Answer:
Time taken for 1 swing = 3.81 second
Explanation:
Given:
Time taken for 1 swing = 2.20 Sec
Find:
Time taken for 1 swing , when triple the length(T2)
Computation:
Time taken for 1 swing = 2π[√l/g]
2.20 = 2π[√l/g].......Eq1
Time taken for 1 swing , when triple the length (3L)
Time taken for 1 swing = 2π[√3l/g].......Eq2
Squaring and dividing the eq(1) by (2)
4.84 / T2² = 1 / 3
T2 = 3.81 second
Time taken for 1 swing = 3.81 second
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static friction between tires and road is 0.50?
Answer:
v= 21.47m/sExplanation:
For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road
we will use the expression relating centripetal force and static friction below
let U represent the coefficient of static friction
Given that
U= 0.50
mass m= 1200-kg
radius r= 94.0 m
Assuming g= 9.81 m/s^2
[tex]U*m*g=\frac{mv^2}{r}[/tex]
[tex]U*g=\frac{v^2}{r}[/tex]
substituting our given data in to expression we can solve for the speed V
[tex]0.5*9.81=\frac{v^2}{94}[/tex]
making v the subject of formula we have
[tex]0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47[/tex]
v= 21.47m/s
hence the maximum velocity of the car is 21.47m/s
When water freezes, it expands about nine percent. What would be the pressure increase inside your automobile engine block if the water in there froze? The bulk modulus of ice is 2.0 × 109 N/m2, and 1 ATM = 1.01 × 105 N/m2.
Answer:
The pressure increase inside the automobile engine block is 1782.18 ATM
Explanation:
Given;
the change in volume of water, ΔV = 9%
the bulk modulus of ice, K = 2 x 10⁹ N/m²
Bulk modulus is given by;
[tex]K = -V\frac{dP}{dV}[/tex]
for pressure increase in the automobile engine block, when the water in there froze;
[tex]dP = K(\frac{dV}{V} )\\\\dP = K(\frac{0.09V}{V} )\\\\dP = 0.09K\\\\dP = 0.09 (2*10^9)\\\\dP = 1.8 *10^{8} \ N/m^2\\\\dP = 1782.18 \ ATM[/tex]
Therefore, the pressure increase inside the automobile engine block is 1782.18 ATM
The pressure increase inside your automobile engine block will be 1782.18 atm. The force involved vertical to the surface of an object per unit area is pressure.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
ΔV is the change in volume of water = 9%
K is the bulk modulus of ice = 2 x 10⁹ N/m²
dP is the change in the presure=?
The bulk modulus is found as;
[tex]\rm K=-V \frac{dp}{dv} \\\\[/tex]
The change in the presure is obtained as;
[tex]\rm dP = K\frac{dV}{V} \\\\ \rm dP = K\frac{0.09V}{V} \\\\ \rm dP = 0.09 K \\\\ \rm dP = 0.09 \times 2 \times 10^9 \\\\ \rm dP = 1.8 \times 10^8 \\\\ \rm dP =1782.18 \ atm[/tex]
Hence the pressure increase inside your automobile engine block will be 1782.18 atm.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
Why would someone choose a for-profit fitness center over a nonprofit center?
For-profits have important social missions that deserve support. For-profits have important social missions that deserve support. , ,
For-profits are more likely to be located in low-income communities.For-profits are more likely to be located in low-income communities. , ,
For-profits have services like homework support and job training. , For-profits have services like homework support and job training. , ,
For-profits usually have nicer equipment than nonprofit centers.For-profits usually have nicer equipment than nonprofit centers. , ,
Someone would choose a for-profit fitness center over a nonprofit center because for-profits usually have nicer equipment than nonprofit centers.
A for-profit institution is an institution that provides a service to society to obtain an economic profit from this service. On the other hand, non-profit institutions are those that provide a service to society and do not intend to make a profit, they are generally institutions that are part of the government, foundations, or non-governmental organizations.
Non-profit institutions can function thanks to donations, volunteers, public resources, and do not charge for providing the service they provide, while for-profit institutions charge for their service and a percentage of that charge is profit for the owner of the institution and another is reinvested in equipment, human resources or others to improve the service.
According to this, for-profits are likely to have better equipment, while non-profit are characterized by supporting missions and offering services to low-income communities.
