Answer:
7.3 newtons to the west
Explanation:
3.7kg × 11a - 3.7kg × ? = 3.7n
how to calculating critical angle for a glass and air interface when there is a total internal reflection between them.
Answer:
total internal reflection
Just before take-off, the plane is speeding up along the ground.
Which statement is true?
Force B is zero.
Force B is greater than force D.
Force D is equal to force B
Answer:
Force B is zero.
Explanation:
If the plane is able to move forward then force D is greater than B. In this case because the plane is moving in the direction of D, then B is assume to be zero.
The force B in the given image will be 0. The correct option is A.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force.
An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. It is not possible for an object to contain a force.
If the plane may go forward, force D must be greater than force B. In this instance, B is taken to be zero because the plane is travelling in the direction of D.
Thus, the correct option is A.
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A 0.500 m length of wire with a cross-sectional area of 3.14 * 10 ^ - 6 meters squared is found to have a resistance of 2.53 * 10 ^ - 3 ohms according to the resisting chart, from what material is the wire made?
PLEASE HELP MEEEEEEE
Answer:
The wire is made of silver (ρ = 1.59×10⁻⁸ ohms/m)
Explanation:
Applying,
R = ρL/A................. Equation 1
Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire
make ρ the subject of the equation
ρ = RA/L............. Equation 2
From the question,
Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m
Substitute these values into equation 2
ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5
ρ = 1.59×10⁻⁸ ohms/m
Hence from the resistivity chart, the wire is made of silver
An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat
per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω
of the heater wire? (Note: 1 cal = 4.186 J)
Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43
Answer:
1 cal/s =4.184w
p=50 cal/s =2093w
v=12v
P = V*I
I =P/V
I = 17.43 A
P =1²*R
R = P/I²
R = 0.68PLEASE 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...
Tiny organisms were collected from thermal vents deep in the Pacific Ocean. The characteristics below describe these primitive organisms.
1) They are all unicellular.
2) They live in harsh environments.
3) Most have a cell wall.
In which domain would these organisms be found? (SC.6.L.15.1)
A.Archaea
B.Bacteria
C.Eukarya
D.Protista
Answer:
A. Archaea
Explanation:
Living organisms have been taxonomically classified into a highest ranking taxa called DOMAIN. The domains are as follows: Eukarya, Prokarya (bacteria) and Archaea. Although both Domains Bacteria and Archaea are unicellular, the Archaea is specifically characterized by their ability to survive in harsh weather conditions like the thermal vents tiny organisms were collected from.
Also, members of the domain Archaea have a unique cell wall different from other organisms like bacteria. Hence, based on the matching characteristics of the collected tiny organisms, they would be found in the Domain ARCHAEA.
If I am going to explore Mars what will I need?
Answer:
uhm oxygen tank, a suit (like an astronaut suit) uh food, and a space ship
Explanation:
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 .
The application of solid-state physics that is the study of the arrangement of atoms in a solid is
O metallurgy.
O quantum mechanics.
O crystallography
electromagnetism.
Answer:
crystallography.......
(c) Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0= 1.68×10-21 J and R0= 3.82×10-10 m. Find the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
Explanation:
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
The formula for calculating the elastic potential energy is:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
By rearrangement and using (K) as the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the formula used to calculate the frequency of the small oscillation is:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom
assuming
m = 1.66 × 10⁻²⁶ kg
Then:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Suppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm
Answer:
0.035 J
Explanation:
Applying,
W = ke²/2.............. Equation 1
Where W = workdone by the stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = 2W/e²............... Equation 2
From the question
Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m
Substitute these values into equation 2
k = (2×2)/(0.15²)
k = 177.78 N/m
Hence, work need to stretch the spring from 33 cm to 35 cm
therefore,
e = 35-33 = 2 cm = 0.02 m
Substitute into equation 1
W = 177.78(0.02²)/2
W = 0.035 J
From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height of the tower is 200 m, how long will it take for the ball to hit the ground? Ignore air resistance.
Answer:
66,7 seconds
Explanation:
the formula for height/distance is : S=v.t
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting
Answer:
the person is sitting 1.5 m from the left end of the board
Explanation:
Given the data in the question;
Wb = 125 N
Wm = 500 N
T₂ = 250 N
Now, we know that;
T₁ + T₂ = Wb + Wm
T₁ + 250 = 125 + 500
T₁ = 125 + 500 - 250
T₁ = 375 N
so tension of the left chain is 375 N.
Now, taking torque about the left end
500 × d + 125 × 2 = 250 × 4
500d + 250 = 1000
500d = 1000 - 250
500d = 750
d = 750 / 500
d = 1.5 m
Therefore, the person is sitting 1.5 m from the left end of the board.
B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.
