Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
Infrared and ultraviolet waves have different frequencies.
Both types of wave can have harmful effects on human beings.
Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves.
Answer:
For infrared and ultraviolet waves have different frequencies. Both types of wave can have harmful effects on human beings. Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves. Medical studies indicate that prolonged IR exposure can lead to lens, cornea and retina damage, including cataracts, corneal ulcers and retinal burns, respectively. To help protect against long-term IR exposure, workers can wear products with IR filters or reflective coatings.When you look at the EM spectrum, UV waves are quite a bit smaller in wavelength than infrared, and x-rays/gamma rays are even smaller. Therefore, UV waves are probably causing more harm than infrared waves, and x-rays/gamma rays are probably doing even more damage.
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Infrared and ultraviolet waves have different frequencies. Infrared waves have lower frequencies and longer wavelengths, while ultraviolet waves have higher frequencies and shorter wavelengths.
Harmful effects of Infrared waves:
Infrared waves have lower frequencies and are often associated with heat radiation. Prolonged exposure to intense infrared radiation can lead to thermal burns and damage to the skin and eyes. Infrared radiation can also cause dehydration and overheating of the body, especially in hot environments. While infrared radiation is not as harmful as ultraviolet radiation, excessive exposure can still lead to health issues.
Harmful effects of Ultraviolet waves:
Ultraviolet waves have higher frequencies and shorter wavelengths, making them more energetic than infrared waves. UV radiation from the sun is a well-known harmful agent. Short-term exposure to intense UV radiation can cause sunburn, skin redness, and eye irritation. Long-term exposure to UV radiation can lead to more serious health problems such as skin aging, cataracts, and an increased risk of skin cancer. UV radiation can also damage DNA in skin cells, leading to mutations and potential carcinogenesis.
It is essential to protect ourselves from both infrared and ultraviolet waves to prevent harmful effects. Using sunscreen and wearing protective clothing can help shield the skin from UV radiation. Limiting exposure to intense sources of infrared radiation, such as hot objects or infrared heaters, can help reduce the risk of thermal burns and overheating. Understanding the differences in the frequencies of these waves allows us to implement appropriate safety measures and protect our health.
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What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.
Answer:
It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.
Explanation:
The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.
The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.
Use the general formulas for gravitational force and centripetal force to derive the relationship between speed (v) and orbital radius (r) for circular orbits. Show the relationship you derived to your instructor before going on.
Solution :
We know that :
Formula for Gravitational force is given by :
[tex]$F_g=\frac{Gmn}{r^2}$[/tex]
where, G is the gravitational constant
M is the mass of the bigger body
m is the mass of the smaller body
r is the distance between the two bodies.
And the formula for the centripetal force is given by :
[tex]$F_c=\frac{mv^2}{r}$[/tex]
where, m is the mass of the rotating body
v is the velocity
r is the radius of rotation of the body.
We know that mathematically, the gravitational force is equal to the centripetal force of the body.
Therefore,
[tex]$F_g=F_c$[/tex]
[tex]$\frac{GMm}{r^2}=\frac{mv^2}{r}$[/tex]
[tex]$\sqrt{\frac{GM}{r}}=v$[/tex]
Hence derived.
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest point? (b) How high does the ball rise? (c) How long after projection will the ball have a velocity of 10m/s upward? A velocity of 10m/s downward? (d) When is the displacement of the ball zero? (e) When is the magnitude of the ball’s velocity equal to half its velocity of projection? (f) When is the ball‘s displacement equal to half the maximum height to which it rises? (g) What is the magnitude and direction of the acceleration while the ball is moving upward? While moving downward? While at the highest point?
All the answers are:
a) The time that will it take to rise to the highest point is 3.06 seconds.
b) The ball will rise to a height of 45.87 meters.
c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.
The time when the ball has 10 m/s downward is 1.02 seconds.
d) The displacement of the ball will be zero at 6.12 seconds.
e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.
f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.
g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
Let's calculate the values for each case.
a) At the highest point, the final velocity is 0, so we can use the following equation.
[tex]v_{f}=v_{i}-gt[/tex] (1)
Where:
v(i) is the initial velocityv(f) is the final velocityg is the acceleration due to gravity (9.81 m/s²)We know that v(i) = 30 m/s.
[tex]0=30-9.81t[/tex]
Solve it for t:
[tex]t=3.06\: s[/tex]
Hence, the time is 3.06 s.
b) At the highest point, the final velocity is 0, so we can use the following equation.
[tex]v_{f}^{2}=v_{i}^{2}-2gh[/tex] (2)
[tex]0=v_{i}^{2}-2gh[/tex]
We know that the initial velocity is 30 m/s.
[tex]0=30^{2}-2gh[/tex]
Solving it for h we have:
[tex]h=\frac{30^{2}}{2*9.81}[/tex]
[tex]h=45.87 \: m[/tex]
Then, the height is 45.87 m.
c) Using equation (1) we can find the time (t).
[tex]10=30-(9.81t)[/tex]
So, the time elapsed to get 10 m/s is:
[tex]t_{upward}=2.04\: s[/tex]
We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.
[tex]t_{upward}=2.04+t[/tex]
[tex]t=1.02\: s[/tex]
This is the time when the ball has 10 m/s downward.
Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.
d) It will be when the ball returns to the ground.
