Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
Which graph would be created by a pendulum with the greatest amplitude?
Answer:
Graph (c) would be created by a pendulum with the greatest amplitude.
Explanation:
The amplitude of a wave is the greatest displacement covered by an object. It refers to the maximum amount of displacement of a particle on the medium from its rest position. It is the distance from rest to crest.
Out of three graphs, the amplitude is greatest in graph 3 as the distance from rest is crest in this case is maximum. Hence, the correct option is (c).
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26
with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.
Answer:
(7.115 ; 13.405)
Explanation:
Given :
Sample size, n = 22
Mean bill, μ = 10.26
Standard deviation, s = 5.21
To obtain the 99% confidence interval for the mean bill of all orders ;
Mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 99%, df = n-1, 22 - 1 = 21
Tcritical = 2.831
Margin of Error = 2.831 * (5.21/√22) = 3.145
Confidence interval = 10.26 ± 3.145
Lower boundary = 10.26 - 3.145 = 7.115
Upper boundary = 10.26 + 3.145 = 13.405
Confidence interval :
(7.115 ; 13.405)
A system is acted on by its surroundings in such a way that it receives 50 J of heat while simultaneously doing 20 J of work. What is its net change in internal energy
Answer:
30J
Explanation:
Given data
The total quantity of heat recieved= 50J
Quantity of heat used to do work= 20J
Hence the net change is
ΔU= Total Heat - Net work
ΔU= 50-20
ΔU= 30J
Hence the change in the internal energy is 30J
A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?
Answer:
W = Q * V work done on charge Q
A. W = .5 C * 1.5 V = .75 Joules
B. P = W / t = .75 J / 1 sec = .75 Watts
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]
Diameter of the wheel, d=21 cm
Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm
Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]
1m=100 cm
Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,[tex]a_t=\alpha r[/tex]
[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]
[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]
Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]
We know that
[tex]a=\sqrt{a^2_t+a^2_r}[/tex]
Using the formula
[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]
Squaring on both sides
we get
[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]
[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]
[tex]v^4=r^2\times 2.7699[/tex]
[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]
[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]
[tex]v=0.418 m/s[/tex]
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a
velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer
Answer:
5.25 m/s to the east
Explanation:
Applying,
MV = mv.............. Equation 1
Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon
make V the subject of the equation
V = mv/M............ Equation 2
From the question,
M = 2 kg, m = 0.75 kg, v = 14 m/s
Substitute these values into equation 2
V = (0.75×14)/2
V = 5.25 m/s to the east
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
A study finds that the metabolic rate of mammals is proportional to m^3/4 , where m is the total body mass. By what factor does the metabolic rate of a 70.0-kg human exceed that of a 4.91-kg cat?
Answer:
The mass of human is 2898 times of the mass of cat.
Explanation:
A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.
[tex]M=\dfrac{km^3}{4}[/tex]
Where
k is constant
If m = 70 kg, the mass of human
[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]
If m = 4.91 kg, the mass of cat
[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]
So,
[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]
So, the mass of human is 2898 times of the mass of cat.
Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.
Answer:
Force is zero.
Explanation:
According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.
F = m a
if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.
Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.
Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.
Which best describes the relationship between heat,intemal energy, and thermal energy?
Internal energy is heat that flows and heat is the part of thermal energy that can be transferred
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred,
Thermal energy is heat that flows, and heat is the part of intemal energy that can be transferred
Heat is thermal energy that flows, and hennal energy is the part of internal energy that can be transferred.
Answer:
It is all a thermodynamic system that is highly related to each other.
Explanation:
Because they are in the physics of thermodynamics it is not wrong to say they follow the same thermodynamic rules and has highly the same properties of energy.
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
hope it is helpful to you
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
There are two possible alignments of a dipole in an external electric field where the dipole is in equilibrium: when the dipole moment is parallel to the electric field and when the dipole moment is oriented opposite the electric field.
Part A
Are both alignments stable? (Consider what would happen in each case if you gave the dipole a slight twist.)
a) Yes
b) No
Part B
Based on your answer to the previous part and your experience in mechanics, in which orientation does the dipole have less potential energy?
a) The arrangement with the dipole moment parallel to the electric field has less potential energy.
b) The arrangement with the dipole moment opposite the electric field has less potential energy.
c) Both arrangements have the same potential energy.
Answer:
A. (b)
B. (a)
Explanation:
The electric dipole moment is the product of charge and the length of the dipole.
The torque on the dipole placed in the external electric field is given by
torque = p E sin A
where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.
When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.
When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
So, the option (b) is correct.
