Complete Question
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?
Answer:
The velocity is [tex]v = 80.82 \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of charge one is [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]
The magnitude of charge two [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]
The distance of separation is [tex]d = 58.0 \ cm = 0.58 \ m[/tex]
Generally the electric potential of the electron at the midway point is mathematically represented as
[tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]
substituting values
[tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]
[tex]V = 1.8103 *10^{-8} \ V[/tex]
Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two
Now the electric potential at that point is mathematically represented as
[tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]
substituting values
[tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]
[tex]V_1 = 3.67*10^{-8} \ V[/tex]
Now the law of energy conservation ,
The kinetic energy of the electron = potential energy of the electron
i.e [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]
where q is the magnitude of the charge on the electron with value
[tex]q = 1.60 *10^{-19} \ C[/tex]
While m is the mass of the electron with value [tex]m = 9.11*10^{-31} \ kg[/tex]
[tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]
[tex]v = \sqrt{6532.4}[/tex]
[tex]v = 80.82 \ m/s[/tex]
g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube
Answer:
The length of the tube is 85 cm
Explanation:
Given;
speed of sound, v = 340 m/s
first harmonic of open-closed tube is given by;
N----->A , L= λ/₄
λ₁ = 4L
v = Fλ
F = v / λ
F₁ = v/4L
Second harmonic of open-closed tube is given by;
L = N-----N + N-----A, L = (³/₄)λ
[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]
Third harmonic of open-closed tube is given by;
L = N------N + N-----N + N-----A, L = (⁵/₄)λ
[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]
The difference between second harmonic and first harmonic;
[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]
The difference between third harmonic and second harmonic;
[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]
Thus, the difference between successive harmonic of open-closed tube is
v / 2L.
[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]
Therefore, the length of the tube is 85 cm
The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
Answer:
Kidney
Explanation:
One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.
Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.
Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.
Why was Bohr's atomic model replaced by the
modern atomic model?
Answer:
Explanation:
Bohr's atomic model was replaced by the modern atomic model because of its limitations, which included :
(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )
(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).
(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.
The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related
Answer:
The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].
Explanation:
Net force on the hoop is given by;
[tex]F_{net} = ma[/tex]
where;
a is linear acceleration
m is the mass
Net torque on the hoop is given by;
[tex]\tau_{net} =I\alpha[/tex]
where;
I is moment of inertia
α is the angular acceleration
But, τ = Fr
[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]
[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]
let the angular acceleration of the smaller hoop = α₁
let the radius of the smaller hoop = r₁
then, the radius of the larger loop, r₂ = 2r₁
let the angular acceleration of the larger hoop = α₂
[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]
Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]
¿Cómo podrías utilizar el sistema de posicionamiento global para proponer recorridos alternativos para llegar a un lugar específico (centro educativo, supermercado, el hogar, el parque, entre otros)?
Answer:
El sistema de posicionamiento global (conocido mundialmente como GPS) podría utilizarse para proponer recorridos alternativos para llegar a un lugar específico, como un parque, a través de la creación de un recorrido guiado por una aplicación móvil con diferentes rutas de acceso al lugar.
Así, por ejemplo, se crearían diferentes rutas de acceso desde un punto A hasta un punto B, teniendo en cuenta factores como: rapidez, congestión vehicular, pago o no de peajes, posibilidad de acceso a pie y determinados factores extra que influyan en la forma de llegar al lugar. Todo ello plasmado en un mapa interactivo en el cual se señalen las rutas disponibles mediante el marcado del mapa en cuestión.
?a wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
A diffraction grating with 200 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At what angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 617.3 nm
Answer
123.5 x 10 ^-3 radian
Explanation:
Given the Width of slit a = 1 x 10⁻³ / 200
a = 5x 10⁻⁶ m .
angle at which first order peak is formed
= λ / a (where λ is wavelength and a is width of slit)
given λ = 617.3 x 10⁻⁹ m
a = 5 x 10⁻⁶
θ = 617.3 x 10⁻⁹ / 5 x 10⁻⁶
= 123.5x 10⁻³ radian .
first order peak is formed at an angle of 123.5 x 10⁻³ radian .
Explanation:
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answer:I don’t know
Explanation:
Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?
