Answer:
0.0665 days
Explanation:
We are given;
The mean distance from the Earth's center to the moon;a1 = 385000 km
The mean distance from the Earth's center to the space craft;a2 = 6965 km
Formula for kepplers third law is;
T² = 4π²a³/GM
However, the proportion of both distances would be;
(T1)²/(T2)² = (a1)³/(a2)³
Where;
T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days
T2 is the period of the space craft orbit.
Making T2 the subject, we have;
T2 = √((T1)²×(a2)³)/(a1)³)
Thus, plugging in the relevant values;
T2 = √(27.322² × 6965³)/(385000)³
T2 = 0.0665 days
An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.
Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3
The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
The formula of the pressure of the static fluid
P1 = P2+ρgh
Where,
P1 - pressure at depth 'h'
P2 - atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] = 1670900 Pa
h - Depth = 160m
ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]
g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]
The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]
[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]
For an ideal gas,
PV =nRT
[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]
[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]
So,
[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]
Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
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Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −4.70 N, respectively. For the resultant electric force, find the following.
(a) the x-component N
(b) the y-component N
(c) the magnitude of the resultant electric force N
(d) the direction of the resultant electric force, measured counterclockwise from the positive x-axis ° counterclockwise from the +x-axis
Answer:
Explanation:
Given two vectors as follows
E₁ = 13.5 i -12 j
E₂ = -7.4 i - 4.7 j
Resultant E = E₁ + E₂
= 13.5 i -12 j -7.4 i - 4.7 j
E = 6.1 i - 16.7 j
a ) X component of resultant = 6.1 N
b ) y component of resultant = -16.7 N
Magnitude of resultant = √ ( 6.1² + 16.7² )
= 17.75 N
d ) If θ be the required angle
tanθ = 16.7 / 6.1 = 2.73
θ = 70° .
counterclockwise = 360 - 70 = 290°
By working with the vector forces, we will get:
a) The x-component is 1.5 Nb) The y-component is -12.2 Nc) The magnitude is 12.9 Nd) The direction is 277.01°.How to find the resultant force?
Remember that we can directly add vector forces, so if our two forces are:
F₁ = <13.5 N, -7.5 N>
F₂ = < -12 N, -4.70 N>
Then the resultant force is:
F = F₁ + F₂ = <13.5 N + (-12 N), -7.5 N + ( -4.70 N) >
F = < 1.5 N, -12.2 N>
so we have:
a) The x-component is 1.5 N
b) The y-component is -12.2 N
c) The magnitude will be:
|F| = √( (1.5 N)^2 + (-12.2 N)^2) = 12.29 N
d) The direction of a vector <x, y> measured counterclockwise from the positive x-axis is given by:
θ = Atan(y/x)
Where Atan is the inverse tangent function, then here we have:
θ = Atan(-12.2 N/1.5 N) = 277.01°
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A copper telephone wire has essentially no sag between poles 36.0 m apart on a winter day when the temperature is −20.0°C. How much longer is the wire on a summer day when the temperature is 34.0°C?
Answer:
The extension is [tex]\Delta L = 0.033 \ m[/tex]
Explanation:
From the question we are told that
The length of the wire on a winter day is [tex]L_w = 36.0 \ m[/tex]
The temperature on the winter day is [tex]T_w = -20.0 ^o C[/tex]
The temperature on a summer day is [tex]T_s = 34.0 ^0 C[/tex]
The the extension of the wire on a summer day is mathematically represented as
[tex]\Delta L = \alpha L_w [T_s - T_w][/tex]
Where
[tex]\alpha[/tex] is the coefficient of linear expansion of copper with a values [tex]\alpha = 17 *10^{-6}[/tex]
substituting value
[tex]\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]][/tex]
[tex]\Delta L = 0.033 \ m[/tex]
38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which launches the box into the air. What is the maximum height reached by the box assuming negligible frictional forces
Answer:
Explanation:
Potential energy stored in the spring = 25 J
This energy is converted into gravitational potential energy . If h be the height attained
gravitational potential energy = mgh
mgh = 25
.15 x 9.8 x h = 25
h = 17 m
Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.
Answer:
5.49 x 10^17 m is the distance between the sun-like star to the earth
Explanation:
Radiation intensity on Earth = 1.0 x 10^-10 W/m^2
Power of radiation of the star = 3.8 x 10^26 W
Recall that the intensity of radiation is given as
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where
[tex]I[/tex] = intensity of radiation
P = power of radiation
A is the area through which the radiation spreads out in all three dimensional direction.
A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2
This area is spread out in the form of a sphere of area
A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]
3.8 x 10^36 = 12.568[tex]r^{2}[/tex]
[tex]r^{2}[/tex] = (3.8 x 10^36)/12.568 = 3.02 x 10^35
r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m this is the distance of the star to the Earth
two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will have the electric field zero?
