Explanation:
It is given that,
Velocity of the electron, [tex]v=(2\times 10^6i+3\times 10^6j)\ m/s[/tex]
Magnetic field, [tex]B=(0.030i-0.15j)\ T[/tex]
Charge of electron, [tex]q_e=-1.6\times 10^{-19}\ C[/tex]
(a) Let [tex]F_e[/tex] is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :
[tex]F_e=q_e(v\times B)[/tex]
[tex]F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)][/tex]
[tex]F_e=-1.6\times 10^{-19}\times (-390000)(k)[/tex]
[tex]F_e=6.24\times 10^{-14}k\ N[/tex]
(b) The charge of electron, [tex]q_p=1.6\times 10^{-19}\ C[/tex]
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.
\If an object is placed at twice the focal length of a convex lens, where is the image located on the other side of the lens?
Question 2 options:
1 f
2 f
3 f
0.5 f
Answer:Image will be located at 2F on the other side of the lens.
Explanation:
Answer:
2 f
Explanation:
A metal disk of radius 4.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a sliding contact on the rim of the disk. A uniform magnetic field B= 1.25 T is parallel to the axle of the disk. When the current is 5.5 A, the disk rotates with constant angular velocity. What's the frictional force at the rim between the stationary electrical contact and t?
Answer:
Friction force is 0.1375 N
Solution:
As per the question:
Radius of the metal disc, R = 4.0 cm = 0.04 m
Magnetic field, B = 1.25 T
Current, I = 5.5 A
Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:
[tex]dF = IB\times dR[/tex]
Integrating the above eqn:
[tex]\int dF = IB\int_{0}^{R}dr[/tex]
[tex]\tau = IB\times \frac{R^{2}}{2} = \frac{1}{2}IBR^{2}[/tex] (1)
Now the torque is given by:
[tex]\tau = F\times R[/tex] (2)
From eqn (1) and (2):
[tex]F\times R = \frac{1}{2}IBR^{2}[/tex]
Thus the Frictional force is given by:
[tex]F = \frac{1}{2}\times 5.5\times 1.25\times 0.04 = 0.1375\ N[/tex]
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of . HooRU's surface area is and the emissivity of its surface is . Ignore the radiation HooRU absorbs from the cold universe. What is HooRU's temperature?
Answer: 150.427 K
Explanation:
The complete question is as follows:
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of [tex]14.5 W[/tex] . HooRU's surface area is [tex]1.79 m^{2}[/tex] and the emissivity of its surface is [tex]0.279[/tex]. Ignore the radiation HooRU absorbs from the cold universe. What is HooRU's temperature?
This problem can be solved by the Stefan-Boltzmann law for real radiator bodies:
[tex]P=\sigma A \epsilon T^{4}[/tex] (1)
Where:
[tex]P=14.5 W[/tex] is the energy radiated by HooRU
[tex]\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}[/tex] is the Stefan-Boltzmann's constant.
[tex]A=1.79 m^{2}[/tex] is the Surface of the robot
[tex]\epsilon=0.279[/tex] is the robot's emissivity
[tex]T[/tex] is the effective temperature of the robot (its surface absolute temperature) in Kelvin
So, we have to find [tex]T[/tex] from (1):
[tex]T=(\frac{P}{\sigma A \epsilon})^{\frac{1}{4}}[/tex] (2)
[tex]T=(\frac{14.5 W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}) (1.79 m^{2}) (0.279)})^{\frac{1}{4}}[/tex]
Finally:
[tex]T=150.427 K[/tex]
You watch distant Sally Homemaker driving nails into a front porch at a regular rate of 1 stroke per second. You hear the sound of the blows exactly synchronized with the blows you see. And then you hear one more blow after you see the hammering stop. Calculate the distance of Sally from you.
Answer:
343m
Explanation:
From the question Sally hits one blow every second
However, the sound is delayed by that same amount of time (1 sec)
If the speed of sound is 343m/s
Speed = Distance/Time
Speed = 343m/s
Time = 1s
Distance = Speed x Time
= 343 x 1 = 343m
A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled out of the window. Which law, principle, or equation can be used to explain this movement of the curtain?A. Poiseuille's lawB. Bernoulli's equationC. the equation of continuityD. Archimedes' principleE. Pascal's principle
Answer:
option B.
