Driving a motor........
chemical energy is converted into kinetic energy.
Falling off of cliff
.........gravitational potential energy is converted into kinetic energy.
Hydroelectric energy generation
.......gravitational potential energy is converted into kinetic energy (i.e. driving a generator), which is then converted into electrical energy.
Nuclear power generation
.........mass is converted into energy, which then drives a steam turbine, which is then converted into electrical energy.
An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
the air pressure at the base of the mountain is 75.0cm of mercury while at the top is 60cm of mercury. Given that theaverage density of air is 1.25kg/m³ and the density of mercury is 13 600 kg/m³and gravity-10N/kg, calculate the height height of the mountain
Answer:
질문?
Explanation:
평균 공기 밀도가 1.25kg/m³이고 수은 밀도가 13600kg/m³이고 g=10N/kg인 경우 산 기슭의 기압은 수은의 75.0cm이고 정상의 수은은 60cm입니다. 산의 높이를 계산?
68.6 bags are required to fill the specified volume of 10"x20"x20".
It is 40 lbs/1.9 gallons, assuming US gallons, for the topsoil.
US gallons equal 3785 ml. 1 lb = 1/2.2 = 0.454 kg or 454 grammes since 1 kilogramme equals 2.2 lbs. 3785 x 1.9 = 7192 ml is equal to 1.9 gallons. 40 lbs = 18160 g. Therefore, the topsoil density is 2.52 g/cc (18160/7192).
The volume of the peat bag is 14x20x30 inches, which is equal to 35.6x50.8x76.2 cm (1 inch = 2/54 cm)=137806 ccs.
In other words, for the 40 lb again, 18160 grams/137806 equals 0.13 g/cc. The peat is therefore significantly lighter than topsoil.
The volume of the latter volume is 120"x240"x20" or 576,000 cubic inches, and the volume of a bag is 14"x20"x30" or 8400 cubic inches, hence 576,000/8400 = 68.6 bags are required to fill the specified volume of 10"x20"x20".
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Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
when do things move faster? Day or Night?
what is the distance time how can we find the speed of an object from its distance time graph
Answer:
speed is the gradient of the graph
Answer:
Speed is the slope of a distance time graph.
Explanation:
Speed= d/t
Slope is equal to rise/run
If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
A light spectrum is formed on the screen using a diffraction grating. The entire apparatus made up of laser, grating and the screen is now immersed in a liquid with refractive index 1.33. Do the bright spots on the screen get closer together, farther apart, remain the same or disappear entirely? Explain
Answer:
the points are closer to each other
Explanation:
The expression for the diffraction of a grating is
d sin θ = m λ
sin θ = m λ / d (1)
where d is the distance between slits and m is the order of diffraction, the most general is to work in the order m = 1, the angle te is the angle of diffraction
When we immerse the apparatus in a medium with refractive index n = 1.33, the light emitted by the laser must comply
v = λ f
where v is the speed of light in the medium, the frequency remains constant
velocity and refractive index are related
n = c / v
v = c / n
we substitute
c / n = λf
λ = [tex]\frac{c}{f} \ \frac{1}{n}[/tex]
λ = λ₀ / m
where λ₀ is the wavelength in vacuum
we substitute is equation 1
d sin θ = m λ₀ / n
sin θ = λ₀/ n d
sin θ = [tex]\frac{1}{n}[/tex] sin θ₀
we can see that the value of the sine is redueced since the refractive index is greater than 1,
consequently the points are closer to each other
A 2000 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
What is the maximum possible acceleration the truck can give the SUV?
At this acceleration, what is the force of the SUV's bumper on the truck's bumper?
Answer:
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
Explanation:
Concepts and reason
The concept required to solve this problem is Newton’s second law of motion.
Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.
Fundamentals
According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:
F = maF=ma
Here, m is mass and a is the acceleration.
(a)
Rearrange the equation F = maF=ma for a.
a = \frac{F}{m}a=
m
F
Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=
m
F
.