Answer:
On that line segment between the two charges, at approximately [tex]0.7\; \rm m[/tex] away from the smaller charge (the one with a magnitude of [tex]5 \times 10^{-19}\; \rm C[/tex],) and approximately [tex]1.3\; \rm m[/tex] from the larger charge (the one with a magnitude of [tex]20 \times 10^{-19}\; \rm C[/tex].)
Explanation:
Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.
Let [tex]k[/tex] denote the Coulomb constant, and let [tex]q[/tex] denote the size of a point charge. At a distance of [tex]r[/tex] away from the charge, the electric field due to this point charge will be:
[tex]\displaystyle E = \frac{k\, q}{r^2}[/tex].
At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.
Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.
When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.
On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.
Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.
Let [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] and [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex]. Assume that the electric field is zero at [tex]r[/tex] meters to the right of the [tex]5\times 10^{-19}\; \rm C[/tex] point charge. That would be [tex](2 - r)[/tex] meters to the left of the [tex]20 \times 10^{-19}\; \rm C[/tex] point charge. (Since this point should be between the two point charges, [tex]0 < r < 2[/tex].)
The electric field due to [tex]q_1 = 5\times 10^{-19}\; \rm C[/tex] would have a magnitude of:
[tex]\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}[/tex].
The electric field due to [tex]q_2 = 20 \times 10^{-19}\; \rm C[/tex] would have a magnitude of:
[tex]\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}[/tex].
Note that at all point in this section, the two electric fields [tex]E_1[/tex] and [tex]E_2[/tex] will be acting in opposite directions. At the point where the two electric fields balance each other precisely, [tex]| E_1 | = | E_2 |[/tex]. That's where the actual electric field is zero.
[tex]| E_1 | = | E_2 |[/tex] means that [tex]\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}[/tex].
Simplify this expression and solve for [tex]r[/tex]:
[tex]\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0[/tex].
[tex]\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0[/tex].
Either [tex]r = -2[/tex] or [tex]\displaystyle r = \frac{2}{3}\approx 0.67[/tex] will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, [tex]0 < r < 2[/tex]. Therefore, [tex](-2)[/tex] isn't a valid value for [tex]r[/tex] in this context.
As a result, the electric field is zero at the point approximately [tex]0.67\; \rm m[/tex] away the [tex]5\times 10^{-19}\; \rm C[/tex] charge, and approximately [tex]2 - 0.67 \approx 1.3\; \rm m[/tex] away from the [tex]20 \times 10^{-19}\; \rm C[/tex] charge.
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. The amount of charge that passes per unit time is called
The amount of charge that passes per unit time is called electric current .
Current has dimensions of [Charge] / [Time] .
It's measured and described in units of ' Ampere ' .
1 Ampere means 1 Coulomb of charge passing a point every second.
23.15. Can an object carry a charge of 2.0 10-19 C?
Answer:
Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)
Where the charge of a single proton is:
C = 1.6x10^-19 C
Now, you need to remember that when we are working with charges, we are working with discrete math:
What means that?
If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)
So the consecutive charge will be:
C = 2*1.6x10^-19 C = 3.2x10^-19 C.
So, because we are working in discrete math, we can not have any object that has charge between 1.6x10^-19 C and 3.2x10^-19 C.
Particularly, 2.0x10^-19 C is in that range, so we can conclude that:
No, an object can not carry a charge of 2.0x10^-19 C.
Which two types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light?
radio waves and infrared light
visible light and X-rays
microwaves and gamma rays
gamma rays and X-rays
The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
WHAT ARE ELECTROMAGNETIC WAVES?Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:
Radio wavesInfraredUltravioletVisible lightX-raysGamma raysmicrowaveEach electromagnetic wave have a specific frequency and wavelength.
However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.
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Answer:
gamma rays and X-rays
Explanation:
d on edge I got 100%
To shoot a swimming fish when an intense light beam from a laser gun you should aim
Answer
aim directly at the image
Explanation
the light from the laser beam will also bend when it hits the air water interface , so aim directly at the fish
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
Two very long parallel wires are a distance d apart and carry equal currents in opposite directions. The locations where the net magnetic field due to these currents is equal to zero are
Answer:
Its not zero anywhere
Explanation:
The magnetic field B at a distance r due to a long conductor carrying current I is given as
B= μol/2pi r
so the net magnetic field due to the current is not zero anywhere
A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter of 50 cm. The speed of the top of the wheel is
Answer:
The speed of the top of the wheel is twice the speed of the car.
That is: 72 m/s
Explanation:
To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,
plus the velocity due to the translation of the center of mass (v = 36 m/s).
The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]
Then the addition of these two velocities equals:
[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]
You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.
a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.