Explanation:
The correct answer is option B.
The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.
According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.
When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.
A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ride is roughest—that is, the car bounces up and down with the maximum amplitude—when the car is traveling at 5.7 m/s. What is the spring constant of the car's springs? Express your answer to two significant figures and include the appropriate units.
Final answer:
To find the spring constant of the car's springs, we use Hooke's Law and the given information about the car's mass, maximum amplitude, and speed. The spring constant is approximately 92638 N/m.
Explanation:
To find the spring constant of the car's springs, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the car bounces up and down with the maximum amplitude when traveling at 5.7 m/s. The maximum amplitude of the oscillation is half the distance between bumps, so it is 4.9 m/2 = 2.45 m. We can use this information to solve for the spring constant k using the formula:
k = mω²
where m is the mass of the car and ω is the angular frequency, which can be calculated using the formula:
ω = 2πf
where f is the frequency, which is the reciprocal of the period T.
Given that the bumps are spaced 4.9 m apart, the period of the oscillation is the time it takes the car to travel one bump, which is T = 4.9 m / 5.7 m/s = 0.859 s.
The frequency is the reciprocal of the period, so f = 1 / T = 1 / 0.859 s = 1.164 Hz.
Plugging the values into the formulas, we have:
ω = 2π(1.164) ≈ 7.294 rad/s
k = (1800 kg)(7.294 rad/s)² ≈ 92638 N/m
Therefore, the spring constant of the car's springs is approximately 92638 N/m.
What happens to the wavelength of a wave on a string when the frequency is doubled
Answer:
[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]
Explanation:
The number of oscillation per unit time is called frequency of a wave while the distance between two consecutive crests and the trough is called its wavelength.
The wavelength of a wave is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
If frequency is doubled, f' = 2f
[tex]\lambda'=\dfrac{v}{f'}[/tex]
[tex]\lambda'=\dfrac{v}{2f}[/tex]
[tex]\lambda'=\dfrac{1}{2}\times \dfrac{v}{f}[/tex]
[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]
So, when the frequency is doubled the wavelength of a wave on the string becomes half.
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon?
The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation.
Explanation:The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation. The formula is F = G × (m1 × m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers. In this case, the Earth is the object experiencing the acceleration and the Moon is the object exerting the gravitational force.
Given the distance between the center of the Earth and the center of the Moon as r = 4700 km (or 4.7 x 10^6 m) and the gravitational constant F = G × (m1 × m2) / r^2[tex]F = G × (m1 × m2) / r^2[/tex], the formula becomes:
[tex]F = G × (m1 × m2) / r^2[/tex]
Since we know the mass of the Earth, we can calculate the acceleration by rearranging the formula to solve for ae:
ae = F / m1
Substituting the values, we get:
[tex]ae = (G × (m1 × m2) / r^2) / m1[/tex]
Simplifying the equation, we are left with:
[tex]ae = G × m2 / r^2[/tex]
Now, we can plug in the known values. The mass of the Moon (m2) is approximately 7.35 x 10^22 kg:
[tex]ae = (6.67 x 10^-11 N · m^2/kg^2) × (7.35 x 10^22 kg) / (4.7 x 10^6 m)^2[/tex]
Calculating this expression will give us the value for the magnitude of the acceleration of the Earth due to the gravitational pull of the Moon.
In afternoon, the decibel level of a busy freeway is 100 dB with 100 cars passing a given point every minute. Late at night, the traffic flow is only 20 cars per minute. What is the late-night decibel level? A. 20 dB b. 50 dB c. 80 dB d. 88 dB e. 93 dB
Answer:
option E
Explanation:
given,
decibel level of a busy freeway = 100 dB
when flow is = 100 car/minutes
traffic flow= 20 car/minute
Intensity factor =[tex]\dfrac{I}{I_0}[/tex]
=[tex]\dfrac{20}{100}[/tex]
=[tex]\dfrac{1}{5}[/tex]
[tex]\Delta \beta = 10 log\dfrac{I}{I_0}[/tex]
[tex]\Delta \beta = 10 log\dfrac{1}{5}[/tex]
[tex]\Delta \beta = - 6.989\ dB[/tex]
The late night decibel level
= 100 - 6.989
= 93
The correct answer option E
Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is the new level of sound β2?
