Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
A football is kicked with a velocity of 18 m/s at an angle of 20°. What is the
ball's acceleration in the horizontal direction as it flies through the air?
Explanation:
It is given that,
The velocity of football is 18 m/s
It is projected at an angle of 20 degrees
We need to find the ball's acceleration in the horizontal direction as it flies through the air.
When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.
In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.
Mention the importance of occupation??
Answer:
ln the contemporary time , farming can be considered as comparitively important occupation as it can feed the population , So agriculture is having a greater importance than any other occupation.
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
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A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
We have that for the Question, it can be said that the average induced emf in the coil is
E=0.028565V
From the question we are told
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Generally the equation for the Average emf induced is mathematically given as
[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]
Hence
[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]
Therefore
[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]
E=0.028565V
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A pump is to deliver 10, 000 kg/h of toluene at 1140C and 1.1 atm absolute pressure from the Reboiler of a distillation tower to the second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the Reboiler and the pump is 7 kN/m2. The density of toluene is 886 kg/m3. How far above the pump must the liquid be maintained to avoid cavitation
Answer:
3.4093
Explanation:
NPSHa = hatm + hel + hf +hva
the elevation head is the hel
friction loss head is hf
NPSHa is the head of vapour pressure of fluid
atmospheric pressure head is hatm
log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]
[tex]A, B, C are fixed[/tex]
log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]
= 4.07827 - 1343.943/333.377
=4.07827 - 4.0313009
= 0.0469691
we take the log
p* = 1.114218
we convert this value to get 111421.8
hvap = 111421.8 * 1/776.14 * 1/9.81
= 14.63
hatm = 1.1 *101325/1 * 1/9.81 *1/776.14
=14.64
hf = 7000/1 * 1/776.14 * 1/9.81
= 0.9193
NPSHa = 2.5
hel = 0.9193 + 2.5 + 14.63 - 14.64
hel = 3.4093
The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.
The vapor pressure can be calculated by,
[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]
Where,
[tex]\mathrm {NPSH_A}[/tex] = available NPSH
[tex]p_i[/tex] = absolute pressure at the inlet = 1.1 atm
[tex]V_i[/tex] = average velocity at the inlet = 10, 000 kg/h
[tex]\rho[/tex] = fluid density = 886 kg/m3.
g = acceleration of gravity = 9.8 m/s²
[tex]p_v[/tex] = vapor pressure of the fluid = ?
Put the values in the equation, we get
[tex]p_v = 1.114\ atm[/tex]
Therefore, the vapor pressure of the liquid is 1.114 atm.
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Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
write any two importance of gravitational force
Answer:
plz mark me as brainliest plz
Explanation:
The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.
A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) about 20 times per second.part a. Determine the number of photons in one 80-mJ pulse.part b. Determine the average power of photons in one 80-mJ pulse during 1 s.
Answer:
a
[tex]n = 1.119 *10^{18} \ photons[/tex]
b
[tex]P = 1.6 \ W[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 2780 nm = 2780 *10^{-9} \ m[/tex]
The energy is [tex]E = 80 mJ = 80 *10^{-3} \ J[/tex]
This energy is mathematically represented as
[tex]E = \frac{n * h * c }{\lambda }[/tex]
Where c is the speed of light with a value [tex]c = 3.0 *10^{8} \ m/s[/tex]
h is the Planck's constant with the value [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
n is the number of pulses
So
[tex]n = \frac{E * \lambda }{h * c }[/tex]
substituting values
[tex]n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }[/tex]
[tex]n = 1.119 *10^{18} \ photons[/tex]
Given that the pulses where emitted 20 times in one second then the period of the pulse is
[tex]T = \frac{1}{20}[/tex]
[tex]T = 0.05 \ s[/tex]
Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as
[tex]P = \frac{E}{T}[/tex]
substituting values
[tex]P = \frac{ 80 *10^{-3}}{0.05}[/tex]
[tex]P = 1.6 \ W[/tex]
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
what are the applications of pascal's principle
Explanation:
The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.Hydraulic jack- used in the braking system of cars.use of a straw- to suck fluids, which goes because of air pressure.The question simply asks, where pressure can be applied. There are many others, such as lift pump.Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <
Answer:
1. Energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared >> visible > ultraviolet
Explanation:
Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.
Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:
1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.
i.e energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.
i.e frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.
i.e wavelength: infrared >> visible > ultraviolet
In terms of lowest to the highest energy,frequency, and wavelength is;
Energy: infrared > visible light > ultraviolet
Frequency: infrared > visible light > ultraviolet
Wavelength: ultraviolet > visible light > infrared
The electromagnetic spectrum:
The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.
The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.Learn more about electromagnetic spectrum here:
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Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over its entire length to a temperature of 600C. Calculate the heat transferred from the plate.
Answer:
Heat transferred = 22.9 watt
Explanation:
Given that:
[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K
[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K
speed v = 2 m/s
length x = 40 cm = 0.4 cm
width = 1 cm = 0.001 m
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
From the tables of properties of air, the following values where obtained.
