Answer:
x_total = 3.07 m
Explanation:
Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.
Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry
tan θ₁ = 1.0 /2.0
θ₁ = tan⁻¹ 0.50
θ₁ = 26.57º
now let's use the law of refraction to find the angle of refraction (in water)
n₁ sin θ₁ = n₂ sin θ₂
the refractive index of air is n₁ = 1 and that of water n₂ = 1.33
sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]
sin θ₂ = [tex]\frac{1}{1.33} \ sin \ 26.57[/tex]
θ₂ = sin⁻¹ 0.3363
θ₂ = 19.65º
now we can find the distance from the entry point to the water to the lenses
tan θ₂ = x₂ / 3
x₂ = 3 tan θ₂
x₂ = 3 tan 19.65
x₂ = 1.07 m
the total distance from the edge of the pool is
x_total = 2 + x₂
x_total = 2 + 1.07
x_total = 3.07 m
What is the law of conservation of energy
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
Cần nâng vật nặng G = 15B lên độ cao 2m bằng mặt phẳng nghiêng dài 12m. Hệ số ma sát giữa vật với mặt phẳng nghiên là f = 0,3. Hãy tính số người có thể tham gia vao công việc, nếu như sức đẩy theo phương song song với mặt phẳng nghiêng của mỗi người là B
a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7,calculate the maximum kinetic energy
Complete question:
a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.
Answer:
the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J
Explanation:
Given;
work function, Ф = 1.8 eV
wavelength of the light, λ = 3 x 10⁻⁷ m
The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.
[tex]E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J[/tex]
[tex]K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J[/tex]
Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
Who stated that man is an animal
16. What type of lens bends light inwards?
concave
convex
Answer:
convex lens because it makes parallel light rays passing through it bends inwards and meets convex at a spot just beyond the lens known as the focal point
Answer:
Explanation:
each of the following conversions contains an error. In each case, explain what the error is and find the correct answer to make a true statement .
a 1000 kg mg (1kg/1000g) = 1g.
b. 50m (1cm/100m)=0.5 cm
c. "Nano" is 10^-9 , so there are 10^-9 nm in a meter.
d. micro is 10^-6, so 1kg is 10^6 ug
Answer:
a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d. micro is 10⁻⁶, so 1kg is 10⁹ ug
Explanation:
a.
The conversion factor is written inverted. The correct statement will be:
[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b.
The values in the conversion factor used are wrong. The correct statement will be:
[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c.
Change of units is the mistake here. The correct statement will be:
"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d.
the conversion will be as follows:
[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]
therefore, the correct statement will be:
micro is 10⁻⁶, so 1kg is 10⁹ ug
Hi there! I'm not quite sure on how to solve this....
[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]
Explanation:
The volume of a cube is V = x^3. Taking the time derivative of this expression, we get
[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]
or
[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]
We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is
[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]
Answer:
Explanation:
[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]
what is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s
Answer:
by using p = mv equation we can find v,
6 = 1.3 v
4.615 = v
what is the value of gravitational constant
Answer:
6.67×10^-11 Nm^2kg^_2
A pilot flies an airplane at a constant speed of 590 km/h in the vertical circle of radius 970 m. Calculate the force exerted by the seat on the 82-kg pilot at point A and at point B.
Answer:
[tex]F = \frac{m {v}^{2} }{r} \\ = \frac{82 \times { (\frac{590 \times 1000}{3600} )}^{2} }{970} \\ = 2270.6 \: newtons[/tex]
A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]
Two insects are at points 1 and 2. which
of the two insects has the greater kinetic energy of rotation
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of +2.0 m/s, while Ender recoils with a velocity of -3.4 m/s. Determine the ratio mBonzo/mEnder of the masses of these two enemies.