The new level of sound β2 = β1 + 10 log10(f).
Suppose that a sound has initial intensity level β1 measured in decibels.
This sound now increases in intensity level by a factor f. What is the new level of sound β2?
Formula to calculate the new sound intensity β2 = β1 + 10 log10(f)
The new level of sound β2 can be calculated using the formula β2 = β1 + 10 log10(f)
Here,β1 is the initial sound intensity levelβ2 is the new sound intensity level if is the factor by which the sound intensity level is increased by.
Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.
The SI unit of intensity, which includes sound intensity, is the watt per square meter (W/m2)
Hence, The new level of sound β2 = β1 + 10 log10(f).
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The new level of the sound, β2, can be found by using the property of logarithms. The formula to calculate the new level is β2 = β1 + 10 log10(f). If one sound is twice as intense as another, it has a sound level about 3 dB higher.
Explanation:The new level of the sound, β2, can be found by using the property of logarithms. When the intensity of a sound increases by a factor of f, the sound level increases by 10 times the logarithm (base 10) of f. So, in this case, if the initial sound has intensity β1, the new level of sound β2 can be obtained by:
β2 = β1 + 10 log10(f)
For example, if the initial sound has intensity β1 = 60 dB and it increases by a factor of f = 2, then:
β2 = 60 + 10 log10(2) = 60 + 3 = 63 dB
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of length 16.8 in. with a cross-sectional area of 3.00 in.2, how far down must the piston be pushed before air will flow into the tire? Assume the temperature is constant.
Answer:
L - h = 12.3672 in
Explanation:
Given
P = 41.0 lb/in² = 41 P.S.I
L = 16.8 in
A = 3.00 in²
h = ?
In order that air flows into the tire, the pressure in the pump must be more than the tire pressure, 41.0 PSI.
We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be 14.6959 P.S.I , we can use the ideal gas equation
P*V = n*R*T
As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have
P₁*V₁ = P₂*V₂ ⇒ V₂ = P₁*V₁ / P₂
where
1 and 2 are initial and final states respectively,
V₁ = A*L = (3.00 in²)*(16.8 in) ⇒ V₁ = 50.4 in³
P₁ = 14.6959 P.S.I
P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²
Inserting various values we get
V₂ = (14.6959 P.S.I)*(50.4 in³) / (55.6959 lb/in²)
⇒ V₂ = 13.2985 in³
Length of pump, measured from bottom, this volume corresponds to is
h = V₂ / A = (13.2985 in³) / (3.00 in²)
⇒ h = 4.4328 in
Piston must be pushed down by more than
L - h = 16.8 in - 4.4328 in = 12.3672 in
Your tutor tells you that an ampere (A) and a volt (V) measure the same thing and that the different terms just make a simple concept confusing. Why should you consider getting a different tutor?
Answer:
Explained
Explanation:
Your tutor is completely wrong here. Ampere and volt are both completely different concept. Ampere measure current in circuit, whereas volt measure the potential difference across two points in circuit. It is due to this difference in the potential that leads to flow of current in the circuit. The potential difference is the cause and current is the effect.
An ampere measures electric current, while a volt measures electrical potential difference, and they are critical in understanding electrical circuits. Ammeters are used to measure current and are wired in series, while voltmeters, which include an internal resistor, measure voltage and are connected in parallel. Confusing these concepts suggests a serious misunderstanding of basic electrical principles, thus getting a different tutor would be recommended.
Explanation:If your tutor tells you that an ampere (A) and a volt (V) measure the same thing and suggests that these terms complicate a simple concept, it would be wise to consider finding a different tutor. This is because an ampere and a volt actually measure two fundamentally different electrical properties. The ampere measures electric current, which is the flow of electric charge, and is analogous to how much water flows through a pipe. In contrast, a volt measures electrical potential difference, or voltage, which is the energy available to move the charge, akin to the height from which water falls. When using tools to measure these properties, we use an ammeter to measure current, placing it in series so that all the current flows through it, and a voltmeter to measure voltage, connecting it in parallel to get a reading without significantly altering the flow of current.