Answer:
tmin= lambda/2
Explanation:
See attached file pls
At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m
Answer:
a) The value of g at such location is:
[tex]g=9.8005171\,\frac{m}{s^2}[/tex]
b) the period of the pendulum with the length is 0.970 m is:
[tex]T=1.9767 sec[/tex]
Explanation:
Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:
[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]
a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:
[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]
b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:
A 1.5 V battery is connected to a 1000 ohm resistor and a 500 ohm resistor in series. The voltage across the 1000 ohm resistor is _____ V.
Answer:
1 volt and 0.5 voltExplanation:
Given data
voltage supplied Vs= 1.5 volts
resistance R1= 1000 ohms
resistance R2= 500 ohms
The total resistance is
Rt= 1000+ 500
Rt= 1500 ohms
The current I is given as
[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]
Voltage across R1
[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]
Voltage across R2
[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]
In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.
(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
For more information on potential difference visit
https://brainly.com/question/14883923
A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?
Answer:
m = 22,877 kg / s
Explanation:
Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat
The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%
Q = 300 60% / 40%
Q = 450 MW
having the amount of heat generated we can use the calorimeter equation,
Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]
m = Q / c_{e} (T_{f} - T_{o})
let's use the maximum temperature change allowed
(T_{f} - T_{o}) = 5
the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water
power and energy are related
W = Q / t
Q = W t
let's calculate
m = 450 10⁶ / (3934 5)
m = 22,877 kg / s
In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainland.
Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?
Answer:
a
[tex]Q = 5.34 *10^{19} \ J[/tex]
b
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
Explanation:
From the question we are told that
The area of Manhattan is [tex]a_k = 87.46 *10^{6} \ m^2[/tex]
The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]
The thickness is [tex]t = 500 \ m \\[/tex]
Generally the volume of the ice is mathematically represented is
[tex]V = a_i * t[/tex]
substituting value
[tex]V = 500 * 3.498*10^{8}[/tex]
[tex]V = 1.75 *10^{11}\ m^3[/tex]
Generally the mass of the ice is
[tex]m_i = \rho_i * V[/tex]
Here [tex]\rho_i[/tex] is the density of ice the value is [tex]\rho _i = 916.7 \ kg/m^3[/tex]
=> [tex]m_i = 916.7 * 1.75*10^{11}[/tex]
=> [tex]m_i = 1.60 *10^{14} \ kg[/tex]
Generally the energy needed for the ice to melt is mathematically represented as
[tex]Q = m _i * H_f[/tex]
Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is [tex]H_f = 3.33*10^{5} \ J/kg[/tex]
=> [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]
=> [tex]Q = 5.34 *10^{19} \ J[/tex]
Considering part b
We are told that the annual energy consumption is [tex]G = 1.2*10^{20 } \ J / year[/tex]
So the time taken to melt the ice is
[tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]
[tex]T = 0.445 \ years[/tex]
converting to days
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct
Explanation:
the missing figure in the Question has been put in the attachment.
Then from the figure we can observe that
the center of the sphere is positive, therefore, negative charge will be induced at A.
As B is grounded there will not be any charge on B
Hence the answer is A is negative and B is charge less.
The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 1/2 in).
If we treat the spring assembly as a single spring, what is the approximate spring constant?
k= ____________
Answer:
The approximate spring constant is [tex]k = 55533.33 \ N/m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 68 \ kg[/tex]
The dip of the car is [tex]x = 1.2 \ cm = 0.012 \ m[/tex]
Generally according to hooks law
[tex]F = k * x[/tex]
here the force F is the weight of the person which is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]m * g = k * x[/tex]
=> [tex]k = \frac{m * g }{x }[/tex]
=> [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]
=> [tex]k = 55533.33 \ N/m[/tex]
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
Two narrow slits are illuminated by a laser with a wavelength of 593 nm. The interference pattern on a screen located x = 4.80 m away shows that the fourth-order bright fringe is located y = 8.20 cm away from the central bright fringe. Calculate the distance between the two slits.
Answer:
The distance is [tex]d = 1.39 *10^{-4} \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 593 \ nm = 593 *10^{-9} \ m[/tex]
The distance of the screen is x = 4.80 m
The location of the fourth order bright fringe is y = 8.20 cm = 0.082 m
The order of the fringe is n = 4
Generally the position of a fringe with respect to the central fringe is mathematically represented as
[tex]y = \frac{ n * x * \lambda }{d}[/tex]
Where d is the distance between the slits, so making d the subject
[tex]d = \frac{\lambda * x * n }{ y }[/tex]
substituting values
[tex]d = \frac{ 593 *10^{-9} * 4.80 * 4 }{ 0.082 }[/tex]
[tex]d = 1.39 *10^{-4} \ m[/tex]
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards
Answer:
His angular velocity will increase.