Answer:
Things you should do for your family
help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softlythings you shouldn't
backanswering them Disobey And anything that's harsh or make it parents sadAn object that gives off electromagnetic waves based on its temperature
demonstrates which phenomenon?
A. Emission spectra
B. Blackbody radiation
C. Quantum mechanics
D. Photoelectric effect
Answer:
B) Blackbody radiation
MARK THIS AS BRAINLIEST PLEAWESSS :)
Answer:
Black body radiation
Explanation:
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
OPTION C is the correct answer.
The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.
What is half-life?The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.
The isotope decay of an atom is given by the equation:
ln [A] = -kt + ln [A]₀
The rate constant, k is:
k = ln 2 / Half life
k = ln 2 / 4.96 x 10³
k = 1.40 × 10⁻⁴ s⁻¹
t = 1.98 x 10⁴
[A]₀ = 3.21 x 10¹⁷
ln [A] = -1.40 × 10⁻⁴ × 1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538
[A] = 2.02 x 10¹⁶ nuclei
Thus the correct option is C.
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A stone of mass 0.2 kg falls with an acceleration of 10.0 m/s. How big is the force that causes this acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ Newtons}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration.
[tex]F= m \times a[/tex]
The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.
m= 0.2 kg a= 10.0 m/s²Substitute the values into the formula.
[tex]F= 0.2 \ kg * 10.0 \ m/s^2[/tex]
Multiply.
[tex]F=2 \ kg*m/s^2[/tex]
Convert the units.
1 kilogram meter per square second (kg*m/s²) is equal to 1 Newton (N)Our answer of 2 kg*m/s² is equal to 2 N[tex]F= 2 \ N[/tex]
The force is 2 Newtons.
2.
A Velocidade Escalar de um automóvel aumenta de 36km/h para 108km/h em 10 segundos.
Determina a sua aceleração media.
Answer:
Aceleración, a = 2 m/s²
Explanation:
Dados los siguientes datos;
Velocidad inicial = 108 km/h
Tiempo = 10 segundos
Velocidad final = 36 km/h
To find the average acceleration;
Conversión:
36 km/h to meters per seconds = 36*1000/3600 = 10 m/s
108 km/h to meters per seconds = 108*1000/3600 = 30 m/s
I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;
[tex] V = U + at[/tex]
Dónde;
V es la velocidad final.
U es la velocidad inicial.
a es la aceleración.
t es el tiempo medido en segundos.
Sustituyendo en la fórmula, tenemos;
[tex] 30 = 10 + a*10 [/tex]
[tex] 30 = 10 + 10a [/tex]
[tex] 10a = 30 - 10 [/tex]
[tex] 10a = 20 [/tex]
[tex] Aceleracion = \frac{20}{10}[/tex]
Aceleración, a = 2 m/s²
In the chemical equation Zn+2HCI ZNCI+H the reaction are
Answer:
In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid. In the chemical equation Zn + 2HCL-> ZnCl2 + H2, the reactants are zinc and hydrochloric acid.
Explanation:
this is the correct
A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird
Answer:
The bird appears farther
Explanation:
This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.
So, the bird appears to the diver to be farther from the surface than the actual bird
The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instead of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,
Answer:
A) vectors: veloicty, force
scalar: speed, work
B) t = 1.75 s, C) v = - 17 2 m / s
Explanation:
We answer each part separately
A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude
vector quantities: the speed of a car number is the magnitude and direction is where it goes
Force, the number is the magnitude and above that applies gives direction
Scalar magnitude: how quickly the number of the speedometer of the car
Temperature, work
B) I = 15 m height to the soil and get to calculate time = 0
y = y₀ + v₀ t - ½ g t²
as the ball is loose its initial velocity is zero
0 = 0 +0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]
t = 1.75 s
C) the velocity to the reach the floor
v = vo - g t
v = 0 - g t
v = - 9.8 1.74
v = - 17 2 m / s
The negative signt iindicates that the speed goes down
A glass of milk has what kind of energy?
A. Chemical Potential Energy
B. Kinetic Energy
C. Elastic Potential Energy
D. Radiant Energy
15. A car is stationary at the top of a hill with the engine
switched off. The brakes are released and the car rolls down
the hill. At which labelled point does the car have the greatest
kinetic energy? *
A
Answer:
kinetic energy
Explanation:
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Can someone help me
Answer:
Explanation:before the phase change the substance is a particle.
A television of mass 8 kg sits on a table. The coefficient of static friction
between the table and the television is 0.48. What is the minimum applied
force that will cause the television to slide?
