The speed of light and the propagation of errors allows to find the results on the questions of the radiation emitted by the laser are:
a) The frequency is: f = 3.7 10¹⁴ Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
a) The speed of a wave is related to its wavelength and frequency.
c = λ f
[tex]f = \frac{c}{\lambda}[/tex]
Where c is the speed of light, λ the wavelength and f the frequency.
They indicate that the wavelength is λ = 800 nm = 800 10⁻⁹ m, the speed of light is a constant c = 2.99 10⁸ m/s.
f = [tex]\frac{2.99 \ 10^8}{800 \ 10^{-9}}[/tex]
F = 3.7 10¹⁴ Hz
b) Planck's equation states that the energy is proportional to the frequency of the radiation.
E = h f
Where E is the energy, h the Planck constant and f the frequency.
E = 6.63 10⁻³⁴ 3.7 10¹⁴
E = 2.46467 10⁻¹⁹ J
The uncertainty or error is the fluctuation that a magnitude may have due to the precision in the measurements, when the magnitude is calculated by some formula, the propagation of these uncertainties must be carried out.
Δm = ∑ [tex]\sum \frac{dm}{dx_i} | \Delta x_I|[/tex]
the expression for energy is:
E = [tex]\frac{hc}{\lambda }[/tex]
[tex]\Delta E = \frac{dE}{d \lambda} |D\lambda |[/tex]
[tex]\Delta E = \frac{h c }{\lambda^2 } |\Delta \lambda |[/tex]
When the error in the measured magnitude is not explicitly indicated, we assume that the error is in the last digit written, therefore
Δλ = ± 1 nm = ± 1 10⁻⁹ m
We calculate.
[tex]\Delta E = \frac{6.63 \ 10^{-34} \ 2.99 \ 10^8 }{(800 \ 10^{-9})^2} 1 \ 10^{-9}[/tex]
ΔE = 3.1 10⁻²² J
the error is given with a significant figure.
ΔE = 3 10⁻²² J = 0.004 10⁻¹⁹ J
The result of the energy is:
E = (2.465 ± 0.004) 10⁻¹⁹ J
In conclusion, using the speed of light and the propagation of errors, we can find the results on the questions of the radiation emitted by the laser are:
a) The frequency is; f = 3.7 1014 Hz
b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J
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Terri is excited about the baby she is carrying in her womb. One day, she tells her
cousin that the child is now the size of a grape. Weeks later, she describes the child as
the size of an apple. A few weeks after that, she describes the child as the size of a
banana. "Seriously, Terri, are you having a baby or shopping for groceries?" asks her
cousin. Terri explains that this is just the easiest way to describe the size of the baby
now that it is a(n)
Zygote
O Fetus
O Embryo
Infant
Answer:
Fetus
Explanation:
the outsiders I got nuggets out of the elder and I get to make it in and I got I got
How is a controlled variable different from a responding variable? O A. A controlled variable changes due to changes in the responding variable during the experiment. B. A controlled variable stays the same throughout an experiment, but a responding variable changes, C. A controlled variable changes throughout an experiment, but a responding variable stays the same. O D. A controlled variable stays the same due to changes in the responding variable during the experient.
Answer:The controlled variable is the one that you keep constant. The responding variable or variables is what happens as a result of the experiment
Explanation:Mark me brainly please
ASAP
A truck is speeding up (with a constant acceleration) from rest to 45 m/s and it takes it 15 seconds to do so. How far will the car travel?
Answer:
675 KM
Explanation:
hAVE A NICE DAY
anybody know the answer??
Answer:
Centimeter
Explanation:
We know the measurement is defined as:
1 meter= 100cm
1 cm= 1/100 meter
1 centimeter= 0.01 meter
Therefore, 0.01 meter is equal to 1 cm or 1 centimeter.
The steepest street in the world is Baldwin Street in Dunedin, NZ. It is inclined at an angle of 380 , with the horizontal. A child slides down the street with a constant velocity on a sled with high friction runners. What is the coefficient of friction between the sled runners and the street?
Newton's second law allows to find the result for the friction coefficient of the street is:
The friction coeficinwete is: μ = 0.78
Newton's second law establishes a relationship between the net force, the mass and the acceleration of the body.
∑ F = m a
Where bold indicates vectors, m is to mass and acceleration.
In the attached we see a free body diagram, it is a diagram of the forces without the details of the body, the x-axis is parallel to plane also shown with the positive in the direction of movement, going down the plane and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight.