\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}
a=
(2300kg+2400kg)
18,000N
=
(4700kg)
18,000N
=3.83m/s
2
(b)
Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
for a and 2400 kg for m in the equation F = maF=ma .
\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}
F=(2400kg)(3.83m/s
2
)
=9120N
Ans: Part a
The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
.
Part b
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².
The force of the SUV's bumper on the truck's bumper is 10000N
What is acceleration?Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.
F = ma
And, a = F/m
Given, the mass of the ruck , m = 2000 Kg
The mass of the SUV, M = 2500 Kg
The total mass of the both = 2000 + 2500 = 4500 Kg
The maximum force on the trick , F = 18000 N
The maximum acceleration of the truck can give the SUV:
[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]
a = 18000/4500
a = 4 m/s²
The force of the SUV's bumper on the truck's bumper will be:
[tex]F_{max} -f= ma_{max}[/tex]
[tex]f= 18000-2000\times 4[/tex]
[tex]f =10000N[/tex]
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Given that the temperature of a body is 527K determine the value in degree C
Answer:
253.85°C
Explanation:
Here is the formula for converting K to °C
527K − 273.15 = 253.85°C
(a) Calculate the force needed to bring a 800 kg car to rest from a speed of 85.0 km/h in a distance of 115 m (a fairly typical distance for a nonpanic stop).
N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).
(force in part (b) / force in part (a))
Answer:
2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2
Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).
Explanation:
If the accuracy in measuring the velocity of a particle increases, the accuracy in measuring its position will:__________.
a. It is impossible to say since the two measurements are independent and do not affect each other.
b. remain the same.
c. increase.
d. decrease.
When the accuracy in measuring the velocity of a particle increases, that of its position decreases. The correct option is d. decrease.
The Heisenberg's uncertainty principle states that there is always an uncertainty in any attempt to measure accurately the value of any two complementary variables simultaneously. This implies that measuring of position and momentum of a particle simultaneously would lead to uncertainty in their values.
So that;
Δx.ΔP ≥ [tex]\frac{h}{2\pi }[/tex]
Where Δx is the uncertainty in the value of its position, ΔP is the uncertainty in the value of momentum and h is the Planck's constant.
This principle simply explains and provides the required answer to the given question. So that according to Heisenberg's uncertainty principle, the accuracy in measuring its position will definitely decrease.
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what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field.
Answer:
The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = [tex]\frac{U_{o}I }{2\pi r}[/tex]
where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.
But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A
So that;
B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]
= 1.8333 x [tex]10^{-5}[/tex]
B = 1.83 x [tex]10^{-5}[/tex] T
Imagine that you and your lab partner are standing on smooth ice holding onto opposite ends of a long rope. There is no friction between your feet and the ice. You tug gently on the rope. (a)Describe the motion of you and your lab partner before, during, and after the tug. Be as specific as you can. Expla
Answer:
Following are the complete solution to the given question:
Explanation:
In this question, before tug they are at rest, thus their initial momentum is 0 [tex]\pi=0[/tex], according to the law of conservation [tex]p_f=0[/tex] thus both should move in the opposite direction whose sum of the momentums is equaled to zero as there is no friction they will continue to move with their velocities until their collision. Its collision doesn't happen it will continue up to an infinite distance.
The motion of both you and your partner depends on the tension in the rope.
We must note that frictional force is the force that acts between two surfaces in contact with each other. The frictional force depends on the nature of the surfaces in contact.
Now, since there is no friction between your leg and the ice. The motion of both you and your partner depends on the tension in the rope.
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A particle moves along X-axis in such a way that X-coordinate varies with time according to expression x= 2-5t+6t2 meters, Calculate the initial velocity of the particle?
A 5
v= dt/ dx =−5+12t
Initial velocity means at t=0, which is −5+0=−5.