[tex]t=2t_{upward}[/tex]
[tex]t=2*3.06[/tex]
[tex]t=6.12\: s[/tex]
The displacement will be zero after 6.12 s.
e) Here we need to find the time when v(f) is 15 m/s
[tex]15=30-gt[/tex]
[tex]t=\frac{15}{9.81}[/tex]
[tex]t=1.53\: s[/tex]
The time when the v(f) is 15 m/s is 1.53 s.
f) Here, we need to find t when h = 45.87/2 m = 22.94 m
We can use the next equation:
[tex]h=v_{i}t-0.5gt^{2}/tex]
[tex]22.94=30t-0.5*9.81*t^{2}/tex]
Solving this quadratic equation, t will be:
[tex]t=0.90\: s/tex]
Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.
g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
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Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one
Answer:
F' = 16 F
Hence, the electric force between charges becomes sixteen times its initial value.
Explanation:
The electric force between the two charges is given by the Colomb's Law:
[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)
where
F = electric force
K = Colomb's Constant
Q₁ = magnitude of the first charge
Q₂ = magnitude of the second charge
R = Distance between charges
Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:
[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]
using equation (1):
F' = 16 F
Hence, the electric force between charges becomes sixteen times of its initial value.
which one of the following is a product of an acid base reaction? A. Base B. Acid C. Salt D. Fire
Answer:
salt
Explanation:
salt is a component for many acid base reactions
Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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Answer:
OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
What is abstraction?Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.
One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
Hence option D is correct.
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Integrated Concepts Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another.
(a) Calculate the potential energy of two singly charged nuclei separated by 1.00×10–12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
(b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
(a) 2.3 x 10^-16 J
(b) 1.1 x 10^7 K
Explanation:
charge, q = 1.6 x 10^-19 C
distance, r = 10^-12 m
(a) Let the potential energy is U.
[tex]U = \frac{k q q}{r^2}\\\\U = \frac{9\times 10^{9}\times 1.6\times 1.6\times 10^{-38}}{10^{-12}}\\\\U = 2.3\times 10^{-16} J[/tex]
(b) Let the temperature is T.
[tex]U = K = \frac{3}{2} kT\\\\2.3 \times 10^{-16} = 1.5\times 1.38\times 10^{-23} T\\\\T = 1.1\times 10^7 K[/tex]
Calculate area moment of inertia for a circular cross-section with 3 mm diameter:
Answer:
circles
A=7.07multiple 10-6 m2
I hope you understand and help
The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].
What is the area?The measurement that represents the size of a region on a plane or curved surface is called an area.
What is cross-section?A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.
Given data:
Diameter = 3 × [tex]10^{-3} m[/tex].
It is known that. Diameter = 2 radius.
The area can be calculated by using the formula:
A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].
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The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
Answer:
The wavelength stays the same.
Explanation:
When the amplitude is increased, the wavelength stays the same.
Here the wavelength doesn't depend upon the amplitude.
A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of the magnetic field if there is a 45o angle between it and the proton's velocity
Answer:
the strength of the magnetic field is 3 x 10⁻⁵ T
Explanation:
Given;
velocity of the cosmic ray, v = 5 x 10⁷ m/s
force experienced by the ray, f = 1.7 x 10⁻¹⁶ N
angle between the ray's velocity and the magnetic field, θ = 45⁰
The strength of the magnetic field is calculated as;
[tex]F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T[/tex]
Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T
I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)
Solve for a
Answer:
a. solve for a
[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]
I hope I helped you ^_^
How to calculate voltage U1 ?
Please help!
Answer:
he is a baby art and design
PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!
Vf^2 = v0^2 + 2a (xf -x0)
Solve for v0
b. solve for Vo
[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]
I hope I helped you ^_^
What would be the consequences if the animated structures suddenly become non-polar
Answer:
the lipid bilayer would not be able to hold its shape in water and the cell membrane would disassemble. Only a lipid monolayer would be possible. The fatty acid "tails' would fall off The phospholipids would flip upside down.
Explanation:
hope this helps
write a use of magnetic force and frictional force each
which option is correct n why?
6. The projectile motion is a good example of
A. one dimensional motion.
B. two dimensional motion.
C. three dimensional motion.
D. four dimensional motion.
2. two dimensional motion
Because it has just 2 dimensions x and y
A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1
The length of the brass at a temperature rise of 100 K is 2.0036 m
From the question given above, the following data were obtained:
Original length (L₁) = 2 m
Temperature rise (ΔT) = 100 K
Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹
Final length (L₂) =?The final length of the brass can be obtained as follow:
α = L₂ – L₁ / L₁ΔT
18×10¯⁶ = L₂ – 2 / (2 × 100)
18×10¯⁶ = L₂ – 2 / 200
Cross multiply
L₂ – 2 = 18×10¯⁶ × 200
L₂ – 2 = 0.0036
Collect like terms
L₂ = 0.0036 + 2
L₂ = 2.0036 m
Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m
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If "FRICTION" means breaking up of relationships, then how can you reduce friction among friends?
A. Respect the opinion of your friends
B. Learn how to be understanding
C. Avoid bullying your friends
D. Force your friends to do things they don't like
Answer:
A
Explanation:
respect the opinion of your friends
Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?
Answer:
Explanation:
(a)Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
(a)Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
write state of matter with 5 example of each
There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).
1. Solid
Examples: Iron, wood, steel, ice, paper
2. Liquid
Examples: Water, mercury, milk, soup, juice
3. Gas
Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen
a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
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What is a measure between the difference in start and end positions?
Answer:
Displacement
General Formulas and Concepts:
Kinematics
Displacement vs Total DistanceExplanation:
Displacement is the difference between the start position and end position.
Total Distance is the entire distance traveled between the start and end position.
Topic: AP Physics 1 Algebra-Based
Unit: Kinematics
Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?
Answer:
[tex]_{18}^{39} } Ar[/tex]
Explanation:
The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.
[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]
Argon is a stable gas and is found in the group 8 on the periodic table of elements.
Answer:
Answer is below
Explanation:
39 18 Ar
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