Additionally, the internal design of a voltmeter includes an internal resistor and when connected in a circuit, it effectively creates a parallel resistance arrangement with the component across which the voltage is being measured. Therefore, to accurately measure and understand electrical circuits, one must recognize the difference between amperes and volts and use the appropriate measurement tools accordingly.
An arsenic surface emits electrons after shining with a light source of the frequency of 5.32×1023 Hz. Calculate the maximum kinetic energy of the ejected electron if the work function of arsenic is 7.67×10.
Answer:
The maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]
Explanation:
We are given that
Frequency of light source=[tex]5.32\times 10^{23}[/tex] Hz
Work function of arsenic=[tex]7.67\times 10[/tex] eV
We have to find the maximum kinetic energy of ejected electron.
We know that the maximum kinetic energy of ejected electron
[tex]K.E=h\nu-w_o[/tex]
Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]
[tex]\nu[/tex] =Frequency of light source
[tex]w_o[/tex]=Work function
Substitute the values in the given formula
Then, the maximum kinetic energy of ejected electron
[tex]K.E=6.63\times 10^{-34}\times 5.32\times 10^{23}-7.67\times 10\times 1.6\times 10^{-19}[/tex]
Because 1 e V=[tex]1.6\times 10^{-19} J[/tex]
[tex]K.E=35.2716\times 10^{-11}-12.272\times 10^{-18}[/tex]
[tex]K.E=3.53\times 10^{-10}[/tex] J
Hence, the maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]
Suppose you are standing such that a 35-foot tree is directly between you and the sun. If you are 7 feet tall and the tree casts a 75-foot shadow, how far away from the tree can you stand and still be completely in the shadow of the tree?
Answer:
60 ft
Explanation:
The tree man and their shadow system forms two similar right angle triangles.
the figure is in the attachment.
let ∠BAC= θ
Therefore, in triangle ABC
tanθ = BC/AC= 35/75= 7/15
now in ΔAEF also
tanθ = EF/AE = 7/75-x= 7/15
solving we get x=60 ft
Now AC and FA are the shadows of the tree and the man respectively.
now FA =75-x=75-60= 15 ft
Therefore, the man must stand at a distance of 60 ft from the tree can you stand and still be completely in the shadow of the tree
A person 7 feet tall can stand 60 feet away from a 35-foot tree casting a 75-foot shadow to be completely in the shadow of the tree.
To solve for how far away from the tree a person can stand and still be completely in the shadow of the tree, we can use similar triangles. Since the tree is 35 feet tall and casts a 75-foot shadow, and a person is 7 feet tall, the height ratio between the tree and the person is 5:1 (35 feet : 7 feet). Therefore, the length of the shadow that the person will cast will also be 5 times shorter than the length of the tree's shadow.
We calculate the length of the person's shadow as follows:
Length of person's shadow = Length of tree's shadow / Height ratio
Length of person's shadow = 75 feet / 5
Length of person's shadow = 15 feet
The total distance from the tree at which the person can stand and still be completely in the shadow is the length of the tree's shadow minus the length of the person's shadow.
Total distance from the tree = Length of tree's shadow - Length of person's shadow
Total distance from the tree = 75 feet - 15 feet
Total distance from the tree = 60 feet
9. Batman (mass=91 kg) jumps straight down from a bridge into a boat (mass=510 kg) in which a criminal is fleeing. The velocity of the boat is initially 11 m/s. What is the velocity of the boat after Batman lands in it?
Answer:
V = 9.33 m/s
Explanation:
Given that,
Mass of the batsman, [tex]m_1=91\ kg[/tex]
Mass of the boat, [tex]m_2=510\ kg[/tex]
Initial speed of the boat, v = 11 m/s
Let V is the velocity of the boat after Batman lands in it. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :
[tex]510\times 11=(91+510)V[/tex]
[tex]V=\dfrac{510\times11}{(91+510)}[/tex]
V = 9.33 m/s
So, the velocity of the boat after Batman lands in it 9.33 m/s. Hence, this is the required solution.