Thus, −v=5n
The velocity of a body is given by the equation v=a+bx, where 'x' is displacement. The unit of b is
Answer:
2bsnsnsnns181991oiwiw
A 2.0 kg frictionless puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.5 s?
Answer:
the distance moved by the puck after 2.5 s is 7.8 m
Explanation:
Given;
mass of the puck, m = 2 kg
initial velocity of the puck, u = 0
applied force, F = 5 N
time of motion, t = 1.5 s
Acceleration of the puck is calculated from Newton's second law of motion;
F = ma
a = F/m
a = 5/2
a = 2.5 m/s²
The distance moved by the puck after 2.5 s is calculated as;
s = ut + ¹/₂at
s = 0 + ¹/₂at²
s = ¹/₂at²
s = 0.5 x 2.5 x (2.5)²
s = 7.8 m
Therefore, the distance moved by the puck after 2.5 s is 7.8 m
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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Several years ago, there was a solar eclipse visible only from a remote part of Mexico (Los Cabos). To view this event, many thousands of tourists went to that remote area. What effect would this event have on the price of pesos in tersm of U.S. dollars
The price of Pesos is increased, and it takes more to buy a given amount of Pesos
According to the demand and supply economic theory, made popular by Adam Smith in 1776, which provides a price determination model based on the relationship between the amount of a good producers of the good are willing to sell at a given price and the amount of that good that the end users wish to purchase, there is an inverse relationship between price and supply, therefore as the supply decreases, the price increases, and therefore, there is a direct relationship between the demand and price, such that as the demand increases, the price of the goods increases because the ratio of supply to demand is decreased
Given that the arrival of thousands of tourists to the remote Los Cabos, will result in the tourists requiring Pesos, the demand of Pesos will rise, which according to the demand and supply theory, will increase the price of Pesos
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The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the propeller SLOWS at a steady rate of 0.9 revolutions per second per second. After 17 revolutions, how fast is the propeller spinning in revolutions per second
Answer: [tex]5.77\ rps[/tex]
Explanation:
Given
Initial angular velocity is [tex]\omega_i=8\ rps[/tex]
rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]
after 17 revolution i.e. [tex]\theta =17\ rev[/tex]
using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]
Insert the values
[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]
an object of volume has 20ml has a mass of 2.5kg what will be its density
Answer:
0.125
Explanation:
Density =mass/volume
Density =2.5/20
Density =0.125
Answer:
D=m/v
=2.5kg/(20/1000000)m^3
2.5kg÷0.000002m^3
1250000kgm^-3
Explanation:
20 is divided by 1000000 because m^3 is its si unit
HELP ME ASAP PLSSS!!
define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
Which of the following statements about magnetism is TRUE?
a) The direction of the magnetic force on a current-carrying wire is parallel to the wire.
b) Magnetic poles always occur in pairs (N and S).
c) Magnetic field lines begin at south poles and end on north poles.
d) Moving charges do not experience a force in magnetic fields.
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.
Answer:
always runs slower than normal.
Explanation:
The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.
According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.
In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.
) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].
Answer:
89.87m/s
Explanation:
Given the velocity function
v(t)=−t²+3t−2
In order to get the displacement function, we will integrate the velocity function as shown:
[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]
at t = 5
[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]
at t = -2
[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]
Required displacement = d(5) - d(-2)
Required displacement = 89.2 - (-0.67)
Required displacement = 89.2 + 0.67
Required displacement = 89.87m/s
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.30 cm?
A.
52.3 kN
B.
62.3 kN
C.
72.3 kN
D.
42.3 N
Answer:
cobina
Explanation:
me 2
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
If Light is travels from air into pure water with an incident angle of 30°. What is the angle of refraction?
Explanation:
For air, n1 = 1.00003; for water, n2 = 1.3330
Given: θ2 = 30 degrees, then
θ1 = arcsin [(n2/n1) sin θ2]
= arcsin [(1.3330/1.0003) sin (40)]
= 58.93 degrees
Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.