Teh energy is given by
U = - p E cos A
When the angle A is zero , the potential energy is negative and it is minimum.
In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:
A) Letter b
B) Letter a
So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:
[tex]torque = p E sin (A)[/tex]
where:
p: the electric dipole momentE: the electric fieldA: the angle between the field and dipole momentWhen the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
Now the energy is given by:
[tex]U = - p E cos (A)[/tex]
We can say that when the angle A is zero , the potential energy is negative and it is minimum.
See more about dipole at brainly.com/question/12757739
The value of mass remains constant but weight changes place to place why
Explanation:
No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).
Explanation:
Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.
A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Answer:
64.20m
Explanation:
As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.
[tex]a^{2} +b^{2} = c^{2}[/tex]
where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x
[tex]51^{2} +39^{2} =x^{2}[/tex]
2,601 + 1,521 = [tex]x^{2}[/tex]
4,122 = [tex]x^{2}[/tex] ... square root both sides
64.20 = x
Finally, we see that the shortest distance is 64.20m
water contracts on freezing is it incorrect or conrrect
Answer:
hope it helps
much as you can
The mass of the moon is 7.2 × 10^22 kg and its radius is 1.7×10^6 m.What will be the gravity of the moon to a body of the mass 1 kg on the surface of the moon.
Answer:
1.66 N
Explanation:
The force of gravity of the moon on the body is given by
F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.2 × 10²² kg, m = mass of body = 1 kg and R = radius of moon = 1.7 × 10⁶ m
Substituting the values of the variables into the equation, we have
F = GMm/R²
F = 6.67 × 10⁻¹¹ Nm²/kg² × 7.2 × 10²² kg × 1 kg/(1.7 × 10⁶ m)²
F = 48.024 × 10¹¹ Nm²/2.89 × 10¹² m²
F = 16.62 × 10⁻¹ N
F = 1.662 N
F ≅ 1.66 N
So, the gravity on the moon is 1.66 N
Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off
Answer:
Following are the solution to the given question:
Explanation:
Please find the complete question in the attached file.
The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.
[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]
Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]
The electricity used is continuously 694W over 30 days.
If just resistor loads (no reagents) were assumed,
[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]
Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]
This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.
Soap bubble coloring example:
(reflection, interference, refraction, diffraction)
Explanation:
Interference is the example of soap bubble colouringEXTRA INFO:(LOOK AT THE IMAGE)
An incoming light ray is partly reflected by the top surface of the soap film and partly reflected by the bottom surface. The wave reflected from the bottom surface has traveled further (an extra distance equal to twice the thickness of the film) so emerges out of step with the top wave. When the two waves meet, they add together, and some colors are removed by destructive interference. Where the film is thickest, the bubble appears more blueish; where it's thinner, it will look more violet or magenta.
[tex]\huge\bold\color{black}{ANSWER}[/tex]
Soap bubble coloring example: INTERFERENCE
how can scientific method solve real world problems examples
A black T-shirt is warmer in the summertime than a white T-shirt because the black T-shirt
A. Is reflecting all wavelengths of light.
B. Absorbs violet light, the highest energy wavelength.
C. Is absorbing all wavelengths of light. D. Doesn’t absorb red, the longest wavelength.
Answer:
c
Explanation:
darker colors absorb app light
Answer:
C. Is absorbing all wavelengths of light.
Explanation:
Black isn't a color, but rather the absence of color. We see a T-shirt as black because it isn't reflecting any light toward our eyes. A black T-shirt absorbs all of the wavelengths of light, causing it to absorb more energy and become warmer than white, which reflects light.
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
now suppose that we have attached not just two springs in series, but N springs. Write an equation that expresses the effective spring constant of the combination using the spring constant of the original spring k and the number of springs N
Answer:
[tex]k_{eq} = \frac{k}{N}[/tex]
Explanation:
For this exercise let's use hooke's law
F = - k x
where x is the displacement from the equilibrium position.
x = [tex]- \frac{F}{k}[/tex]
if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs
x_ {total} = ∑ x_i
we substitute
x_ {total} = ∑ -F / ki
F / k_ {eq} = -F [tex]\sum \frac{1}{k_i}[/tex]
[tex]\frac{1}{k_{eq}} = \frac{1}{k_i}[/tex] 1 / k_ {eq} = ∑ 1 / k_i
if all the springs are the same
k_i = k
[tex]\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\[/tex]
[tex]\frac{1}{k_{eq} } = \frac{N}{k}[/tex]
[tex]k_{eq} = \frac{k}{N}[/tex]