When Batman lands in the boat, it causes a change in the boat's velocity due to the force exerted by Batman. Using the principle of conservation of momentum, we can find the final velocity of the boat. The final velocity of the boat is 10.0 m/s, moving in the opposite direction at an angle of 26.6° to a line drawn across the river.
Explanation:When Batman jumps into the boat, he exerts a force on the boat due to his mass and acceleration. According to Newton's third law of motion, the boat exerts an equal and opposite force on Batman. This force causes a change in the velocity of the boat.
To find the final velocity of the boat, we can use the principle of conservation of momentum. The initial momentum of the boat-criminal system is equal to the final momentum of the system after Batman lands in the boat. Using the equation for momentum, we can find the final velocity of the boat.
After plugging in the given values, we find that the velocity of the boat after Batman lands in it is 10.0 m/s. The boat moves in the opposite direction, at an angle of 26.6° to a line drawn across the river.
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What temperature has the same value in both the fahrenheit and kelvin scales?
Answer:
574.25 °
Explanation:
To know this, we need to get the expressions to calculate °F and °K.
In the case of Kelvin:
°K = °C + 273
in fahrenheit:
°F = 9/5(°C + 32)
Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.
°C = °K - 273 (1)
°C = 5/9(°F - 32) (2)
Using 1 and 2 into a same expression:
°K - 273 = 5/9(°F - 32)
With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:
°K - 273 = 5/9(°K - 32)
°K = 0.555(°K - 32) + 273
°K = 5/9K - 17.78 + 273
°K - 5/9K = 255.22
4/9°K = 255.22
°K = 255.22 * 9 / 4
°K = 574.25 °
And this is the temperature in which both Kelvin and Fahrenheit are the same.
A 500 Hz whistle is moved toward a listener at a speed of 10 m/s. At the same time, the listener moves at a speed of 20 m/s in a direction away fromm the whistle. What is the apparent frequency heard by the listener? (the speed of sound is 340 m/s)
A) 473 hz
B) 376 hz
C) 485 hz
D) 533 hz
E) 547hz
Answer:
Frequency will be 485 Hz
So option (c) will be the correct option
Explanation:
We have given frequency f = 500 Hz
Velocity of observer [tex]v_o=20m/sec[/tex]
Velocity of source [tex]v_s=10m/sec[/tex]
When both observer and source are moving then frequency is given by
[tex]f'=\frac{v-v_o}{v-v_s}\times f[/tex] , here v is the velocity of sound
So [tex]f'=\frac{340-20}{340-10}\times 500=485Hz[/tex]
So option (c) will be the correct option
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the student to overcome gravity is most nearly 1. P = 87.5 W. 2. P = 5, 600 W. 3. P = 1.1 W. 4. P = 875 W. 5. P = 560 W.
Answer:
Power, P = 560 W
Explanation:
Given that,
Weight of the student, F = 700 N
Distance, d = 8 m
Time taken, t = 10 s
To find,
The average power expended by the student.
Solution,
Let P is the power. The work done per unit time is called its power. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
[tex]P=\dfrac{F\times d}{t}[/tex]
[tex]P=\dfrac{700\ N\times 8\ m}{10\ s}[/tex]
P = 560 W
So, the average power expended by the student to overcome gravity is most nearly 5670 watts.
The average power expended by the student to overcome gravity =
( 5.) P = 560 W
The formula for Power is given by the equation (1)
[tex]Power = \dfrac{Work}{Time}[/tex]...........(1)
Also the formula for Work is given by equation (2)
[tex]Work = Force \times Distance[/tex]............(2)
(Considering magnitude of Displacement)
From equation (1) and (2) we can get
[tex]Power = \dfrac{Force \times Distance}{Time}[/tex].......(3)
Also [tex]Speed= \dfrac{ Distance}{Time}[/tex]....(4)
From equation (2) , (3) and (4)
So [tex]Power= Force \times Speed[/tex]
Given
Weight of the Student = 700 N
Height of the rope = 8 m
Time taken to climb the rope = 10 s.
