Answer:
The acceleration in the block is 2.1 m/s²
Explanation:
Given that,
Mass = 50 kg
Angle = 54°
Force = 40 N
Coefficient of friction = 0.33
We need to calculate the acceleration in the block
Using balance equation
[tex]F_{net}=F_{f}-F\cos\theta[/tex]
[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]
[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]
[tex]a=2.1\ m/s^2[/tex]
Hence, The acceleration in the block is 2.1 m/s²
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C
Answer:
Objects that are closer together have a stronger force of gravity between them.
Explanation:
For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.
You are performing an experiment that requires the highest-possible magnetic energy density in the interior of a very long current-carrying solenoid. Which of the following adjustments increases the energy density?a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed. b. Increasing the number of turns per unit length on the solenold. c. Increasing the cross-sectional area of the solenoid. d. None of these. e. Increasing the current in the solenoid.
Answer:
The correct choice is B & E.
Explanation:
A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.
Cheers!
You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?
Answer:
In parallel
Explanation:
Ctotal = C1 + C2 + ... + Cn
"If a beam of monochromatic light is passed though a slit of width 15 μm and the second order dark fringe of the diffraction pattern is at an angle of 5.2o from the central axis, what is the wavelength of the light?"
Answer:
λ= 5.4379 10⁻⁷ m = 543.79 nm
Explanation:
The phenomenon of diffraction is described by the expression for destructive diffraction is
a sin θ = (m + 1/2) λ
λ = a sin θ / (m + 1/2)
let's reduce the magnitudes to the SI system
a = 15 um = 15 10⁻⁶ m
m = 2
θ = 5.2º
Let's calculate
λ = 15 10⁻⁶ sin 5.2 / (2 +1/2)
λ = 5.4379 10⁻⁷ m
Let's reduce to nm
λ= 5.4379 10⁻⁷ m = 543.79 nm
You shine unpolarized light with intensity 52.0 W/m2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 15.0 W/m2. Find the intensity of the light that emerges from the first polarizer.
Answer:
The intensity of light from the first polarizer is [tex]I_1 = 26 W/m^2[/tex]
Explanation:
The intensity of the unpolarized light is [tex]I_o = 52.0 \ W/m^2[/tex]
Generally the intensity of light that emerges from the first polarized light is
[tex]I_1 = \frac{I_o}{2 }[/tex]
substituting values
[tex]I_1 = \frac{52. 0}{2 }[/tex]
[tex]I_1 = 26 W/m^2[/tex]
A charged particle moving through a magnetic field at right angles to the field with a speed of 25.7 m/s experiences a magnetic force of 2.98 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 4.64 m/s at an angle of 29.2° relative to the magnetic field.
Answer:
The magnetic force would be:
[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]
Explanation:
Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:
F = q v x B
(where the bold represents vectors)
the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:
[tex]F=q\,v\,B\,sin(\theta)[/tex]
Therefore replacing the known quantities for the first case:
[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]
Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:
[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]
A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any objects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3
Answer:
v = - 14.08 m / s
Explanation:
The definition of center of mass is
[tex]x_{cm}[/tex] = 1 /M ∑sun [tex]x_{i} m_{i}[/tex]
where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.
The velocity of the center of mass can be found by deriving this expression with respect to time
[tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi
let's find the total mass
M = m₁ + m₂ + m₃ + m₄
M = 1.45 + 2.81 +3.89 + 5.03
m = 13.18 kg
let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0
0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)
v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄
let's substitute the given values
v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03
They ask us for the calculations for a time t = 2.83 s
v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03
v = - 14.08 m / s
The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
The given parameters;
[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]
The velocity of the center mass of the particles is calculated as;
[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]
The velocity of particle 1 at time, t = 2.83 s;
[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]
The velocity of particle 2 at time, t = 2.83 s;
[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]
The velocity of particle 3 at time, t = 2.83 s;
[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]
The velocity of the particle 4 at time, t = 2.83 s;
[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]
Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
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How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 587 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 11th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source?