Sin θ = [tex]\frac{W_x}{W}[/tex]
cos θ = [tex]\frac{W_y}{W}[/tex] / W
Wₓ = W sin θ
[tex]W_y[/tex] = W cos θ
We write Newton's second law for each axis.
y-axis
N- [tex]W_y[/tex] = 0
N = mg cos θ
x-axis
Wₓ - fr = ma
Since they indicate that the body goes down at a constant speed, the acceleration is zero.
W sin θ = fr
The friction force is the macroscopic representation of the interactions between the two surfaces and the formula.
fr = μ N
we substitute.
fr = μ mg cos θ
mg sin θ = μ cos θ
μ = tan θ
Let's calculate.
μ = tan 38.0
μ = 0.78
In conclusion using Newton's second law we can find the results for the friction coefficient of the street is:
The frivtion coefficient is: μ = 0.78
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A 10 kg box is motionless on the floor. If the coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.3 (between the box and the floor), find the force required to start the block in motion
Please I need help ❤️
The force required to start the block in motion is equal to 39.2 Newton.
Given the following data:
Mass of box = 10 kgCoefficient of static friction = 0.4 Coefficient of kinetic friction = 0.3To determine the force required to start the block in motion:
Note: The force that is required to start the block in motion must be greater than or equal to the force of static friction.
Mathematically, the force of static friction is given by the formula;
[tex]Fs = uFn = umg[/tex]
Where;
Fs represents the force of static friction.μ represents the coefficient of friction.[tex]F_n[/tex] represents the normal force.g is the acceleration due to gravity.m is the mass of an object.Substituting the given parameters into the formula, we have;
[tex]F_s = 0.4 \times 10 \times 9.8\\\\F_s = 4 \times 9.8\\\\F_s = 39.2\;Newton[/tex]
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If you have a rocket that has a mass of 20kg and is accelerating at 25 m/s what force
would that rocket experience?
a . 20N
b . 40000N
c. 500N
d. 25N
If Earth's mass doubled, your weight would..
a. increase
b. decrease
c. stay the same
Answer:
a. increase
Explanation:
Doubling the mass of the Earth would double your weight, since gravitational force is directly proportional to mass, but doubling the radius (which doubles if the diameter doubles) would decrease your weight by a factor of 1/4, since gravitational force is inversely proportional to the square of the radius.
how many cubic objects of volume 2cm cube can be started in a room of dimension 2m by 3m by 4m
Answer:
12,000,000 boxes
Explanation:
the volume of the room can be found by using the equation for volume of a rectangular box:V=LxWxH
where:
L=2m
W=3m
H=4m
(it doesn't really matter which is which since it is multiplication)
when we multiply our values (2m*3m*4m) we get 24cubic meters
now we need to convert cubic meters to cubic centimeters
each cubic meter is 1,000,000 cubic centimeter we multiply 24 by 1,000,000 and we get: 24,000,000 cubic centimeters (cc)
dividing 24,000,000 by 2 (since each box is 2cc) we get 12,000,000
so, we know we can fit 12,000,000, 2 cubic centimeter boxes in this room
The metal hammer has a mass of 1500 kg and it hits the baseplate
with a speed of 8.0 m/s.
0 Calculate the kinetic energy of the hammer as it hits the baseplate.
Answer:
2300.68
Explanation:
The metal hammer has a mass of 1500 kg and it hits the baseplate
with a speed of 8.0 m/s.
0 Calculate the kinetic energy of the hammer as it hits the baseplate. there for it will create the kinetic energy of 2300.68 as it hits the baseplate.
A boy standing on a bridge 150m above the canyon floor throws a stone downward with a speed of 20.0m/s.
A) with what velocity will it hit the ground?
B) How long will it take to descend?
Answer:
57.97m/s
12.1s
Explanation:
Use one of the key equations of accelerated motion and sub in the values
vf^2=vi^2+2aΔd
vf^2=(20)^2+(2*9.8)(150)
vf^2 = 3340
vf = 57.79m/s (square rooted both sides)
now use another equation
Δd = (vi+vf/2)Δt
and rearrange for Δt
Δt = (2Δd-vf)/vi
sub in values
Δt = (2*150-57.79)/20
Δt = 12.1s
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which of the following changes will result in a weaker electromagnet
Answer:
Changing the metal core for a different metal will make the electromagnet stronger or weaker.
Explanation:
Hope this helped, nya~ :3
what is the model size to actual size
Answer:
You have to divide your real measurement to the model. For example, if your real measurement is 5m and your model is 20cm, divide 5/20 =1/4 and that means the ratio would be 1:4.
Explanation:
if an astronaut has a mass of 80kg on earth, what is their mass on the moon?