So from equation (3) we can get
Average Power expended by the student to overcome gravity = (700)[tex]\times[/tex] (8)/(10) = 560 W.
Hence option 5 is correct.
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A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the opposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
The change in angular velocity of the bicycle tire is -5.40 rad/s, indicating a change in direction. The average angular acceleration during this time period was -2.84 rad/s2, also reflected by the negative sign representing a change in direction.
Explanation:This question pertains to physics, specifically focusing on angular velocity and angular acceleration. For the bicycle tire spinning counterclockwise at 2.70 rad/s to stop and spin in the clockwise direction at the same angular velocity, its total change in angular velocity (∆ω) is a combination of the initial and final angular velocities, given by ∆ω = ω_final - ω_initial. Since the tire goes from spinning counterclockwise (considered positive) to clockwise (considered negative), ∆ω = (-2.70 rad/s) - (2.70 rad/s) = -5.40 rad/s.
Next, to find the average angular acceleration (∆av), we use the formula ∆av = ∆ω / ∆t. Substituting in the time given (1.90 s) along with our calculated ∆ω, ∆av = -5.40 rad/s / 1.90 s = -2.84 rad/s2. Note that the negative sign indicates a change in direction.
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Two uniform spheres,each of mass 0.260kg are fixed at points A and B
A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.
B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.
Complete Question:
Two uniform spheres,each of mass 0.260kg are fixed at points A and B
A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.
B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.
What factors determine whether a collision between two reactant molecules will result in a reaction?
Answer:
Minimum required energy
The correct orientation
Explanation:
A collision is described as successful or effective if the particles:
Collide with the right amount of energy. This is why, generally increasing the temperature of a reaction leads to an increased number of successful collisions. Higher temperature means a higher average kinetic energy. It is this minimum energy requirement (sometimes called activation energy) that must be met to achieve a successful collisionCollide in the correct orientation. Meeting the minimum energy requirement is not enough, particles must also collide in the correct orientation.The success of a collision between reactant molecules in causing a reaction depends on the molecules colliding with enough energy to overcome the activation energy, proper orientation during the collision, and a higher collision frequency influenced by reactant concentration.
Explanation:To answer your question about the factors that determine whether a collision between two reactant molecules results in a reaction, we need to consider several key factors. First, reactant molecules must collide with sufficient energy; this energy must be equal to or higher than the activation energy of the reaction. Second, the molecules must have proper orientation during the collision to allow the necessary rearrangement of electrons and atoms. Lastly, the collision frequency is also essential, which is influenced by the concentration of the reactants; higher concentration generally leads to more collisions and an increased likelihood of reaction.
The deformation of molecules during a collision can also be a contributing factor, as more violent collisions tend to lead to greater deformation and enhance the probability of reaction. Additionally, external factors such as temperature and the presence of catalysts can increase the kinetic energy of the molecules or reduce the activation energy, both of which can lead to a higher rate of successful collisions and reactions.
A certain baryon (a subatomic particle) has a charge q > 0. The baryon moves with a kinetic energy K in a circular path of radius R in the presence of a uniform magnetic field of magnitude B. (Use any variable or symbol stated above as necessary.) What is the speed of the bayron? Assume the speed is low compared to the speed of light. What is the mass of the bayron?
Answer:
mass of the baryon = [tex]\frac{(qRB)^{2} }{2K}[/tex]
speed of the baryon = [tex]\frac{2K}{qRB}[/tex]
Explanation:
For any body to move in a circular path, there must be a centripetal force which is directed towards the center of the circle. Here, the centripetal force is provided by the magnetic Lorentz force that acts on the baryon. Therefore we can equate the magnitudes of centripetal force and magnetic Lorentz force.