O A. 38 N
O B. 62 N
O C. 78 N
D. 55 N
The television has weight (8 kg) g = 78.4 N, and the magnitude of the normal force between the table and television would be the same, 78.4 N. This mean the maximum magnitude of static friction between the table and television is
0.48 (78.4 N) ≈ 37.6 N ≈ 38 N
and this is the minimum required force needed to get the television to slide.
make ansentance rkdloebebjekeoejbe
Answer:
the man has returned from his trip
Answer:
just did by typing this lol
At a particular instant, a proton, far from all other objects, is located at the origin. The proton is traveling with velocity (-3 x 106,0,0)m/s. Consider the electric and magnetic fields at observation point (9 x 10-10,2 x 10-10,0)m caused by this proton.
What is the electric field at the observation point?
What is the magnetic field at the observation point?
Answer:
The electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
The magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
Explanation:
Calculating the electric field at the observation point:We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m[/tex]
The equation used to calculate the electric field follows:
[tex]\vec{E}=\frac{kq\hat{r}}{r^2}[/tex]
OR
[tex]\vec{E}=\frac{kq}{r^2}\frac{\vec{r}}{|\vec{r}|}[/tex]
We know:
[tex]|\vec{r}|=r[/tex]
[tex]q=1.6\times 10^{-19}C[/tex]
So, the equation becomes:
[tex]\vec{E}=\frac{kq(\vec{r})}{r^3}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\vec{E}=\frac{(9\times 10^9}(1.6\times 10^{-18})(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})}{\left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{E}=(1.84\times 10^{18}(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})\\\\\vec{E}=1.65\times 10^{9}\hat{i}+3.68\times 10^{8}\hat{j}[/tex]
Hence, the electric field at the observation point is [tex]<1.65\times 10^{9}, 3.68\times 10^{8}, 0> N/C[/tex]
Calculating the magnetic field at the observation point:The magnetic field due to moving charge is given by:
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \hat{r}}{r^2}[/tex]
OR
[tex]\vec{B}=\frac{\mu_o}{4\pi}\frac{q\vec{v}\times \vec{r}}{r^3}[/tex] .....(2)
We are given:
[tex]\vec{r}=(9\times 10^{-10}\hat{i}+2\times 10^{-10}\hat{j})m\\\\\vec{v}=-3\times 10^6\hat{i}[/tex]
Putting values in equation 2, we get:
[tex]\vec{B}=\frac{\mu_o\times (1.6\times 10^{-19})\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]}{4\pi\times \left [ \sqrt{(9\times 10^{-10})^2+(2\times 10^{-10})^2} \right ]^3}\\\\\vec{B}=(20.42)\left [ (-3\times 10^6\hat{i})\times (9\times 10^{-10}\hat{i}+2\times 10^{-3}\hat{j}) \right ]\\\\\vec{B}=-1.23\times 10^{-14}\hat{k}[/tex]
Hence, the magnetic field at the observation point is [tex]<0, 0, -1.23\times 10^{-14}> T[/tex]
6. A transverse periodic wave on a string with a linear density of 0.200 kg/m is described by the following equation: y = 0.08 sin(469t – 28.0x), where x and y are in meters and t is in seconds. What is the tension in the string? A) 3.99 N B) 32.5 N C) 56.1 N D) 65.8 N
Answer:
T = 56.11 N
Explanation:
Given that,
The equation of a wave is :
y = 0.08 sin(469t – 28.0x),
where x and y are in meters and t is in seconds
The linear mass density of the wave = 0.2 kg/m
The speed of wave is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Also,
[tex]v=\dfrac{\omega}{k}[/tex]
We have,
[tex]k=469\ and\ \omega=28[/tex]
Put all the values,
[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]
Put all the values,
[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]
So, the tension in the string is 56.11 N.
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
Coherent light with wavelength 603 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.
Required:
For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen?
Answer:
λ = 4.023 10⁻⁷ m
Explanation:
The double-slit interference phenomenon is described by
d sin θ = (m + ½) λ destructive interference
d sin θ = m λ constructive interference
we can use trigonometry
tan θ = y / L
how these experiments occur for small angles
tan θ = sin θ/cos θ = sin θ
sin θ = y / L
we substitute
d y / L = (m + ½) λ destructive interference
d y / L = m λ constructive interference
with the expression for constructive interference we look for the separation of the slits
d = m λ L / y
d = 1 603 10⁻⁹ 3 /4.84 10⁻³
d = 3.738 10⁻⁴ m
Now let's analyze the case where the distance for constructive and destructive interference occurs at the same point y = 4.84 mm = 4.84 10⁻³m
d y / L = (m + ½) λ
λ = [tex]\frac{ d \ y}{L\ (m+ 1/2) }[/tex]
the first strip is for m = 1
let's calculate
λ = [tex]\frac{3.738 \ 10^{-4} 4.84 \ 10^{-3} }{ 3 \ ( 1 + 0.5) }[/tex]
λ = 4.023 10⁻⁷ m