Answer:
a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.
Explanation:
a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.
By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
Where:
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.
The equation is now simplified and the coefficient of friction is consequently cleared:
[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]
[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]
The final height of the crate is:
[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]
[tex]y_{2} = 0.8\,m[/tex]
If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:
[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\mu_{k} \approx 0.548[/tex]
Then, the magnitude of the friction force is:
[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:
[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]f = 55.851\,N[/tex]
The magnitude of the force of friction is 55.851 newtons.
b) The energy equation of the situation is:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]
Now, the final speed is cleared:
[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]
[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]
Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:
[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]
The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.
A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?
Answer:
A. 0.0374C
B. 0.012F
C. 18 ohms
Explanation:
See attached file
Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
Image is missing, so i have attached it
Answer:
19.04 × 10⁻⁴ T in the +x direction
Explanation:
We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.
Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)
We are given;
I = 2.4 A
R = 5 cm = 0.05 m
μ₀ is a constant = 4π × 10⁻⁷ H/m
B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)
B = 9.6 × 10⁻⁴ T
To get the direction of the field from each wire, we will use Flemings right hand rule.
From the diagram attached:
We can say the field at P from the top wire will point up/right
Also, the field at P from the bottom wire will point down/right
Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.
If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.
Let the angle be θ
Thus;
sin(θ) = (1.3/2)/5
θ = sin⁻¹(0.13) = 7.47⁰
The x component of each field would be:
9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T
Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction
The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".
Magnetic fieldThe region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.
According to the question,
Current, I = 2.4 A
Radius, R = 5 cm or,
= 0.05 m
Constant, μ₀ = 4π × 10⁻⁷ H/m
We know the relation,
The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]
By substituting the values in the above relation, we get
= [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]
= 9.6 × 10⁻⁴ T
Thus the above answer is appropriate.
Find out more information about magnetic field here:
https://brainly.com/question/14411049
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.
As a skydiver falls, his potential energy ___ and his kinetic energy __
increases,increases
increases,decreases
decreases,increases
decreases, decreases
Answer:
Hey there!
PE=mgh, so as height decreases, so does the potential energy.
KE=mv^2, so as velocity increases, kinetic energy increases.
Thus, the correct answer would be Decreases, Increases.
Let me know if this helps :)
6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?
Answer:
t = t₀ / 2
Explanation:
In this exercise we must use Newton's second law
F = m a
a = F / m
now we can use kinematics
as in object part of rest (v₀ = 0)
v =a t₀
t₀ = v / a
these results are with the first experiment
now repeat the experiment, but F = 2F₀
a = 2F₀ / m = 2 a₀
v = 2 a₀ t
t = v / 2a₀
t = t₀ / 2
The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].
Given the following data:
Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = VTo find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]F = \frac{M(V-U)}{t}[/tex]
Where:
F is the force.V is the final velocity.U is the initial velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]
When the experiment is repeated, the magnitude of the force is doubled:
[tex]F = 2F[/tex]
Now, we can find the time interval that is required to reach the same final speed (V):
[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]
Substituting the value of F, we have:
[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]
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1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Answer:
y = 1/2at^2
we could also write it as-
y = (at^2)/2
2y = at^2
2y/a = t^2
√2y/a = t
hope it helps
a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density
Answer:
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
Explanation:
given data
frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz
we consider here intensity I = 0.2 W/m²
solution
we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s
and take energy density is E
so here
E × C = I
E = [tex]\frac{I}{C}[/tex] ................1
here C = 3 × [tex]10^{8}[/tex] m/s
so photon density is
photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex] ...............2
photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
A rock weighing 20 N (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. What is the tension in the cord?
Answer:
The tension in the cord provides by centripetal force
T = Fc
= mv^2/r
= 2kg ( 6)^2/2
=36 N
please i dont understand
13. A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string. What is the wavelength of the second wave?
Answer:
It will be half that if the first wave
Explanation:
Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it
A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2
Given.
force = 16.88 N is
radius = 0.340m
an angular acceleration = 1.20rad/s^2
the formula for torque is
F*r = I*a
where I is moment of inertia
16.88*.34 = I*1.2
I = 4.78Kg-m^2
so rotational inertia I = 4.78Kg-m^2
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
Find more information: brainly.com/question/23460034
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?