Magnitude of Centripetal force = [tex]\frac{mv^{2} }{R}[/tex]
Magnitude of Magnetic Lorentz force = qvB
where,
m = mass of the baryon
v = velocity of the baryon
Thus,
[tex]\frac{mv^{2} }{R}[/tex] = qvB
[tex]\frac{mv}{R}[/tex] = [tex]qB[/tex] (cancelling v on both sides)
v = [tex]\frac{qRB}{m}[/tex]
We know, Kinetic energy, K = [tex]\frac{1}{2} mv^{2}[/tex]
Substituting v in the above equation we get,
K = [tex]\frac{1}{2} m(\frac{qRB}{m} )^{2}[/tex]
K = [tex]\frac{(qBR)^{2} }{2m}[/tex] (simplifying)
Thus,
m = [tex]\frac{(qRB)^{2} }{2K}[/tex]
We already got that
v = [tex]\frac{qRB}{m}[/tex]
substituting the value of m in this equation gives,
[tex]v = \frac{qRB}{\frac{(qRB)^{2} }{2K}}[/tex]
v = [tex]\frac{2K}{qRB}[/tex] (simplifying)
Thus,
mass of the baryon = [tex]\frac{(qRB)^{2} }{2K}[/tex]
velocity of the baryon = [tex]\frac{2K}{qRB}[/tex]
The intensity level is 65 db at a distance 5.00 m from a barking dog. What would be a reasonable estimate for the intensity level if two identical dogs very close to each other were barking? You can ignore any interference effects.
Answer:
68 db
Explanation:
Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.
Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,
β(db)=10×㏒(2)
=3 db
So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db
The intensity level of two identical dogs barking together would be approximately 68 dB, as an increase in
sound intensity leads to an increase in the decibel level.
Explanation:The intensity level of a sound is determined by its amplitude and is measured in decibels (dB). The relationship between intensity and decibel level is logarithmic, which means that an increase of 10 dB corresponds to a tenfold increase in intensity. Hence, if two identical dogs are barking very close to each other their combined sound would be twice as intense as one dog barking alone. However, this does not mean that the decibel level would double. In fact, when the intensity of a sound is doubled, the increase in decibel level is about 3 dB. Thus, if a single dog barking has an intensity level of 65 dB, two dogs barking together would have an estimated intensity level of 68 dB.
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A ceiling fan is turned on and a net torque of 2.1 N · m is applied to the blades. The blades have a total moment of inertia of 0.19 kg · m2. What is the angular acceleration of the blades?
The angular acceleration of a ceiling fan with a net torque of 2.1 N · m and blades having a moment of inertia of 0.19 kg · m² is approximately 11.05 rad/s².
Explanation:The subject of this question is Physics, specifically rotation dynamics. In rotation dynamics, angular acceleration is defined by the equation α = τ / I, where α is the angular acceleration, τ is the net torque, and I is the moment of inertia. In the given problem, a net torque of 2.1 N · m is applied to a ceiling fan with blades having a total moment of inertia of 0.19 kg · m². Substituting the given values into the equation, we can solve for the angular acceleration: α = 2.1 N · m / 0.19 kg · m² = approximately 11.05 rad/s². Hence, the angular acceleration of the ceiling fan's blades is approximately 11.05 rad/s².
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The angular acceleration of the ceiling fan blades, given a net torque of 2.1 N·m and a moment of inertia of 0.19 kg·m², is calculated using Newton's second law for rotation. The resulting angular acceleration is 11.05 rad/s².
Explanation:The student's question is related to the field of Physics, specifically rotational kinematics and dynamics. The key concept here is Newton's second law for rotation, which can be expressed as τ = Iα, where τ represents the net torque, I is the moment of inertia, and α is the angular acceleration.
Given that the net torque (τ) is 2.1 N·m and the moment of inertia (I) is 0.19 kg·m², we can find the angular acceleration (α) by rearranging the equation α = τ / I. Substituting the given values into this equation: α = 2.1 N·m / 0.19 kg·m² = 11.05 rad/s². Hence, the angular acceleration of the blades of the ceiling fan is 11.05 rad/s².
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A mass is oscillating on the end of a spring. The distance, y, of the mass from its equilibrium point is given by the formula y=4zcos(8πwt) where y is in centimeters, t is time in seconds, and z and w are positive constants.
a. What is the furthest distance of the mass from its equilibrium point?
b. How many oscillations are completed in 1 second?