Answer:
The wavelength is [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the first light is [tex]\lambda _ 1 = 587 \ nm[/tex]
The order of the first light that is being considered is [tex]m_1 = 10[/tex]
The order of the second light that is being considered is [tex]m_2 = 11[/tex]
Generally the distance between the fringes for the first light is mathematically represented as
[tex]y_1 = \frac{ m_1 * \lambda_1 * D}{d}[/tex]
Here D is the distance from the screen
and d is the distance of separation of the slit.
For the second light the distance between the fringes is mathematically represented as
[tex]y_2 = \frac{ m_2 * \lambda_2 * D}{d}[/tex]
Now given that both of the light are passed through the same double slit
[tex]\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1[/tex]
=> [tex]\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1[/tex]
=> [tex]\lambda_2 = \frac{m_1 * \lambda_1}{m_2}[/tex]
=> [tex]\lambda_2 = \frac{10 * 587 *10^{-9}}{11}[/tex]
=> [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars
Answer:
Star A has a higher surface temperature than star B.
Explanation:
The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:
[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)
Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.
A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.
Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The volume of the block is 306 cm³
The density of the block is 7.07 g/cm³
Explanation:
Given;
weight of block in air, [tex]W_a[/tex] = 21.2 N
Weight of block in water, [tex]W_w[/tex] = 18.2 N
Mass of the block in air;
[tex]W_a = mg[/tex]
21.2 = m x 9.8
m = 21.2 / 9.8
m = 2.163 kg
mass of the block in water;
[tex]W_w = mg[/tex]
18.2 = m x 9.8
m = 18.2 / 9.8
m = 1.857 kg
Apply Archimedes principle
Mass of object in air - mass of object in water = density of water x volume of object
2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block
0.306 kg = 1000 kg/m³ x Volume of block
Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]
Volume of the block = 3.06 x 10⁻⁴ m³
Volume of the block = 306 cm³
Determine the density of the block
[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]
An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pipe? The speed of sound is 343 m/s in air.
Answer:
The length is [tex]l = 8.6 \ m[/tex]
Explanation:
From the question we are told that
The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] , [tex]f_2 = 240 \ Hz[/tex]
The speed of sound in the air is [tex]v_s = 343 \ m/s[/tex]
Generally the frequency of a given harmonic is mathematically represented as
[tex]f_n = \frac{n v }{2l}[/tex]
Here n defines the position of the harmonics
Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as
[tex]220 = \frac{n v}{2l}[/tex]
and
[tex]240 = \frac{(n+1) v}{2l}[/tex]
So
[tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]
=> [tex]\frac{v}{2l} = 20[/tex]
=> [tex]l = 8.6 \ m[/tex]
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)
Answer:
(a)106.4C
b)0.5676mm
Explanation:
(a)To get the charge that have passed through the starter then The current will be multiplied by the duration
I= current
t= time taken
Q= required charge
Q= I*t = 140*0.760 = 106.C
(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)
diameter of the conductor is 4.20 mm
But Radius= diameter/2= 4.20/2=
The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then
radius of the conductor is 0.0021m.
We can now calculate the area of the conductor which is
A = π*r^2
= π*(0.0021)^2 = 13.85*10^-6 m^2
We can proceed to calculate the current density below
J = 140/13.85*10^-6 = 10108303A/m
According to the listed reference:
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s
Therefore , the distance traveled is:
x = v*t = 0.7468 * 0.760 = 0.5676mm
(a) The charge passes through the starter motor is 106.4C.
(b) An electron travel along the wire while the starter motor is on 0.5676mm.
ElectronAnswer (a)
I= current
t= time taken
Q= required charge
Q= I*t
Q= 140*0.760
Q= 106.C
Answer (b)
The n electron travel along the wire while the starter motor is on:
Diameter of the conductor is 4.20 mm
Radius= diameter/2= 4.20/2
Radius =2.1mm
Radius of the conductor is 0.0021m.