Answer:
a) d sin θ = m λ₀ / n
b) d sin θ = (m + ½) λ₀ / n
c) d = 2,439 10⁻⁷ m
Explanation:
For the interference these rays of light we must take as for some aspects,
* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2
* when the ray penetrates the lens the donut length changes by the refractive index
λ = λ₀ / n
now let's write the destructive interference equation for these lightning bolts
d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n
d sin θ = m λ₀ / n
b) now nc> np
in this case there is no phase change in the reflected ray and the equation for destructive interference remains
d sin θ = (m + ½) λ₀ / n
c) select the value of nc = 2.05 of the ZnO
we calculate the thickness of the film (d)
d = m λ / (n sin 90)
in this type of interference the observation is normal, that is, the angle is 90º)
d = 1 500 10-9 / (2.05 1)
d = 2,439 10⁻⁷ m
d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference
A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose separation is 0.170 nm. Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the first order.
Answer:
N=3176.5rulling
Explanation:
We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with
wavelengths whose mean is 540 nm
Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)
Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then
Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)
the formular below can be used to can be used to calculate our minimum number of lines
N= λ /(m Δλ)
Where N is number of fillings i.e number of lines
λ= wavelength
Δλ= difference in wavelength
m=1
Then if we substitute the values we have
,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]
N =3176.5rulling
Therefore, minimum number of lines = =3176.5rulling
A stellar object is emitting radiation at 3.55 mm. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?
Answer:
E = 6.45 x 10⁻¹¹ J
Explanation:
First we need to find total number of photons detected in 1 hour. Therefore,
No. of Photons = n = (3.2 x 10⁸ photons/s)(1 h)(3600 s/1 h)
n = 11.52 x 10¹¹ photons
Now, the energy of these photons can be given by the formula:
E = nhc/λ
where,
E = Total Energy of the Photons = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of radiation = 3.55 mm = 3.55 x 10⁻³ m
Therefore,
E = (11.52 x 10¹¹)(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.55 x 10⁻³ m)
E = 6.45 x 10⁻¹¹ J
explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage
Answer:
polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.
the cup helps it become more insulated
An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.
Answer:
[tex]1.357\times 10^{-12}[/tex]
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]
[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]
= [tex]1.357\times 10^{-12}[/tex]
Therefore for computing the capacitance we simply applied the above formula.
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?
Answer:
The angle is [tex]\theta = 15.48^o[/tex]
Explanation:
From the question we are told that
The distance of the dartboard from the dart is [tex]d = 3.66 \ m[/tex]
The time taken is [tex]t = 0.455 \ s[/tex]
The horizontal component of the speed of the dart is mathematically represented as
[tex]u_x = ucos \theta[/tex]
where u is the the velocity at dart is lunched
so
[tex]distance = velocity \ in \ the\ x-direction * time[/tex]
substituting values
[tex]3.66 = ucos \theta * (0.455)[/tex]
=> [tex]ucos \theta = 8.04 \ m/s[/tex]
From projectile kinematics the time taken by the dart can be mathematically represented as
[tex]t = \frac{2usin \theta }{g}[/tex]
=> [tex]usin \theta = \frac{g * t}{2 }[/tex]
[tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]
[tex]usin \theta = 2.23[/tex]
=> [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]
[tex]\theta = tan^{-1} [0.277][/tex]
[tex]\theta = 15.48^o[/tex]
What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L
Answer:
2.78 cm³/s
Explanation:
From the question,
Q = v/A'.................... Equation 1
Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.
Given: v = 100 km/h, A' = 10 km/L
Substitute this value into equation 1
Q = 100/10
Q = 10 L/h.
Now, we convert L/h to cm³/s.
Since,
1 L = 1000 cm³, and
1 h = 3600 s
Therefore,
Q = 10(1000/3600) cm³/s
Q = 2.78 cm³/s
A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the electric field established between the plates
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
The magnitude of the electric field developed that lies between the plates should be considered as the 2,000 V/m.
Calculation of the Electric Field:Since
The distance that lies between 2 parallel plates should be d = 0.04 m
The potential that lies between first and second plate should be like +40V and -40V
So, The magnitude of the electric field should be
E = ΔV / d
here,
ΔV represents the change in potential that lies between 2 parallel plates.
d represents the distance between the plates.
So,
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
And,
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m.
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What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
action and reaction are equal in magnitude and opposite in direction.Then why do they not balance each other.
Explanation:
Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
A customs inspector was suspecting that some of the 12 plastic spheres, which were shipped out of the country, had something in them. Each sphere weighted the same and had hard walls everywhere. Inspector thought that it was possible to hide something inside each sphere. He was correct, and was able to use a simple experiment in determining which sphere had diamonds inside. How did he do it?
Answer:
use a hammer to hit it
Explanation:
if u hit it u will be able to hear the shattered noise
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
9.09 × 10³ capacitors must be connected in parallel.
How to calculate the number of capacitors connected in parallel?
Given C = 1.00μF = 1 × 10⁻⁶ F
q = 1.00 C
V = 110 V
The equivalent capacitance is given by
Ceq = q/V
where q = total charge on all the capacitors
V = potential difference
For N number of identical capacitors in parallel,
Ceq = NC
Therefore,
NC = q/V
N = q/VC
Putting on the values in the above formula,
N = 1/ (110)(1 × 10⁻⁶)
= 1 / 110 × 10⁻⁶
= 9.09 × 10³
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