Answer:
he will see the sticker because its behind a window bruh and thats a big daddy stack of greens
Explanation:
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.
Learn more about electromagnetic waves at: https://brainly.com/question/8553652
Answer:
gamma rays and X-rays
Explanation:
d on edge I got 100%
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
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
"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 wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
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
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]
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
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.
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
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.
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.
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°
If you want to learn more about vectors, you can read:
https://brainly.com/question/7802337
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
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
The maximum amount of pulling force a truck can apply when driving on
concrete is 10,560 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
Answer:
Explanation:
Let the weight of the truck be W . reaction force R = W
Maximum frictional force = μ R
= .8 x W
So for movement of truck
Pulling force = frictional force
10560 = .8W
W = 13200 N
weight of heaviest truck required = 13200 N .
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 find its approximate location at time t = 1.02.
Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
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
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]
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.
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
If the string is 7.6 m long, has a mass of 34 g , and is pulled taut with a tension of 15 N, how much time does it take for a wave to travel from one end of the string to the other
Answer:
The wave takes 0.132 seconds to travel from one end of the string to the other.
Explanation:
The velocity of a transversal wave ([tex]v[/tex]) travelling through a string pulled on both ends is determined by this formula:
[tex]v = \sqrt{\frac{T\cdot L}{m} }[/tex]
Where:
[tex]T[/tex] - Tension, measured in newtons.
[tex]L[/tex] - Length of the string, measured in meters.
[tex]m[/tex] - Mass of the string, measured in meters.
Given that [tex]T = 15\,N[/tex], [tex]L = 7.6\,m[/tex] and [tex]m = 0.034\,kg[/tex], the velocity of the tranversal wave is:
[tex]v = \sqrt{\frac{(15\,N)\cdot (7.6\,m)}{0.034\,kg} }[/tex]
[tex]v\approx 57.522\,\frac{m}{s}[/tex]
Since speed of transversal waves through material are constant, the time required ([tex]\Delta t[/tex]) to travel from one end of the string to the other is described by the following kinematic equation:
[tex]\Delta t = \frac{L}{v}[/tex]
If [tex]L = 7.6\,m[/tex] and [tex]v\approx 57.522\,\frac{m}{s}[/tex], then:
[tex]\Delta t = \frac{7.6\,m}{57.522\,\frac{m}{s} }[/tex]
[tex]\Delta t = 0.132\,s[/tex]
The wave takes 0.132 seconds to travel from one end of the string to the other.
The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Answer:
The approximate displacement of the object is 23 m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]
= 23 m
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].
To know more air bubble volume,
https://brainly.com/question/10509397
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
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An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
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
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
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