PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 677 nm
B) 218 nm
C) 345 nm
D) 428 nm
OPTION D is the correct answer.
Refer to the attachment for complete calculation...
calculate the voltage that is being applied across a 10W bulb if a current of 0.2A flows through it
Answer:
below
Explanation:
from P= I * V
v = p/I
v = 10/0.2
v = 50 volts
Kilometer is a unit of length where as kilogram is a unit of mass
By George, you've nailed it, Stacy !
That's a fact, uh huh.
Truer words were never written.
Your statement is one of unquestionable veracity.
The pure truthiness of it cannot be denied.
Was there a question you wanted to ask ?
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
A) 1.67 x 10^-4 s^-1
B) 5.43 x 10^-4 s^-1
C) 1.40 x 10^-4 s^-1
D) 2.22 x 10^-4 s^-1
OPTION C is the correct answer.
Answer in your PE notebook
I have learned that
I have realized that
I will apply
Answer:
physical science is important
hety
civil engineering
A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.
6.0E23
1.0E24
2.4E25
3.6E25
4.8E26
Answer:
The number of photons in the pulse is 1.04 x 10²⁴
Explanation:
Given;
frequency of the emitted photons, f = 2.83 x 10¹¹ Hz
duration of the incident light, t = 3 ms = 3 x 10⁻³ s
power of the incident light, P = 65,000 W
The energy of each photon emitted is calculated as;
E = hf
where;
h is Planck's constant, = 6.626 x 10⁻³⁴ Js
E = 6.626 x 10⁻³⁴ x 2.83 x 10¹¹
E = 1.875 x 10⁻²² J
let the number of photons in the pulse = n
n(E)= Power x time
[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]
Equal masses of two different liquids are put into identical beakers.
Liquid 1 is heated for 100s and liquid 2 is heated for 200s by heaters of the same power.
The temperature of both liquids increases by the same amount.
Which statement is correct?
A Both liquids receive the same amount of energy.
B. Liquid 1 receives more energy than liquid 2.
C. Both liquids have equal thermal capacity.
D. The thermal capacity of liquid 1 is less than the thermal capacity of liquid 2.
Answer:
C
Explanation:
Because they both received the same temperature
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?
Answer:
The total momentum of the cars before the collision is 61,000 kg.m/s
The total momentum of the cars after the collision is 61,000 kg.m/s
The velocity of the cars after the collision is 27.727 m/s
Explanation:
Given;
mass of the first car, m₁ = 1000 kg
initial velocity of the car, u₁ = 25 m/s
mass of the second car, m₂ = 1200 kg
initial velocity of the second car, u₂ = 30 m/s
The common velocity of the cars after collision = v
The total momentum of the cars before collision is calculated as;
P₁ = m₁u₁ + m₂u₂
P₁ = (1000 x 25) + (1200 x 30)
P₁ = 61,000 kg.m/s
The total momentum of the cars after collision is calculated as;
P₂ = m₁v + m₂v
where;
v is the common velocities of the cars after collision since they stick together.
P₂ = v(m₁ + m₂)
To determine "v" apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(1000 x 25) + (1200 x 30) = v(1000 + 1200)
61,000 = 2,200v
v = 61,000/2,200
v = 27.727 m/s
The total momentum after collsion = v(m₁ + m₂)
= 27.727(1000 + 1200)
= 61,000 kg.m/s
Thus, momentum before and after collsion are equal.
calculate the correct fuse that should be used for a 230V,1KW electric hair dryer.
Answer: 4 A
Explanation:
Given
Voltage [tex]V=230\ V[/tex]
Power [tex]P=1\ kW[/tex]
Power is given by [tex]P=VI\\[/tex]
Insert the values
[tex]\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A[/tex]
The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.
according to the law of conservation of vhange , what must always be true in a nuclear reaction?
Answer:
The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same
Explanation:
Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.
A ratio that compares the width and length of a garden is what type of model?