Answer:
ratio of the masses of the two enemies = 1.7
Explanation:
Applying the law of conservation of momentum,
Momentum fo Bonzo = Momentum of Ender
mv = m'v'................. Equation 1
Where m = mass of Bonzo, v = velocity of Bonzo, m' = mass of Ender, v' = velocity of Ender
m/m' = v'/v............... Equation 2
Where m/m' = ratio of the masses of the two enemies
Given: v = 2.0 m/s, v' = 3.4 m/s
Substitute into equation 2
m/m' = 3.4/2
m/m' = 1.7
Hnece, ratio of the masses of the two enemies = 1.7
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two significant figures and include the appropriate units.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
Calculate the magnitude of minimum acceleration at which the thief can descend
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat
Solution :
Given :
Mass of the baseball, m = 200 g
Velocity of the baseball, u = -30 m/s
Mass of the baseball after struck by the bat, M = 900 g
Velocity of the baseball after struck by the bat, v = 47 m/s
According to the conservation of momentum,
[tex]Mv+mu=Mv_1+mv_2[/tex]
(900 x 47) + (200 x -30) = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
36300 = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)
[tex]9v_1 = 363 - 2v_2[/tex]
[tex]v_1=\frac{363 - 2v_2}{9}[/tex]
The mathematical expression for the conservation of kinetic energy is
[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]
[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex] ................(ii)
[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]
[tex]21681 = 9v_1^2+2v_2^2[/tex]
Substituting (i) in (ii)
[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]
[tex](363-2v_2)^2+18v_2^2=195129[/tex]
[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]
[tex]22v_2^2-145v_2-63360=0[/tex]
Solving the equation, we get
[tex]v_2=96 \ m/s, -30 \ m/s[/tex]
The negative velocity is neglected.
Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get
[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]
= 19
Thus, only impulse of importance is used to find final velocity.
Tres ladrillos idénticos están atados entre sí por medio de cuerdas y penden de una balanza que marca en total 24 N. ¿Cuál es la tensión de la cuerda que soporta al ladrillo inferior? ¿Cuál es la tensión en la cuerda que se encuentra entre el ladrillo de en medio y el superior?
ayuda!!!!!
answer 8N 8N
la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8
la tensión entre el ladrillo superior y el medio también es 8N
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
what are two injuries that heal slowly and with difficulty
Determine the tension in the string that connects M2 and M3.
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Wavelength of blue photons 495 nm, what is the frequency? and what is the energy?
Answer:
1.F: About 6*10^14 Hz
2.E: About 4*10^ -19 J
Explanation:
Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz). By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)
=6.05*10^14 /s
Energy: The energy photon contains can be calculate by this formula-- E=hf
f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.
So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20= 4.051*10^-19 J
Why does the moon appear dark from space?
But why does it appear bright when observed from earth, especially when it is full moon?
Answer:
The moon is actually quite dim.
Explanation:
compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.
Answer:
It reflects the light send from the sun.
Explanation:
If the moon is between you and the sun, you will see the back of it which doesnt reflect light.
Suppose the water at the top of Niagara Falls has a horizontal speed of 2.73 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 52.9 ° angle below the horizontal?
Answer:
required vertical distance below the edge is 0.6648 m
Explanation:
Given the data in the question;
Horizontal speed of water falls v = 2.73 m/s
direction of water falls 52.9° below the horizontal
The vertical velocity must be such that;
tanθ = v[tex]_y[/tex] / v[tex]_x[/tex]
Now, vertical speed of water falls;
v[tex]_y[/tex] = v[tex]_x[/tex] × tanθ
we substitute
v[tex]_y[/tex] = 2.73 × tan(52.9°)
v[tex]_y[/tex] = 2.73 × 1.322237
v[tex]_y[/tex] = 3.6097
Now, at the top of falls, initial speed u = 0
v² - u² = 2as
s = ( v² - u² ) / 2as
we substitute
s = ( 0² - (3.6097)² ) / (2 × 9.8)
s = 13.029934 / 19.6
s = 0.6648 m
Therefore, required vertical distance below the edge is 0.6648 m
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
Page
E QON
1 What is force ? Write its unit and mention
any
three effects of the force.
Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.
1) Can change the state of an object : For example, pushing a heavy stone in order to move it.
2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.
3) May change the direction of motion of an object.
13. What type of lens bends light outwards and away from a point?
concave
Answer:
No,it isn't concave. The correct answer is convex lens.
Explanation:
A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.
Convex lens is the answer.
See the attached diagram.
The attraction of a person's body toward the Earth is called weight. The reaction to this force is
Answer:
Person's body pulling on the earth
Explanation:
The weight of an object is the attraction of a person's body toward the Earth. The weight of an object is given by :
Weight (W) = mass (m) × acceleration due to gravity (g)
We know that, for an action there is an equal and opposite reaction.
So, there must be a reaction force for the weight of an object. The reaction force must be the person's body pulling on the earth.
