm
BP = 8 cm
(a)
Electric field at P due to the charge at A
[tex]E_{A}=\frac{kq_{A}}{AP^{2}}[/tex]
[tex]E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}[/tex]
EA = 5.625 x 10^8 N/C
Electric field at P due to the charge at B
[tex]E_{B}=\frac{kq_{B}}{BP^{2}}[/tex]
[tex]E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}[/tex]
EB = 0.70 x 10^8 N/C
The resultant electric field at P due to both the charges is
E = EA+ EB = (5.625 + 0.7) x 10^8
E = 6.325 x 10^8 N/C towards left
(b) mass of electron, m = 9.1 x 10^-31 kg
Let a be the acceleration of electron
Force on electron, F = charge of electron x electric field
F = q x E
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}[/tex]
a = 1.11 x 10^20 m/s^2
Things you can do with an environmental engineering major and a broadcasting minor?
Answer:
With an Environmental Engineering and a broadcasting minor
You can work as an On Air personality that host programs that provide your audience with documentaries about the environments and project carried out by Environmental Engineer
and also you can work as a journalist that explore the world making research that will preserve the environment and leveraging the media as a broadcaster to provide this research findings as a video for you audience
Explanation:
In order to get a better understanding let define some terms
Environmental Engineer :
Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.
Broadcasting:
Broadcasting is the distribution of audio or video content to a dispersed audience via any electronic mass communications medium, but typically one using the electromagnetic spectrum (radio waves), in a one-to-many model.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.5 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.)
Answer:
In computing the volume of a cube,
Maximum possible error = +/-1350cm³
Relative error = 0.05
Percentage error = 5%
In computing the surface area of a cube,
Maximum possible error = +/-180cm²
Relative error = 0.0333
Percentage error = 3.33%
Explanation:
A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.
The volume of a cube = x³
Where x= length of the edge of a cube
X = 30cm +/- 0.5cm
Differentiate V with respect to x (V = Volume of a cube)
dV/dx = 3 x²
dV = 3 x² . dx
dV= 3 × 30² × (+/-0.5)
= 2700(+/-0.5)
= +/-1350cm³
Maximum possible error =
+/- 1350cm³
Relative error = Maximum error /surface area
= ΔV/V
Recall that V = x³
V= (30)³
A = 27000cm³
Substitute the values for and V into the formula for Relative error
Relative error = 1350 / 270000
Relative error = 0.05
% error = Relative error × 100
= 0.05× 100
= 5%
Surface Area of a cube = 6x²
A = 6x²
Differentiate A with respect to x
dA/dx= 12x
dA = 12x . dx
dA= 12 × 30 (0.5)
= +/- 180cm²
Maximum possible error =
+/- 180cm²
Relative error = Maximum error / total area
= dA/dx
Recall that A = 6x²
A = 6(30)²
A = 5400cm²
Substitute the values for and A into the formula for Relative error
Relative error = 180/ 5400
Relative error = 0.0333(4 decimal place)
% error = Relative error × 100
= 0.0333 × 100
= 3.33%
Final answer:
Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.
Explanation:
Differentials for Cube:
Maximum possible error in volume: 30*(0.5) = 15 cm³Relative error in volume: 15/30 = 0.5Percentage error in volume: (0.5)*100% = 50%Surface Area:
Maximum possible error in surface area: 6*(30)*(0.5) = 90 cm²Relative error in surface area: 90/(6*30) = 0.25Percentage error in surface area: (0.25)*100% = 25%Approximate the work required to lift a 2.5-kg object to a height of 6.0 meters. A student applies a force to a cart to pull it up an inclined plane at a constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart. Eddy, whose mass is 65-kg, climbs up the 1.6-meter high stairs in 1.2 s. Approximate Eddy's power rating.