Answer:
physical
PLEASE MARK ME AS A BRAINLIEST
Answer: Mathematical
Explanation: I took the quiz
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth
Answer:
0 N
Explanation:
This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N
Calculating Acceleration
Initial
velocity
Time to travel
0.25 m
Final
velocity
Acceleration
Time to travel
0.50 m
# of
washers
11
(m/s)
V2
(m/s)
ti
(s)
t₂
(s)
a = (v2 - v4)/(t2-tı)
(m/s)
1
0.11
0.28
2.23
3.13
0.19
2
0.13
0.36
1.92
2.61
The acceleration of the car with two washers added to the string would be
I can not even read this question.
What are you trying to even say?
The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0.13 m/s.Final velocity = 0.36 m/s.Initial time = 1.92 seconds.Final time = 2.61 seconds.What is an acceleration?An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.
How to calculate average acceleration.In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.
Mathematically, average acceleration is given by this formula:
[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]
Where:
V is the final velocity.U is the initial velocity.[tex]t_i[/tex]initial time measured in seconds.[tex]t_f[/tex] final time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]
a = 0.33 [tex]m/s^2[/tex]
Read more on acceleration here: brainly.com/question/24728358
Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?
Answer:
Freezing Point - Lower
Boiling Point - Higher
Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures
What is binding energy?
A.' The attractive forces between the protons in the nucleus and the
electrons
B. The energy required to force two nuclei to undergo nuclear fusion
C. The amount of energy stored in the strong nuclear forces of the
nucleus
D. The amount of energy required to overcome an activation energy
barrier
Please help me out.
Answer:
the answer is B i hope it helps :)
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
B. The energy required to force two nuclei to undergo nuclear fusion. ✅
They are usually expressed in terms of [tex]\sf\purple{kJ/mole}[/tex] of nuclei or [tex]\sf\pink{MeV's/nucleon}[/tex].[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS
For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
Did you add the links of the websites you used for this project to the Student Worksheet?
Did you use the scaled sizes to create models of the Sun and the planets?
Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
Did you label each component of your model?
Did you add an attention-catching title above or below your model?
Did you write your name on the back of your poster board?
Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.
Answer
I hope this help....
Explanation:
Answer:
Hope this helps
Explanation:
Can somebody please help
Answer:
Explanation:
part A: C
part B: B
how to calculating critical angle for a glass and air interface when there is a total internal reflection between them.
Answer:
total internal reflection
an object is 70 um long and 47.66um wide. how long and wide is the object in km?
Answer:
length = 7*10^(-8)km
width = 4.666*10^(-8) km
Explanation:
We know that:
1 μm = 1*10^(-6) m
and
1km = 1*10^3 m
or
1m = 1*10^(-3) km
if we replace the meter in the first equation, we get:
1 μm = 1*10^(-6)*1*10^(-3) km
1 μm = 1*10^(-6 - 3)km
1 μm = 1*10^(-9)km
Now with this relationship we can transform our measures:
Length: 70 μm is 70 times 1*10^(-9)km, or:
L = 70*1*10^(-9)km = 7*10^(-8)km
And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:
W = 46.66*1*10^(-9)km = 4.666*10^(-8) km
The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?
[pleas ee helpppp)
I= 0.39 A
OPTION B is the correct answer.
Which two statements are true for reversible reactions that reach dynamic
equilibrium?
I A. The products of the forward and backward reactions remain
constant at equilibrium.
B. The products of the forward reaction form more quickly than its
reactants.
C. The rate of the forward reaction is greater than the rate of the
backward reaction.
- D. The rate of the forward reaction is equal to the rate of the
backward reaction at equilibrium.
Answer:
Explanation:
In a reversible reaction which has reached dynamic equilibrium , rate of forward reaction is equal to rate of backward reaction .
Following is a reversible chemical reaction .
A + B = C + D
Rate of forward reaction = k₁ x [ A ] x [ B ]
Rate of backward reaction = k₂ x [ C ] x [ D ]
k₁ x [ A ] x [ B ] = k₂ x [ C ] x [ D ]
[ A ] x [ B ] = k₂ / k₁ [ C ] x [ D ]
[ A ] x [ B ] = k [ C ] x [ D ]
The products of the forward and backward reactions remain
constant at equilibrium.
