Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.
(A) The wheel has angular velocity at time [tex]t[/tex] according to
[tex]\omega=\omega_0+\alpha t[/tex]
so that after 2.50 s, the wheel will have attained an angular velocity of
[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]
(B) The angular displacement of the wheel is given by
[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]
After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to
[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .
What type of device forms images by changing the speed at which light travels?
Answer:
A lens
Explanation:
A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.
The type of device that forms images by changing the speed at which light travels is the lens.
What is refraction through the lens?
A lens bends a light beam at an aimed perspective and converges or diffuses bundles of rays by taking benefit of refraction taking vicinity while the mild travels from air into glass or plastic. For that purpose, the aspect geometry of a lens has a spherical parent, which can be kind of divided into sorts.
A lens bends a mild beam at an aimed perspective and converges or diffuses bundles of rays through taking gain of refraction taking area whilst the mild travels from air into glass or plastic. For that motive, the facet geometry of a lens has a round parent, which may be kind of divided into sorts.
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Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.
Answer:
The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta _i = 26.50 ^o[/tex]
The angle of refraction angle for sheet 1 is [tex]\theta _{r_1}} = 31.70 ^o[/tex]
The angle of refraction for sheet 3 is [tex]\theta _{r_3}} = 36.70 ^o[/tex]
According to Snell's law
[tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]
Where [tex]n_1 \ and \ n_2[/tex] are refractive index of sheet 1 and sheet 2
=> [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]
Also when sheet 3 in on top of sheet 2
[tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
substituting for [tex]n_2[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
=> [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]
when sheet 1 in on top of sheet 3
[tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3
substituting for [tex]n_3[/tex]
[tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
=> [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]
substituting values
[tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]
=> [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]
=> [tex]\theta_{r_s } = 23.1 ^o[/tex]
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
An experiment invilves three charges objects: A, B, and C. Object A repels object B and attracts onject C. object C ir repelled by ebonite charged with fur. What is the charge on the object?
Answer:
A and B is positive charge
C_negative
Explanation:
because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract
A truck has a bed that is 4.50 metres long,and 2.50 metres wide, and 1.50 metres high. What is maximum volume of sand can the truck carry within this dimensions?
Answer:
since the bed is a cuboid, we find the volume by L×W×H
4.50 × 2.50 × 1.50 = 16.875m³
HOPE THIS HELPS
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.Part aIf the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.part bFind Pnet, the net power radiated by the person when in a room with temperature Troom=20∘C
Answer:
The thermal power emitted by the body is [tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated is [tex]P_{net} = 460 \ W[/tex]
Explanation:
From the question we are told that
The length of the assumed hum[tex]T_{room} = 20 ^oC[/tex]an body is L = 2.0 m
The circumference of the assumed human body is [tex]C = 0.8 \ m[/tex]
The Stefan-Boltzmann constant is [tex]\sigma = 5.67 * 10^{-8 } \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
The temperature of skin [tex]T_{body} = 30^oC[/tex]
The temperature of the room is
The emissivity is e=0.6
The thermal power radiated by the body is mathematically represented as
[tex]P_t = e * \sigma * T_{body}^4[/tex]
substituting value
[tex]P_t = 0.6 * 5.67*10^{-8} * (303)^4[/tex]
[tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated by the body is mathematically evaluated as
[tex]P_{net} = P_t * A[/tex]
Where A is the surface area of the body which is mathematically evaluated as
[tex]A = C* L[/tex]
substituting values
[tex]A = 0.8 * 2[/tex]
[tex]A = 1.6 m^2[/tex]
=> [tex]P_{net} = 286.8 * 1.6[/tex]
=> [tex]P_{net} = 460 \ W[/tex]
Q) Considering the value of ideal gas constant in S.I. unit, find the volume of 35g O2 at 27°C and 72
cm Hg pressure. Later, if we keep this pressure constant, the r.m.s velocity of this oxygen molecules
become double at a certain temperature. Calculate the value of this temperature.
Answer:
V = 0.0283 m³ = 28300 cm³
T₂ = 1200 K
Explanation:
The volume of the gas can be determined by using General Gas Equation:
PV = nRT
where,
P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa
V = Volume of Gas = ?
n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol
R = General Gas Constant = 8.314 J/ mol.k
T = Temperature of Gas = 27°C + 273 = 300 k
Therefore,
(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)
V = 2718.678 J/95992.12 Pa
V = 0.0283 m³ = 28300 cm³
The Kinetic Energy of gas molecule is given as:
K.E = (3/2)(KT)
Also,
K.E = (1/2)(mv²)
Comparing both equations, we get:
(3/2)(KT) = (1/2)(mv²)
v² = 3KT/m
v = √(3KT/m)
where,
v = r.m.s velocity
K = Boltzamn Constant
T = Absolute Temperature
m = mass of gas molecule
At T₁ = 300 K, v = v₁
v₁ = √(3K*300/m)
v₁ = √(900 K/m)
Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?
v₂ = 2v₁ = √(3KT₂/m)
using value of v₁:
2√(900 K/m) = √(3KT₂/m)
4(900) = 3 T₂
T₂ = 1200 K
Identify the five categories of stressors.
Answer:
The five kinds of stressors are:
Acute time-limited
Brief naturalistic
Stressful events sequences
Chronic
Distant
Explanation:
yeah
A gas in a closed container is heated with 12J of energy, causing the lid of the container to rise 3m with 5N of force. What is the total change in energy?