Answer:
Explanation:
(a)mass [tex]m= 2.5 kg [/tex]
height [tex]h=6 m[/tex]
work required to raise
[tex]W=mgh[/tex]
[tex]W=2.5\times 9.8\times 6=147 J[/tex]
(b)Force [tex]F=20.8 N[/tex]
mass of cart [tex]m=3 kg[/tex]
length of track [tex]s=0.636 m[/tex]
[tex]Work\ done=F\cdot s[/tex]
Work done[tex]=20.8\cdot 0.636=13.22 J[/tex]
(c)mass of eddy [tex]m_e=65 kg[/tex]
height climbed [tex]h=1.6 m[/tex]
time [tex]t=1.2 s[/tex]
Energy required [tex]E=mgh=65\times 9.8\times 1.6=1019.2 J[/tex]
[tex]power=\frac{E}{t}=\frac{1019.2}{1.2}=849.33 W[/tex]
The work done in lifting the 2.5 kg object is 147 J, in moving the 3-kg cart is 13.2 J, and the power used by Eddy in climbing stairs is 851.7 W.
Explanation:To solve these problems, we need to apply principles of physics, specifically related to work, energy, and power. For the first question, we use the concept of gravitational potential energy, which is calculated by multiplying together the object's mass, the acceleration due to gravity (~9.81 m/s² on Earth), and the height to which the object is lifted. Thus for the 2.5-kg object, the work done or energy required to lift it to a height of 6.0 meters is: W = m * g * h = 2.5 kg * 9.8 m/s² * 6.0 m = 147 J.
Next, for the 3-kg cart, since the cart moves at constant speed, we can say the work done is equal to the change in potential energy. Therefore, the work done is W = m * g * h = 3.0 kg * 9.8 m/s² * 0.450 m = 13.2 J.
Fianlly, for Eddy's case, power is defined as the work done per unit time. If Eddy lifts his own mass to the height of 1.6 m, the work done (again considering as change in gravitational potential energy) is W = m * g * h = 65 kg * 9.8 m/s² * 1.6 m = 1022 J. Given that he does this work in 1.2 seconds, the power expended would be P = W / t = 1022 J / 1.2 s = 851.7 W.
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The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha particles directed at the foil
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted. While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive. And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.Answer:
Option 1
Explanation:
The correct answer is option 1
The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.
In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.
The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.
Later this heavy mass was known Nucleus.
Hence, most alpha particles passed through the foil.
How long would it take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes?
Answer:
The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
Explanation:
Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).
Mathematically, it is represented as
Q = it.......................... Equation 1
Where Q =quantity of charge, I = current, t = time
making t the subject of formula in equation 1,
t = Q/I ....................... Equation 2
Given: Q = 3.5 C, I = 5 mA.
Conversion: We convert from mA to A
I.e 5 mA = (5 × 10⁻³) A = 0.005 A.
Substituting these values into equation 2
t = 3.5/0.005
t = 700 seconds
or
(700/60) minutes = 11.67 minutes
or
(700/3600) hours = 0.194 hours.
Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
700 second will take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes.
Partical charge and current based problem:What information do we have?
Charges of partical = 3.5 C
Current = 5 mA = (5 × 10⁻³) A
Charge = (Curernt)(time)
Q = iT
T = Q / i
Time taken = 3.5 / (5 × 10⁻³)
Time taken = 700 seconds or 11.67 minutes
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A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 16.5 m and rotates through an angle of 49.0 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
Answer:
4.5m/s
Explanation:
Linear speed (v) = 42.5m/s
Distance(x) = 16.5m
θ= 49.0 rad
radius (r) = 3.67 cm
= 0.0367m
The time taken to travel = t
Recall that speed = distance / time
Time = distance / speed
t = x/v
t = 16.5/42.5
t = 0.4 secs
tangential velocity is proportional to the radius and angular velocity ω
Vt = rω
Angular velocity (ω) = θ/t
ω = 49/0.4
ω = 122.5 rad/s
Vt = rω
Vt = 0.0367 * 122.5
Vt =4.5 m/s
The tangential speed of a point on the 'equator' of the baseball is approximately 1.7983 m/s, based on the provided radius of the ball and its angular speed.
Explanation:The question is essentially asking for the tangential speed of the baseball, which describes the speed of a point on its outer edge (or 'equator') as the baseball spins or rotates. The tangential speed can be calculated using the formula v = rω, where v is the tangential speed, r is the radius, and ω is the angular speed. The angular speed is essentially the rate at which an object is rotating or spinning, measured in radians per second.