Learn more in: https://brainly.com/question/13538740
The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)
Answer:
Following are the answer to this question:
Explanation:
Formula:
[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]
Calculating point A:
when the value is [tex]0.38[/tex]
[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]
[tex]=2.632[/tex]
[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]
[tex]=542,802.6[/tex]
Calculating point B:
when the value is [tex]0.75[/tex]
[tex]\to D(PC)=\frac{1}{0.75}[/tex]
[tex]=1.33[/tex]
[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]
[tex]=275,020[/tex]
Calculating point C:
when the value is [tex]0.28[/tex]
[tex]\to D(PC)=\frac{1}{0.28}[/tex]
[tex]=3.571[/tex]
[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]
[tex]=736660.7[/tex]
Calculating point D:
when the value is [tex]0.42[/tex]
[tex]\to D(PC)=\frac{1}{0.42}[/tex]
[tex]=2.38[/tex]
[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]
[tex]=490910.7[/tex]
Calculating point E:
when the value is [tex]0.31[/tex]
[tex]\to D(PC)=\frac{1}{0.31}[/tex]
[tex]=3.226[/tex]
[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]
[tex]=665370.97[/tex]
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answer:
The value is [tex]N = 3.708*10^{7} \ \ atoms[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]
The temperature is [tex]T = 26 + 273 = 299 \ K[/tex]
The volume is 1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]
Generally according to the ideal gas law we have that
[tex]PV = NkT[/tex]
here k is the Boltzmann constant with a value [tex]k = 1.38 *10^{-23} \ J/K[/tex]
=> [tex]N = \frac{PV}{ k T}[/tex]
=> [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]
=> [tex]N = 3.708*10^{7} \ \ atoms[/tex]
In Young's 2-slit interference experiment, the wavelength of laser light can be determined. The two slits are separated by 0.16 mm. The screen is 1.4 m from the slits. It is observed that the second bright band is located 11 mm from the center of the pattern. Given this information, what is the wavelength of the laser light?
a. 1258 nm
b. 419 nm
c. 500 nm
d. 629 nm
Answer:
d. 629 nm
Explanation:
slit separation d = .16 x 10⁻³ m
distance of screen D = 1.4 m
distance of second bright band = 11 x 10⁻³
distance of second bright band = 2 x band width
= 2 x λ D /d
Putting the values given ,
11 x 10⁻³ = 2 x λ x 1.4 / .16 x 10⁻³
λ = 1.76 x 10⁻⁶ / 2.8
= .6285 x 10⁻⁶
= 628.5 x 10⁻⁹
= 629 nm approx .
A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 2.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring
As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
W = ∆K ==> -0.18 J = 0 - 1/2 (2.5 kg) v ²
where v is the box's speed when it first comes into contact with the spring. Solve for v :
v ² ≈ 0.14 m²/s² ==> v ≈ 0.38 m/s
What did the results of photoelectric-effect experiments establish?
Answer:
Option A
Electrons are emitted if low intensity, high-frequency light hits a metal surface.
Explanation:
From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.
This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.
It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?
Explanation:
F = ma
85 N = m (15 m/s²)
m ≈ 5.7 kg
15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?
Answer:
69.58 m tall
Explanation:
Pls see attached file
A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.
Required:
Calculate the electric energy dissipated in this process.
Answer:
Explanation:
change in magnetic flux = .16 x .16 x .510 - 0
= .013056 weber .
rate of change of flux = change in flux / time
= .013056 / 40.5 x 10⁻³
= .32237
voltage induced = .32237 V
electrical energy dissipated = v² / R where v is voltage , R is resistance
= .32237² / 6.35
= 16.36 x 10⁻³ J .
Someone help me with these questions please!
Answer:
a 25 and b 25
2. 26
60n
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 29.0and 58.0, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 110 after it passes through the stack.
If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?
Answer:
I₂ = 143.79
Explanation:
To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,
Let's start with the set of three curling irons
Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes
I₁ = I₀ / 2
this light reaches the second polarized and must comply with the Mule law
I₂ = I₁ cos² tea
The angle between the first polarized and the second is Tea = 29.0º
I₂ = I / 2 cos² 29
The light that comes out of the third polarized is
I₃ = I₂ cos² tea
the angle between the third - second polarizer is
tea = 58-29
tea = 29th
I3 = (I₀ / 2 cos² 29) cos² 29
indicate the output intensity
I3 = 110
we clear
I₀ = 2I3 / cos4 29
I₀ = 2 110 / cos4 29
I₀ = 375.96 W / cm²
Now we have the incident intensity in the new set of three polarizers
back to the for the first polarizer
I₁ = I₀ / 2
when passing the second polarizer
I₂ = I1 cos² 29
I2 = IO /2 cos²29
let's calculate
I₂ = 375.96 / 2 cos² 29
I₂ = 143.79
One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:
A. the fringe system disappears
B. the bright fringes get brighter and the dark ones get darker
C. the fringes just get dimmer
D. the dark fringes just get brighter
E. the dark fringes get brighter and the bright ones get darker
Answer:
B. the bright fringes get brighter and the dark ones get darker
Explanation:
Let us consider when the intensities are equal, we use the equation
[tex]I_{max} = I_{1} + I_{2} + 2\sqrt{I*I}[/tex] for light fringes and,
[tex]I_{min} = I_{1} + I_{2} - 2\sqrt{I*I}[/tex] for dark fringes
where [tex]I_{1}[/tex] and [tex]I_{1}[/tex] are the light intensities from the first and second slits respectively.