Answer:
27J
Explanation:
From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.
Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;
U = Q + (F×d);
U- change in internal energy
Q is the energy received by the system and is positive when energy is received by the system.
Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.
U=12+(3×5)= 27J
cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)
Answer:
Its initial position was 471 m.
Explanation:
We have,
Final position of the object is 327 m
Displacement of the object is -144 m
It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,
[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]
So, its initial position was 471 m.
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.
Answer:
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
Explanation:
The Free Body Diagram associated with the experiment is presented as attachment included below.
Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.
Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:
[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]
Where:
[tex]W[/tex] - Weight of the eraser, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]\theta[/tex] - Angle of the incline, measured in degrees.
The maximum allowable static friction force is:
[tex]f = \mu_{s} \cdot N[/tex]
Where:
[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
Likewise, the kinetic friction force is described by the following model:
[tex]f = \mu_{k} \cdot N[/tex]
Where:
[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:
[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]
[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]
But [tex]N = m\cdot g \cos \theta[/tex], so that:
[tex]\mu = \tan \theta[/tex]
Now, coefficients of static and kinetic friction are, respectively:
[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]
[tex]\mu_{s} \approx 0.716[/tex]
[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]
[tex]\mu_{k} \approx 0.596[/tex]
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
Which symbol is used to show vector quantities
Answer: arrows
Explanation:
A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.
What is the vector quantity unit?
The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.
The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.
Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).
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A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
[tex]37.8=\frac{1}{2}(3)v^2[/tex]
[tex]v^2=\frac{37.8\times 2}{3}[/tex]
[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]
v=5.02 m/s
c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]
[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]
Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?
Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex] (1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)
Required:
a. Qualitatively, indicate the polarization of the plastic.
1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.
b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.
1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.
c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.
1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.
Answer:
(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center
(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.
Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.
(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
A whistle of frequency 516 Hz moves in a circle of radius 64.3 cm at an angular speed of 17.9 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle
Answer:
(a) 498.6 Hz
(b) 534.6 Hz
Explanation: Please see the attachments below
A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
A 1KW electric heater is switched on for ten minutes
How much heat does it produce?
Explanation:
P=W/T ==> 1000w = Q/600 ==> Q=600000j
If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .
What is thermal energy ?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .
As given in the problem , we have to find out the heat produced by the 1 - kilo watts electric heater if it is switched on for ten minutes ,
The heat produced by the electric heater = Power × time
= 1000 × 600 Joules
= 600 kilo - Joules
Thus , the heat produced by the electric heater would be 600 - kilo Joules .
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An electric point charge of Q = 22.5 nC is placed at the center of a cube with a side length of a = 16.3 cm. The cube in this question is only a mathematical object, it is not made out of any physical material. What is the electric flux through all six sides of the cube?
Answer:
The electric flux is [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
Explanation:
From the question we are told that
The magnitude of the electric point charge [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]
The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]
The number of sides is [tex]N= 6[/tex]
The electric flux according to Gauss law is mathematically evaluated as
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]
[tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.40 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
[tex]vf^2 = vi^2 - 2*g*x[/tex]
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
[tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
[tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT magnetic field if the maximum value of the induced emf is to equal 8.0 V
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.
Calculation of the minimum frequency:Since
number of turns, N = 200 turns
cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
the magnitude of magnetic field strength, B = 30 x 10⁻³ T
the maximum value of the induced emf, E = 8 V
Now
Maximum induced emf should be
E = NBAω
here,
ω is angular velocity (ω = 2πf)
Now
E = NBA2πf
here,
f is the minimum frequency
So,
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz.
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We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power is [tex]P_p = 4.6 \ W[/tex]
Explanation:
From the question we are told that
The power delivered is [tex]P_{s} = 1.15 \ W[/tex]
Let it resistance be denoted as R
The resistors are connected in series so the equivalent resistance is
[tex]R_{eqv} = R+ R = 2 R[/tex]
Considering when it is connected in series
Generally power is mathematically represented as
[tex]P_s = V * I[/tex]
Here I is the current which is mathematically represented as
[tex]I = \frac{V}{2R}[/tex]
The power becomes
[tex]P_s = V * \frac{V}{2R}[/tex]
[tex]P_s = \frac{V^2}{2R}[/tex]
substituting value
[tex]1.15 = \frac{V^2}{2R}[/tex]
Considering when resistance is connected in parallel
The equivalent resistance becomes
[tex]R_{eqv} = \frac{R}{2}[/tex]
So The current becomes
[tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]
And the power becomes
[tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]
substituting values
[tex]P_p = 4 * 1.15[/tex]
[tex]P_p = 4.6 \ W[/tex]
describe Piaget's four stages of cognitive development. Include the major hallmarks of each stage.
Answer:
Explanation:
Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.
Preoperational. Development of language and make-believe play takes place.
Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.
Formal Operational. Adolescences can also evaluate the logic of verbal statements without referring to real-world circumstances.
Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
What is cognitive development?The way youngsters think, investigate, and figure things out is referred to as cognitive development.
Piaget defined four stages of cognitive development:
1. Sensorimotor. From birth through the age of 18-24 months.
2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)
3. Operational concrete. 7 to 11 years old
4. Formal operational. From adolescence to adulthood
Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
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A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
What percent of our solar system's mass is in the sun?
Answer:
99.8
Explanation:
most massive the sun is at the center of the universe