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = [tex]I[/tex]'ω'
where
[tex]I[/tex]' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = [tex]mr'^{2}[/tex]
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].
From
[tex]I[/tex]'ω' = [tex]I[/tex]ω
since [tex]I[/tex] is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?
Answer: 56°
Explanation:
Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.
The formula is;
[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]
[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]
= 56.30993247
= 56°
Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion
A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.
Answer:
B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.
C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.
E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.
#FreeMelvin
Bob and Lily are riding on a merry-go-round. Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform. When the merry-go-round is rotating at a constant angular speed, Bob's angular speed is:_____.
a. the same as Lily's.
b. larger than Lily's.
c. exactly half as much as Lily's.
d. exactly twice as much as Lily's.
e. smaller than Lily's.
Answer: the same as Lily's
Explanation:
Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.
Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.
Based on the above analysis, Bob's angular speed will be thesame as that of Lily.
A bicycle has wheels that are 60 cm in diameter. What is the angular speed of these wheels when it is moving at 4.0 m/s
Answer:
13.33 rad/s
Explanation:
Applying,
v = ωr......................... Equation 1
Where v = linear speed, ω = angular speed and r = radius.
Note that,
r = d/2................. Equation 2
Where d = diameter of the wheel.
Substitute equation 2 into equation 1
v = ωd/2............... Equation 3
make ω the subject of the equation
ω = 2v/d................ Equation 4
Given: v = 4 m/s, d = 60 cm = 0.6 m
Substitute these values into equation 4
ω = 2(4)/0.6
ω = 13.33 rad/s
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?
Answer:
1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216
Explanation:
In the double-slit interference phenomenon it is explained for constructive interference by the equation
d sin θ = m λ
1) the first order maximum occurs for m = 1
sin θ = λ / d
θ = sin⁻¹ λ / d
let's reduce the magnitudes to the SI system
λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm
d = 0.440 mm = 0.440 10⁻³ m ³
let's calculate
θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)
θ = sin⁻¹ (1.18 10⁻³)
θ = 0.00118 rad
2) the second order maximum occurs for m = 2
θ = sin⁻¹ (m λ / d)
θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)
θ = 0.00236 rad
3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains
I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )
where the function sinc = sin x / x
and b is the width of the slits
we caption the values
x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)
x = 2.21
I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²
remember angles are in radians
I / I₀ = cos² (3.0945) [0.363] 2
I / I₀ = 0.9978 0.1318
I / I₀ = 0.1738
4) the maximum second intensity is
I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)
x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)
x = 4.41
I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²
I / Io = cos² 6.273 0.216
I / Io = 0.216
.
what conventions are used in SI to indicate units
Answer:
Conventions used in SI to indicate units are as follows:
Only singular form of units are used. for example: use kg and not kgs.Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.Units named after a scientist should be written in small letters. for example: newton, henry.Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°kJust wondering if I did this right
Yeah
All they are all correct
There is a river in front of you that flows due South at 3.0m/s. You launch a toy boat across the river with the front of the boat pointed due East. When you tested the boat on a still pond, the boat moved at 4.0m/s. Now as it moves to the opposite bank, it travels at some speed relative to you, sitting in your chair. What is this speed
Answer:
5.0 m/sExplanation:
If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.
Let Vr be the relative speed. According to the theorem;
[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]
Hence this relative speed is 5.0 m/s
a ball is kicked on level ground with a speed of 30 m/s at angle of 40 degrees above horizontal g Find the minimum velocity of the ball during its flight
Answer:
The minimum velocity of the ball during its flight is 22.98 m/s.
Explanation:
The velocity of the ball v = 30 m/s
The angle it makes with the horizontal ∅ = 40°
The minimum velocity of the ball during flight will be the horizontal axis component of the velocity, as acceleration is zero on this axis.
[tex]V_{x}[/tex] = v cos ∅
[tex]V_{x}[/tex] = 30 cos 40°
[tex]V_{x}[/tex] = 30 x 0.766 = 22.98 m/s