Answer:
Explanation:
Using the lens formula
1//f = 1/u+1/v
f is the focal length of the lens
u is the object distance
v is the image distance
For convex lens
The focal length of a convex lens is positive and the image distance can either be negative or positive.
Given f = 20cm and u = 10cm
1/v = 1/f - 1/u
1/v = 1/20-1/10
1/v = (1-2)/20
1/V = -1/20
v = -20/1
v = -20 cm
Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image
For concave lens
The focal length of a concave lens is negative and the image distance is negative.
Given f = -20cm and u = 10cm
1/v = 1/f - 1/u
1/v = -1/20-1/10
1/v = (-1-2)/20
1/V = -3/20
v = -20/3
v = -6.67 cm
Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1 cm across, and you estimate that the distance from the window shade to the wall is about 4 m.
Estimate:
a. The average wavelength of the sunlight (in nm)
b. The diameter of the pinhole (in mm).
Given that,
Central maximum = 1 cm
Distance from the window shade to the wall =4 m
We know that,
The visible range of the sun light is 400 nm to 700 nm.
(a). We need to calculate the average wavelength
Using formula of average wavelength
[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]
Put the value into the formula
[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]
[tex]\lambda_{avg}=550\ nm[/tex]
(b). We need to calculate the diameter of the pinhole
Using formula for diameter
[tex]w=\dfrac{2.44\lambda L}{D}[/tex]
[tex]D=\dfrac{2.44\lambda L}{w}[/tex]
Put the value into the formula
[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]
[tex]D=0.537\ mm[/tex]
Hence, (a). The average wavelength 550 nm.
(b). The diameter of the pinhole is 0.537 mm.
A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?
Answer:
The new time period is [tex]T_2 = 3.8 \ s[/tex]
Explanation:
From the question we are told that
The period of oscillation is [tex]T = 5 \ s[/tex]
The new length is [tex]l_2 = 0.76 \ m[/tex]
Let assume the original length was [tex]l_1 = 1 m[/tex]
Generally the time period is mathematically represented as
[tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]
Now I is the moment of inertia of the stick which is mathematically represented as
[tex]I = \frac{m * l^2 }{12 }[/tex]
So
[tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]
Looking at the above equation we see that
[tex]T \ \ \ \alpha \ \ \ l[/tex]
=> [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]
=> [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]
=> [tex]T_2 = 3.8 \ s[/tex]
Calculate the answers to the appropriate number of significant
12.21 x 9.19 =
how many stars are in our solar system?
Answer:
there are over 100 billion stars in our galaxy.
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
or, Do they branch of each others
Answer:
The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a wholeExplanation:
Answer:
quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.
Explanation:
Explain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.
Answer:
Answer in explanation
Explanation:
Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.
The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.
Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.
This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.
An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?
Answer:
1.6×10²⁰
Explanation:
An ampere is a Coulomb per second.
1 A = 1 C / s
The amount of charge after 5 seconds is:
5.0 A × 5 s = 25 C
The number of electrons is:
25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons
g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
Define the following, and give the letter which we will abbreviate them by:
Center of curvature:
Vertex:
Focal Point:
Radius of curvature:
Focal length:
Answer:
As in explanation.
Explanation:
A) Centre of Curvature: This is defined as the point in the center of the sphere from which the mirror was sliced. It is represented by the letter "C"
B) Vertex: It is defined as the point on the mirror's surface where the principal axis meets the mirror. It is represented by the letter A.
C) Focal Point: This is defined as the Midway point between the vertex and the center of curvature. It is represented by the letter "F"
D) Radius of Curvature: This is defined as the distance from the vertex to the center of curvature. It is represented by the letter "R"
E) Focal Length: This is defined as the distance from the mirror to the focal point. It's represented by the letter "f"
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
Learn more about RLC circuit:
https://brainly.com/question/372577?referrer=searchResults
An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]