Answer:
130 N basically
Explanation:
please help me please help
Answer:
Answer → Distance is 5 × 10⁴ km
Explanation:
Gravitational formular
[tex]{ \tt{F = \frac{GMm}{ {r}^{2} } }} \\ [/tex]
F is the gravitational forceG is the universal gravitational constantr is the separation distanceM & m are the massesFor the first case;
[tex]{ \tt{F _{1} = \frac{GMm}{2.5 \times {10}^{4} } }} \\ [/tex]
For the second case;
[tex]{ \tt{F _{2} = \frac{GMm}{r _{2}} }} \\ [/tex]
but F2 = ½F1
Therefore, F1 = 2F2
Hence:
[tex]{ \tt{ \frac{GMm}{2.5 \times {10}^{4} } =2 \times \frac{GMm}{r _{2} } }} \\ \\ { \tt{ \frac{1}{2.5 \times {10}^{4} } = \frac{2}{r _{2} } }} \\ \\ { \tt{r _{2} = 2 \times 2.5 \times {10}^{4} }} \\ \\ { \underline{ \tt{ \: \: r _{2} = 5 \times {10}^{4} \: km \: \: }}}[/tex]
Answer: Distance is 5 × 10⁴ km
Explanation:
A swimming pool has dimension of 30m×10m×3m. When it is filled with water , what is the thrust on the bottom and sides
Answer:
3,900m³
Explanation:
If the momentum of the NASA space shuttle as it leaves the atmosphere is 3.75 x 10^8 kg ·m/s and its mass is 75000 kg, what is its speed?
Hi there!
We know that:
Momentum (p) = mass (kg) × velocity (m/s)
Or:
P = mv
We can rearrange the equation to solve for speed:
P/m = v
Plug in the given values:
(3.75 × 10⁸)/75000 = 5000 m/s
A car of mass 1500 kg is moving with a velocity of 30 m/s. If the velocity-time graph for this car
is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be
Answer:
v=30m/s
Explanation:
m = 1500kg
v = 30m/s
t = 25s
The velocity time graph is parallel to the time axis,
that means The Velocity is constant. ('constant' means remains same).
Therefore, after 25s, v=30m/s
Rope A is tied to block 1, and rope B is attached to both block 1 and block 2 as shown in the diagram. Block 1 has a mass of 4.2 kg and block 2 has a mass of 2.6 kg. You lift both blocks straight up. Calculate the magnitude of tension in each of the ropes when the blocks
Move at constant velocity of 1.5 m/s [up]
Find the magnitude of tension in each rope when the blocks are accelerating at 1.2 m/s^2 [up].
The maximum tension the strings can withstand is 90. N. Knowing this, determine the maximum acceleration of the blocks that would not break the rope.
Hi there!
Part 1:
If the blocks are moving at a constant velocity:
∑F = 0
Begin by summing the forces acting on each block. Let the upward direction be positive.
∑F₁ = Ta - M₁g - Tb
∑F₂ = Tb - M₂g
Sum the forces:
∑F = Ta - M₁g - Tb + Tb - M₂g
∑F = Ta - M₁g - M₂g = 0
Solve for Tension A:
Ta = M₁g + M₂g (Let g = 9.8 m/s²)
Ta = 4.2(9.8) + 2.6(9.8) = 66.64 N
Now, solve for tension B using the summation of ∑F₁:
0 = Tb - M₂g
Tb = (2.6* 9.8) = 25.48 N
Part 2:
We can use the same method, but incorporate the acceleration:
∑F = Ta - M₁g - M₂g
(M₁ + M₂)a = Ta - M₁g - M₂g
(M₁ + M₂)a + M₁g + M₂g = Ta
(4.2 + 2.6)(1.2) + 4.2(9.8) + 2.6(9.8) = 74.8 N
∑F₂ = Tb - M₂g
M₂a + M₂g = Tb = 28.6 N
Part 3:
Since the top string experiences most of the tension, we can use its equation to calculate the maximum acceleration:
∑F = Ta - M₁g - M₂g
(M₁ + M₂)a = Ta - M₁g - M₂g
a = (90 - M₁g - M₂g)/(M₁ + M₂)
a = 3.435 m/s²
A rock climber’s shoe loosens a rock, and her climbing buddy at the bottom of the cliff notices that the rock takes 3.20 s to fall to the ground. How high up the cliff is the rock climber?
Answer:
50.176 [m].