Given from the prompt, we know that ω = 49.0 rad and r = 3.67 cm = 0.0367 m (because we need the radius in meters). Inserting these values into the formula we get: v = (0.0367 m)(49.0 rad/s) = 1.7983 m/s. So, a point on the 'equator' of the baseball is moving at a tangential speed of approximately 1.7983 m/s.
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A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the motorcycle tires and the walls is µ. How does the minimum µ needed to prevent the motorcycle from slipping downwards change with the motorcycle’s speed, s?
a) µ ∝ s0b) µ ∝ s−1/2c) µ ∝ s−1d) µ ∝ s−2e) none of these
Answer:
option D
Explanation:
given,
coefficient of friction between wall and tire = µ
speed of motorcycle = s
friction force = f = μ N
where normal force will be equal to centripetal force
[tex]N = \dfrac{mv^2}{r}[/tex]
for motorcycle to not to slip weight should equal to the centripetal force
now,
[tex]m g =\mu \dfrac{mv^2}{r}[/tex]
[tex]\mu =\dfrac{rg}{s^2}[/tex]
where "rg" is constant
[tex]\mu\ \alpha \ \dfrac{1}{s^2}[/tex]
[tex]\mu\ \alpha \ s^{-2}[/tex]
Hence, the correct answer is option D
Based on Planet Z's size, orbital distance, and rotation rate, which of the following properties is it likely to have?
erosion due to liquid water
polar ice caps
a surface crowded with impact craters
seasons
active volcanoes
strong winds and violent storms
an atmosphere produced by outgassing
active tectonics
Answer:
Planet Z will have the following properties;
Active Volcanoes
Active Tectonics
An Atmosphere produced by outgassing
Explanation:
The little terrestrial worlds have heat shorter than the much bigger terrestrial worlds, so the bigger worlds tend to have active volcanism and tectonics. These active volcanism and tectonics are likely to erase ancient craters. The active volcanism and tectonics would create an atmosphere by producing gases.
It is know that the Terrestrial worlds that are not far from the star have higher surface temperature.
Fast rate of rotation can cause winds and strong storms but here it is slower compared to earth. Also, a tilt of axis causes seasons.
The properties the star have are active volcanoes, active tectonics and an atmosphere produced by outgassing.
Final answer:
Planet Z's characteristics such as geological activity, seasons, and atmosphere can be inferred from its size, orbital distance, and rotation rate. A planetary mass similar to Earth's suggests active geology and an atmosphere from outgassing, while a proper distance from the sun allows for liquid water and polar ice caps. The planet's rotation influences the presence of seasons and potential for strong winds and violent storms.
Explanation:
Based on Planet Z's size, orbital distance, and rotation rate, it is possible to infer several characteristics that this planet might have. The level of geological activity on a planet is often proportional to its mass, suggesting that planets similar in size to Earth and Venus are more likely to exhibit geological activity such as active volcanoes or tectonics. Similarly, a planet's distance from its sun can influence the presence of liquid water, with those at optimum distances having the potential for erosion due to liquid water and possibly polar ice caps. A slower rotation might lead to more extreme temperature differences between day and night, which could impact atmospheric conditions and lead to strong winds and violent storms due to the larger temperature gradient.
Planetary rotation also contributes significantly to the development of seasons; hence, how Planet Z rotates will affect whether it experiences seasons. A planet that has active geology and volcanism will likely have an atmosphere that is at least partially produced by outgassing, as seen on Earth, and could also support active tectonics. Lastly, if the planet is not geologically active, it may have a surface crowded with impact craters, similar to the Moon and Mercury, which have less geological activity to renew their surfaces.
Compared to a 1-kg block of solid iron, a 2-kg block of solid iron has twice as much
A)volume.
B)inertia.
C)mass.
D) all of the above
E) none of the above
The 2-kg block of solid iron has twice the mass of the 1-kg block.
Explanation:The correct option is C) mass.
Mass is a measure of the amount of matter in an object, and it is not affected by its shape or size. Since both blocks are made of solid iron, they have the same density. Therefore, the 2-kg block of solid iron has twice the mass of the 1-kg block.
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