A = π*r^2
A= π*(0.0021)^2
A= 13.85*10^-6 m^2
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )
Vd =0.0007468m/s
Vd =0 .7468 mm/s
The distance traveled is:
x = v*t
x= 0.7468 * 0.760
x = 0.5676mm
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Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal contact and the combination is thermally insulated. The final temperature of both objects is
Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]
But heat capacity of object B is twice that of object A
[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]
Therefore, the final temperature of both objects is 400 K.
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?
Assuming constant speeds, the P-wave covers a distance d in time t such that
9000 m/s = d/(60 t)
while the S-wave covers the same distance after 1 more minute so that
5000 m/s = d/(60(t + 1))
Now,
d = 540,000 t
d = 300,000(t + 1) = 300,000 t + 300,000
Solve for t in the first equation and substitute it into the second equation, then solve for d :
t = d/540,000
d = 300,000/540,000 d + 300,000
4/9 d = 300,000
d = 675,000
So the earthquake center is 675,000 m away from the seismic station.
In a certain process a gas ends in its original thermodynamic state. Of the following, which is possible as the net result of the process?
A. It is adiabatic and the gas does 50 J of work.
B. The gas does no work but absorbs so J of energy as heat.
C. The gas does no work but rejects 50 J of energy as heat.
D. The gas rejects 50 J of energy as heat and does 501 of work.
E. The gas absorbs 50 of energy as heat and does 50」ot work.
Answer:
E. The gas absorbs 50 of energy as heat and does 50」ot work
Explanation:
This is following the law of thermodynamics that energy is neither created nor destroyed
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in the primary if the secondary load resistance is 250 W?
Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, [tex]N_P[/tex] = 50 turns
number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns
the secondary load resistance, [tex]R_S[/tex] = 250 Ω
Determine the turns ratio;
[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]
Now, determine the reflected resistance in the primary winding;
[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]
Therefore, the reflected resistance in the primary winding is 6250 Ω
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the collision. What is the wavelength of the scattered photon?
0.6764*10^-10m
Explanation:
Using E= hc/wavelength
(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV
So subtract the calculated energy from the given energy of scattered photons
9.11-0.75=18.36 keV
To find wavelength
Wavelength= hc/ E
[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m
What is the displacement of the object after 3 seconds?
Answer:
3 meters
Explanation:
A mass m = 0.7 kg is released from rest at the origin 0. The mass falls under the influence of gravity. When the mass reaches point A, it is a distance x below the origin 0; when the mass reaches point B it is a distance of 3 x below the origin 0. What is vB/vA?
Answer:
[tex]v_B/v_A=\sqrt{3}[/tex]
Explanation:
Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:
[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]
Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:
[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]
Let's work in a similar way with the two different positions at those different times, and for which we have some information;
[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]
Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:
[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Answer:
A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.
Equal currents of magnitude I travel into the page in wire M and out of the page in wire N. The direction of the magnetic field at point P which is at the same distance from both wires is
Answer:
The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
Distinguish between physical and chemical changes. Include examples in your explanations.
Answer:
Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.
Instead the physical changes implies that if you can return to the same substance through a reverse process.
Explanation:
A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.
A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.
A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.
• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.
• In a chemical change always a chemical reaction takes place.
• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.
• In a physical change no new chemical species form.
• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.
• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.
Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.
To know more about:
https://brainly.com/question/7279398
A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?
Answer:
3.5 m/s
Explanation:
There are 3 forces on the cyclist:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing left.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
∑F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
a = -(10 m/s²)(0.60)
a = -6 m/s²
Velocity reached at end of 11 m:
v² = v₀² + 2aΔx
v² = (12 m/s)² + 2 (-6 m/s²) (11 m)
v = √12 m/s
v ≈ 3.5 m/s
A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples does it produce when a 3.25 m focal length eyepiece is used? ✕
Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