Answer:
(a.) 4z
(b.) 4w
Explanation:
From the equation y=4zcos(8πwt), where z and w are positive constants.
Comparing this equation to the equation of a wave y = Acos(Wt), where A is the amplitude (largest distance from equilibrium) and W is the angular frequency (W=2πf)
(a.) Comparing our wave equation with the given equation, we see that A = 4z in this case (furthest distance of the mass from equilibrium)
(b.) Comparing similarly we can see from our given equation that angular frequency W =8πw we also know that W = 2πf from our wave equation, therefore 2πf = 8πw
Solving for f we have f = 8πw÷2π
f = 4w (Proves our second answer because the frequency is the number of oscillations completed per second)
What experiments did thomas edison conduct with electricity
Answer:
Vacuum lamp experiment
Explanation:
Edison began his work on electricity the year 1878, where he made use of platinum wire to make filaments for his lamp as the melting point of Platinum is.
However, later he concluded that Platinum pores absorbed air while heating thus reducing its melting point.
So he came up with the idea of vacuum lamp and developed quite a good vacuum lamp with the help of carbon.
Being highly resistive in nature, carbon would end up burning at quite a rapid rate.
Along with his staff, in the year 1879, Edison conducted his first fruitful experiment by using vacuum lamp with carbon filament.
Which person is responsible for the building and operation of stage machinery and scenery; has charge of lighting crews and industrial scheduling; has charge of moving scenery in and out of the theatre; and establishes policies and directives for scene shifting?
Answer:
Technical director.
Explanation:
The answer is the Technical director.
This is very important for every company or organization to manage the activity in a technical way. Because if we did not follow the good technique and did not use a proper technical method then definitely the task will not be completed in the specified time limit.
The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h, in feet, is currently given by the equation h=-1/2 t^2+6t-9, when will the tide first be at 6 ft?
Explanation:
Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :
[tex]h=-\dfrac{1}{2}t^2+6t-9[/tex]
When the tide first be at 6 ft, put h = 6 ft in above equation as :
[tex]-\dfrac{1}{2}t^2+6t-9=6[/tex]
[tex]-t^2+12t-18=0[/tex]
On solving the above equation to find the value of t. It is equal to :
t = 3.551 seconds
or
t = 8.449 seconds
So, the tide of 6 ft is at 3.551 seconds and 8.449 seconds. Hence, this is the required solution.
What phase does the moon have to be in for a solar eclipse
Answer:
new moon
Explanation:
A solar eclipse take place at new moon phase, when the moon passes between the earth and the sun and its shadows fall on the Earth's surface which by definition a solar eclipse.
The moon must be in the new moon phase for a solar eclipse to occur.
Explanation:The moon must be in the new moon phase for a solar eclipse to occur. During the new moon phase, the moon is located between the Earth and the Sun, causing the moon to cast a shadow on Earth and block the Sun's light.
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What two kinds of crust are involved in a subduction zone
Answer:
Oceanic crust and continental crust
Explanation:
A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.
Final answer:
In subduction zones, the denser oceanic crust is pushed beneath another plate, which may be either oceanic or continental. This process results in the formation of ocean trenches and volcanic activity. Subduction zones play a crucial role in the cycle of plate tectonics.
Explanation:
The two kinds of crust involved in a subduction zone are the oceanic crust and the continental crust. In a subduction zone, one tectonic plate, usually the denser oceanic plate, is forced beneath another plate, which can be either oceanic or continental. Often it is the older and colder oceanic plate that subducts beneath the younger and warmer one. The subduction process is marked by features such as ocean trenches, volcanic island arcs, and is associated with powerful earthquakes and volcanic eruptions.
Subduction zones are integral to the Earth's plate tectonics cycle, balancing the creation of new crust at mid-ocean ridges by destroying the old crust as it descends into the mantle. Such areas of crust destruction are most commonly found around the edges of the Pacific Ocean. The descending plate undergoes changes due to high pressures and temperatures, leading to the release of water and the melting of the crust, forming magma that can result in volcanic activity.