[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]
To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:
reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]
reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]
reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]
reynolds number = 52688.11204
Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]
Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]
Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]
Prandtl number = 0.754963
The nusselt number for this turbulent flow over the flat plate can be computed as follows:
Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]
[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]
h = 9.992 W/m.k
Recall that:
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]
Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)
Heat transferred = 22.9 watt
A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Answer:
b
Explanation:
an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the device
Answer: R=24.2Ω
Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
[tex]P=\frac{V^{2}}{R}[/tex]
The resistance of the system is:
[tex]P=\frac{V^{2}}{R}[/tex]
[tex]R=\frac{V^{2}}{P}[/tex]
[tex]R=\frac{110^{2}}{500}[/tex]
R = 24.2Ω
For the device, resistance is 24.2Ω.
A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater =1.333). He will see her soles as being:____
a. right at the water surface.
b. 53.3 cm above the water surface.
c. exactly 80.0 cm above the water surface.
d. 107 cm above the water surface.
e. an infinite distance above the water surface.
Answer:
d. 107 cm above the water surface.
Explanation:
The refractive index of water and air = 1.333
The real height of the girl's sole above water = 80.0 cm
From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.
The boy in the water will therefore see her feet as being
80.0 cm x 1.333 = 106.64 cm above the water
That is approximately 107 cm above the water
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck's kinetic energy
Answer:
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
Explanation:
Given:
Mass of truck(m) = 2,100 kg
Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s
Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s
Find:
Change in kinetic energy (ΔKE)
Computation:
Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]
Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]
Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]
Change in kinetic energy (ΔKE) = 1,050[121.9648]
Change in kinetic energy (ΔKE) = 128063.04
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A in a circular orbit of radius 5000 km and period 4.0 hours. What is the radius of Planet R47A
[tex]2.6×10^6\:\text{m}[/tex]
Explanation:
The acceleration due to gravity g is defined as
[tex]g = G\dfrac{M}{R^2}[/tex]
and solving for R, we find that
[tex]R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)[/tex]
We need the mass M of the planet first and we can do that by noting that the centripetal acceleration [tex]F_c[/tex] experienced by the satellite is equal to the gravitational force [tex]F_G[/tex] or
[tex]F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)[/tex]
The orbital velocity v is the velocity of the satellite around the planet defined as
[tex]v = \dfrac{2\pi r}{T}[/tex]
where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as
[tex]\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}[/tex]
Solving for M, we get
[tex]M = \dfrac{4\pi^2 r^3}{GT^2}[/tex]
Putting this expression back into Eqn(1), we get
[tex]R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}[/tex]
[tex]\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}[/tex]
[tex]\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:= 2.6×10^6\:\text{m}[/tex]
Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential
Answer:
1. the sphere of the radius a
Explanation:
Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.
So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .
With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r
(for r ≥ R ) .
On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.
So Since b > a , the sphere of radius a will have the higher potential.
Also recall Because E = 0 inside a conductor, the potential
The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.
Answer:
μ = 1
F = P√2
Explanation:
The parabola equation is: y = ½ x².
The slope of the tangent is dy/dx = x.
The angle between the tangent and the x-axis is θ = tan⁻¹(x).
At x = 1, θ = 45°.
Draw a free body diagram of the block. There are three forces:
Weight force P pulling down,
Normal force N pushing perpendicular to the surface,
and friction force Nμ pushing up tangential to the surface.
Sum of forces in the perpendicular direction:
∑F = ma
N − P cos 45° = 0
N = P cos 45°
Sum of forces in the tangential direction:
∑F = ma
Nμ − P sin 45° = 0
Nμ = P sin 45°
μ = P sin 45° / N
μ = tan 45°
μ = 1
Draw a new free body diagram. This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.
Sum of forces in the tangential direction:
∑F = ma
F − Nμ − P sin 45° = 0
F = Nμ + P sin 45°
F = (P cos 45°) μ + P sin 45°
F = P√2
If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices
Answer:
0.05cos10t
Explanation:
X(t) = Acos(wt+φ)
The oscillation angular frequency can be calculated using below formula
w = √(k/M)
Where K is the spring constant
But we were given body mass of 5.0 kg
We know acceleration due to gravity as 9.8m)s^2
The lenghth of spring which stretches =10 cm
Then we can calculate the value of K
k = (5.0kg*9.8 m/s^2)/0.10 m
K= 490 N/m
Then if we substitute these values into the formula above we have
w = √(k/M)
w = √(490/5)
= 9.90 rad/s=10rads/s(approximately)
Its position as a function of time can be calculated using the below expresion
X(t) = Acos(wt+φ)
We were given amplitude of 5 cm , if we convert to metre = 0.05m
w=10rads/s
Then if we substitute we have
X(t)=0.05cos(10×t)
X(t)= 0.05cos10t
Therefore,Its position as a function of time=
X(t)= 0.05cos10t
What do Earth scientists do?
Answer:
Study Earth as a whole
Explanation:
ex. oxygen around Earth, layers, formations, temperature, mountains and how they form etc.
Answer:
Geologists study rocks and help to locate useful minerals. Earth scientists often work in the field—perhaps climbing mountains, exploring the seabed, crawling through caves, or wading in swamps. They measure and collect samples (such as rocks or river water), then they record their findings on charts and maps.
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (in V) is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms? V †
Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;
[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]
where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²
[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]
Therefore, the average emf induced in the coil is 175 mV