Explanation:
1) the required height can be calculated using the formula:
h=V₀*t+1/2* g*t₂, where V₀ - initial velocity of the rock, t - elapsed time, g - 9.8 m/s²;
2) if V₀=0, t=3.2, g=9.8, then
h=0.5*9.8*3.2²=1024*49*10⁻³=50.176 [m].≈50[m].
Mushroom is a fungi.
Mushrooms are filamentous fungi that produce large, often edible fruiting bodies. They live on organic material, thriving on compost, fallen leaves and damp wood and any other dead plant or animal matter. Their role in causing decay is important in maintaining ecological cycles.
What is the best reason astronomers have come up with to explain why sunspots are cooler and look darker
Answer:
Sunspots are places where the strong magnetic fields in the Sun resist the upward motion of bubbling hot gases from underneath
Explanation:
answer with proper explanation
Answer:
B
Explanation:
Moment = force * perpendicular distance
Since the beam is balanced(equilibrium), the 2 moments will be equal.
30 * P = 70 * Q
The 2 combinations which achieve this equality is P = 21 and Q = 9
2. Un niño hace girar con la mano una pelota de hule que se encuentra sujeta mediante un cordón de
0.75 m de longitud. Si la pelota da 0.7 vueltas a cada segundo, entonces,
a) ¿Cuál es el periodo de la pelota?
b) ¿Cuál es la velocidad lineal de la pelota?
Cuando el niño hace girar 0.7 vueltas por segundo una pelota de hule que se encuentra sujeta mediante un cordón de 0.75 m de longitud, tenemos que:
a) El periodo de la pelota es 1.43 segundos.
b) La velocidad lineal de la pelota es 3.3 m/s.
a) El periodo de la pelota está dado por:
[tex] T = \frac{2\pi}{\omega} [/tex]
En donde:
ω: es la velocidad angular
Dado que la pelota da 0.7 vueltas (revoluciones) cada segundo, la velocidad angular es:
[tex] \omega = \frac{0.7 \:rev}{s}*\frac{2\pi rad}{1 \:rev} = 4.40 rad/s [/tex]
Entonces, el periodo es:
[tex]T = \frac{2\pi}{\omega} = \frac{2\pi}{4.40 rad/s} = 1.43 s[/tex]
b) La velocidad lineal de la pelota se puede calcular usando la siguiente ecuación:
[tex] v = \omega r [/tex]
En donde:
r: es el radio de la circunferencia = longitud del cordón = 0.75 m
[tex]v = \omega r =4.40 rad/s*0.75 m = 3.3 m/s[/tex]
Por lo tanto, la velocidad lineal de la pelota es 3.3 m/s.
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For a boiling kettle draw a sankey diagram to show where The energy goes
Answer:
draw the picture / (.) (.) \ Patrick star
which of these MANNER of death? (select all that apply)
1-blood loss
2-cessation of brain function
3-accidential death
4-stabbing
5-pulmonary arrest
6-homicidal death
7-natural death
8-shooting
Answer:
3, 4, 6, 7, 8 are manners of death.
the small lightweight particles outside the nucleus of an atom are called___.
The neutron has no charge, and a mass of slightly over 1 amu. Some scientists propose the neutron is made up of a proton and electron-like particle. The electron is a very small particle located outside the nucleus.
Electron: -1
Neutron: 0
Proton: +1
What do you mean by 100w power?
The best evidence for dark matter in the milky way comes from the observation that the rotation curve
7. A taxi is travelling at 15 m/s. Its driver accelerates with acceleration 3 m/s2 for 4 s. What is its new velocity
Hi there!
To solve, we can use the following equation:
vf = vi + at
vi = initial velocity (m/s)
a = acceleration (m/s²)
t = time (s)
vf = final velocity (m/s)
Plug in the given values:
vf = 15 + 3(4)
vf = 15 + 12 = 27 m/s
Answer:
[tex]\boxed {\boxed {\sf 27 \ m/s}}[/tex]
Explanation:
We are asked to find the new or final velocity of a taxi.
We will use the following kinematic equation:
[tex]v_f=v_i+at[/tex]
The car was initially traveling at 15 meters per second. It accelerated at 3 meters per second squared for 4 seconds.
[tex]v_i=[/tex]15 m/s[tex]a=[/tex]3 m/s ²t= 4 sSubstitute the values into the formula.
[tex]v_f= 15 \ m/s + (3 \ m/s^2* 4 \ s)[/tex]
Multiply the numbers in parentheses.
3 m/s/s * 4 s = 3 m/s *4 = 12 m/s[tex]v_f= 15 \ m/s +12 \ m/s[/tex]
Add.
[tex]v_f=27 \ m/s[/tex]
The new velocity of the taxi is 27 meters per second.