Explanation:
First, we convert the energy from eV to Joules:
[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]
[tex]= 4.64×10^{-19}\:\text{eV}[/tex]
We know from definition that
[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]
so the wavelength of the photon is
[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
when a boron is added to a pure semi conducter it becomes
Answer:
it becomes a p-type conductor
Explanation:
answer from gauth math
Calculate the equivalent of 30 degrees Celsius and 50 degrees Celsius on a Kelvin
[tex]\boxed{\sf 1°C=273K}[/tex]
Sol:-1
[tex]\\ \sf\longmapsto 30°C[/tex]
[tex]\\ \sf\longmapsto 273+30[/tex]
[tex]\\ \sf\longmapsto 303K[/tex]
Sol:-2
[tex]\\ \sf\longmapsto 50°C[/tex]
[tex]\\ \sf\longmapsto 50+273[/tex]
[tex]\\ \sf\longmapsto 323K[/tex]
As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.
Answer:
approaches infinity
Explanation:
There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.
The relativistic momentum can be written as:
[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
where
u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.
m = mass of the object
c = speed of light.
So, as u tends to c, we will have:
[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.
Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.
Under normal circumstances: _________
a. Fetal Hb binds to oxygen more tightly than Mb binds.
b. Fetal Hb binds oxygen more tightly in the absence of 2,3-BPG.
c. Fetal Hb does not bind to oxygen.
d. Adult Hb has the lowest affinity for oxygen of the 3.
e. More than one of these statements is correct.
Answer:
Fetal Hb binds oxygen more tightly than adult Hb (not option a)
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
Solving for [tex]v_{0y}[/tex],
[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]
[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]
b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as
[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]
or
[tex]v_{0x} = 150\:\text{m/s}[/tex]
name a device that converts mechanical energy into electrical energy.
Answer:
Electric generator is the device that converts mechanical energy into electrical energy
which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
11. From this lab we can conclude that a) the heat transferred when objects are rubbed together creates an energy that can cause objects to move towards or away from each other. b) objects such as balloons and sweaters have a natural affinity towards each other. They will attract each other whether they are rubbed together or not. c) charges exert forces on other charges. do) charges do not exert forces on other charges.
Answer:
c) charges exert forces on other charges.
Explanation:
When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.
A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
Was this answer helpful?
71
0
SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
View solution
>
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
find the equivalent resistance of this circuit
Answer:
Req = 564 Ω
Explanation:
The equivalent resistance between R1 and R2:
1/R =1/R1 + 1/R2
1/R =1/960 + 1/640
1/R = 1/384
R = 384
Now, the equivalent resistance between R and R3:
Req = 384 + 180
Req = 564 Ω
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
What is the percentage of the population that wanted both the swimming pool and the soccer complex? Use your knowledge
of the addition rule and the Venn diagram to answer.
Answer:
The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.
Explanation:
Edmentum
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
Si un resorte de constante elástica 1300 n/m se comprime 12 cm ¿Cuanta energía almacena? Y si estira 12cm ¿Cuanta energía almacena?
La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.
La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:
[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]
En donde:
k: es la constante del resorte = 1300 N/m
x: es la distancia de compresión o de elongación = 12 cm = 0,12 m
Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.
Por lo tanto, la energía almacenada es:
[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]
Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.
Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults
Espero que te sea de utilidad!
Answer:
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Explanation:
La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)
Donde:
[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.
[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.
[tex]x_{f}[/tex] - Posición final del resorte, en metros.
Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.
Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.
La cantidad de energía a almacenar es:
[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]
[tex]U_{e} = 9,360\,J[/tex]
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
PleasePlease help me solve these articles with me
how does laser works ?
Explanation:
Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put
Answer:
300J
Explanation:
Work done = Force x the distance travelled in the direction of the force
=300 x 1
=300J
A student has a large number of coins of different diameters, all made of the same metal. She wishes to find the density of the metal by a method involving placing the coins in water.
a) State the formula needed to calculate the density.
b) Describe how the measurements of the required quantities are carried out.
Answer:
a)density = mass /volume
b)to find volume put water into a container .measure the level of water , put the coins into the beaker containing water , measure the level of water again, subtract the new volume withe the first one . the result is the volume of coins
The angle of the resultant vector is equal to
the inverse tangent of the quotient of the x-component divided by the y-component of the resultant vector
the inverse cosine of the quotient of the y-component divided by the x-component of the resultant vector.
the inverse cosine of the quotient of the x-component divided by the y-component of the resultant vector.
the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.
The y-component of a vector is the product of the magnitude of the vector and the sine of the angle of the vector to the horizontal. The x-component of a vector is the product of the magnitude of the vector and the cosine of the angle of the vector to the horizontal.The angle of this resultant vector is also known as the direction of the vector.
Mathematically, the direction of a resultant vector is given as;
[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]
Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
Lear more here: https://brainly.com/question/3224643
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
A velocity of ship A relative to ship B is 10m/s in the direction N45E . If the velocity of B is 20m/s in the direction N60W . Find the velocity of ship A and direction.
Answer:
ewjefkljlajwawk;dlqa;wdka:WDKkjlhgzkljwidaJLdkjALIw
Explanation:
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
Data related to Meena’s and Malini’s journey is given below, plot a graph of their
respective journey on a graph paper. You have already plotted Meena’s Journey during the
summer vacation. On the same graph paper, now plot Malini’s Journey.
PDF task 2
Please do this for me urgent I can give you extra points if someone answers this in less than 1 hour.
Answer:
download the pdf
Explanation:
The question is in the photo.
Answer:
heyaa thereeee
see temperature is rising in interval of
0 to 4 minutes
8 to 10 minutes
but 8 to 10 is NOT in options
so answer is option a) 0 to 4 minutes
:))))
In the following experiments, identify the independent and dependent variable.
Answer:
in what experements
Explanation:
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
Identify the factors that affect the intensity of radiation detected from a radioactive source. Select one or more: The color of the source Type of emission from the source Distance of the detector from the source Type of materials between the source and the detector
The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.
The factors that affect the intensity of radiation are
Type of emission from the source :This can be alpha, gamma, beta or electromagnetic rays etc
Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.
Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.
Read more on Radiation Intensity here: https://brainly.com/question/10148635
What will be the acceleration of a body moving with uniform velocity?
Answer:
so the acceleration of the body would be zero because there is no change in velocity
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
[tex]v=0.9833\ c[/tex]
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
[tex]$\text{Density} = \frac{m}{lwh}$[/tex]
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 [tex]kg/m^3[/tex]
So,
[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]
[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]
l = 0.024 m
Then for relativistic length contraction,
[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]
[tex]$\frac{v^2}{c^2}= 0.967$[/tex]
[tex]$\frac{v}{c}=0.9833$[/tex]
[tex]v=0.9833\ c[/tex]
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