For the first case where the light from the two slits have the same intensities, we can say both have intensity [tex]I[/tex]
[tex]I_{max} = I + I + 2\sqrt{I*I}[/tex] = [tex]2I + 2I = 4I[/tex]
[tex]I_{min} = I + I - 2\sqrt{I*I} = 2I - 2I = 0[/tex]
For the case where one of the intensities has half the intensity of the other.
one has intensity [tex]I[/tex] and the other one has intensity [tex]\frac{I}{2}[/tex]
inserting, we have
[tex]I_{max} = I + \frac{I}{2} + 2\sqrt{I*\frac{I}{2}} = 2.932I[/tex]
[tex]I_{min} = I + \frac{I}{2} - 2\sqrt{I*\frac{I}{2}} = 0.068I[/tex]
this shows that the bright fringes get brighter and the dark ones get darker.
6. If you wanted to develop a telescope, what kind of lenses would you use for the objective lens (the lens that collects the light) and the eyepiece? Explain your reasoning. Draw a picture with ray tracing of your setup.
Answer:
objetive: a converging lens for large diameter lenses
eyepiece you must select a lens with a small focal length and the diameter is not important
The selected lenses should decrease chromatic aberration.
Explanation:
A telescope is an instrument that collects light from very distant objects, therefore very weak.
Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.
For the eyepiece you must select a lens with a small focal length and the diameter is not important
the telescope magnification is
m = f_objective / F_ocular
The selected lenses should decrease chromatic aberration.
In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Explanation:
Meaning of power in physics
Answer:
The rate of doing work is called power.
Answer:
The amount of energy transported or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power.
OAmalOHopeO
Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood
Answer:
metal conducts heat better than wood.
Explanation:
Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.
There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,
1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N
Explanation:
Pressure is the same for both plungers.
P = P
F / A = F / A
F / (¼ π d²) = F / (¼ π d²)
F / d² = F / d²
5 N / (0.05 m)² = F / (1 m)²
F = 2000 N
None of the options are correct.
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the collision. What is the wavelength of the scattered photon?
0.6764*10^-10m
Explanation:
Using E= hc/wavelength
(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV
So subtract the calculated energy from the given energy of scattered photons
9.11-0.75=18.36 keV
To find wavelength
Wavelength= hc/ E
[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?
a. It will become four times as large.
b. It will double.
c. It will be cut in half.
d. It will become eight times as large.
e. It will be cut to one-quarter its original size.
Answer:
c : it wil be cut in half.
The pattern is formed on a distant screen so we can use the Fraunhofer difracction for a single slit. The formula of the width of the central bright spot is given by Δx = (2λz)/a, where λ is the wavelength and a is the width of the slit. So if the inicial width (a_1) is doubled (a_2= 2 x a_1),the width of the central spot will be cut in half Δx = (2λz)/a_2 = (2λz)/2xa_1 .
A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?
Answer:
24
Explanation:
The mass = 3 kg
at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.
PE = mgh
but ME = PE + KE = constant (law of energy conservation)
KE is the kinetic energy
since KE is zero at this point, then,
ME = mgh
where m is the mass
g is acceleration due to gravity = 9.81 m/s^2
h is the height = O
ME = 3 x 9.81 x O
ME = 29.43-O
At point A the total ME is due to its PE and its kE
PE at this point = mgh = 3 x 9.81 x A = 29.43-A
KE = [tex]\frac{1}{2}mv^{2}[/tex]
velocity = v at this point, therefore,
KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]
therefore,
ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
Equating ME for the points O and A, we have
29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]
(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]
O - A = 0.051[tex]v^{2}[/tex] this is the distance between point O and A
For point B
PE = 29.43-B
KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex] (velocity is equal to [tex]5v[/tex] at this point)
therefore,
ME = 29.43-B + 37.5[tex]v^{2}[/tex]
Equating the ME for points A and B, we have
29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]
29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]
(A - B)29.43 = 36[tex]v^{2}[/tex]
A - B = 1.22[tex]v^{2}[/tex] this is the distance between points A and B
The distance between points A & B divided by the distance between points 0 & A will be
1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
[tex]d\sin\theta=m\lambda[/tex]
d = 1/N, N = number of lines per mm
[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface
Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced
Answer:
4.635 *10^12 Neutrinos
Explanation:
Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.
We proceed as follows;
In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.
From the question, we are given that
Power P = 9.9 watts = 9.9 J/s
Watts is same as J/s
The number of proton-proton cycles required to generate E energy is N = E / E '
Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question
Converting Mev to J, we have
= 26.7 x1.6 x10 -13 J
To get the number N which is the number of proton-proton cycle required, we have;
N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12
Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.
Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos
The two metallic strips that constitute some thermostats must differ in:_______
A. length
B. thickness
C. mass
D. rate at which they conduct heat
E. coefficient of linear expansion
Answer:
E. Coefficient of linear expansion