Hence option A and D are correct statement .
A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.
Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?
Answer:
a) μ = 0.0136, b) F = 22.8 N
Explanation:
This exercise must be solved in parts. Let's start by using conservation of moment.
a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved
initial instant. Before the crash
p₀ = m v₀
final instant. After inelastic shock
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m + M} \ v_o[/tex]
We look for the speed of the block with the bullet inside
v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]
v = 0.448 m / s
Now we use the relationship between work and kinetic energy for the block with the bullet
in this journey the force that acts is the friction
W = ΔK
W = ½ (m + M) [tex]v_f^2[/tex] - ½ (m + M) v₀²
the final speed of the block is zero
the work between the friction force and the displacement is negative, because the friction always opposes the displacement
W = - fr x
we substitute
- fr x = 0 - ½ (m + M) vo²
fr = ½ (m + M) v₀² / x
the friction force is
fr = μ N
μ = fr / N
equilibrium condition
N - W = 0
N = W
N = (m + M) g
we substitute
μ = ½ v₀² / x g
we calculate
μ = ½ 0.448 ^ 2 / 0.75 9.8
μ = 0.0136
b) Let's use the relationship between work and the variation of the kinetic energy of the block
W = ΔK
initial block velocity is zero vo = 0
F x₁ = ½ M v² - 0
F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]
F = ½ 2.50 0.448² / 0.0110
F = 22.8 N
A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack
of groceries?
the answer is 396 joules :D
If the distance between the center of two objects is quadrupled. The gravitational
force between the two objects will change by a factor of:
1) 16
2) 0.25
3) 4
4) 0.0625
Answer:
F' = F/16
Explanation:
The gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
If the distance between the center of two objects is quadrupled, r' = 4r
New force will be :
[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}[/tex]
So, the new force will change by a factor of 16.
a 1600 kg car rounds a curve of radius 71 m banked at an angle of 15, What is the magnitude of the friction force required for the car to travel at 86 km/h
Answer:
The frictional force required for the car to travel is 8,365.01 N
Explanation:
Given;
mass of the car, m = 1600 kg
radius of the curved road, r = 71 m
banking angle, θ = 15⁰
velocity of the car, v = 86 km/h = 86/3.6 = 23.89 m/s
The two forces acting on the are:
1. the parallel force to the banked plane
2. the centripetal force pushing the car up the banked plane
To keep the car traveling at 86 km/h;
frictional force + parallel force to the plane = centripetal force pushing the car up the banked plane
The parallel force to the banked plane:
F = mgsinθ
F = 1600 x 9.8 x sin(15⁰)
F = 4,057.98 N
The centripetal force pushing the car up the banked plane:
[tex]F_c= (\frac{mv^2}{r} )cos(\theta)\\\\F_c = (\frac{1600 \times 23.89^2}{71} )cos(15^0)\\\\F_c = 12,422.99 \ N[/tex]
The frictional force required for the car to travel:
[tex]F_k = F_c - F\\\\F_k = 12,422.99 \ N - 4,057.98 \ N\\\\F_k = 8,365.01 \ N[/tex]
Therefore, the frictional force required for the car to travel is 8,365.01 N
Which type of energy is stored in a battery?
A. Nuclear energy
B. Electromagnetic energy
C. Chemical energy
D. Electrical energy
SUBMI
Answer:
c
Explanation:
in food and batteries chemical energy is stored :) hope this helped
Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface
Answer:
Shiny surface.
Explanation:
We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.
Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.
This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,
identify the word being referred to choose your answer from the words below
Answer:
1:Rotation
2:Axis
3:Aphelion
4:orbit
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
a) 8.03 x 10^16 nuclei
b) 4.01 x 10^16 nuclei
c) 2.02 x 10^16 nuclei
d) 1.61 x 10^17 nuclei
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
c) 2.02 x 10^16 nuclei