Answer:
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
Explanation:
We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.
P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.
Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node
Which is the definition of refraction?
1)the blocking of light waves vibrating in a particular plane
2) the bending of a light wave as it passes at an angle from one medium to another
3) a false or distorted image causing the gradual distortion of light through hot air
the redirection of light by tiny particles as it passes through a medium
Answer:
2) The bending of a light wave as it passes at an angle from one medium to another .
Hope it is helpful to you ☺️
Two objects are interacting, but stay stationary. Which best describes what is happening to the action and react
forces?
There are no forces acting on the objects.
The forces are equal and opposite each other.
One object is exerting more force than the other.
O Their forces are occurring at different times.
Answer:b the force are equal and opposite each other
Explanation:
A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .
Hurry !! First one to answer gets points !
Answer:
paleontologist
Explanation:
Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.
What is Fossil?A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.
An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.
Therefore, the correct option is B.
Learn more about Fossils, here:
https://brainly.com/question/5431129
#SPJ7
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
convert 2.4 milimetres into metre
Answer: 2.4 millimeters = 0.0024 meters
Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.
How many neutrons are in an isotope of selenium-83
A. 34
B. 83
C. 49
D. 117
Answer:
C.49 is yr ans...
hope it helpsstay safe healthy and happy....A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?
Answer:
The answer is "151.25 J and -547.64 J".
Explanation:
[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]
Using formula:
[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]
[tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]
[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]
Calculating the Work by net force
[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]
The above work is converted into thermal energy.
Now,
[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
Learn more about work done at: https://brainly.com/question/25573309
#SPJ6
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]
Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]
We know that,
Force, F = ma
Put all the values,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]
So, the required force is 6666.7 N.
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
Answer:
[tex]\sigma=0.014\ C/m^2[/tex]
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]
So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].
It takes 20 Joules of Work to push 4 coulombs of charges Across the filament of a bulb.'find the potential difference Across the filament
Answer:
V = 5 Volts
Explanation:
Given the following data;
Work done = 20 Joules
Charge = 4 Coulombs
To find the potential difference;
Mathematically, the work done in moving a charge is given by the formula;
W = qv
Where;
W is the work done
q is the quantity of charge
v is the potential difference
Substituting we have;
20 = 4 * v
V = 20/4
V = 5 Volts
A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
Question 8 of 10
What was the name of the book that Ibn al-Haytham wrote?
A. Weather and Air Flow
B. Book of Optics
C. Light and Vision
D. Book of Sound
Answer:b
Explanation:
1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?
Answer:
1) attached below
2) assumption that the earth is spherical
Explanation:
1) Four illustrations of a globe
attached below
2) Reason for distortions at areas that do not fall on lines of tangency or secancy
The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true
2. How do the phytochemicals present in various foods help us?
Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.
collisions may take place between
Answer:
Collisions may take place between the reactants.
Explanation:
The collision frequency must be greater than the frequency factor for the reaction. A collision between the reactants must occur.
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s
Answer:
8m/s
Explanation:
Vavg= 16-0/2=8m/s
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
Typhoon signal number 2 is raised. What is the speed of the expected typhoon?
the simple answer is from 61kmph to 120kmph
Explanation:
no explanation is needed
Hi, so i have to find T1, can some1 help?
30.1 N
Explanation:
Given:
[tex]W_1 = 16\:\text{N}[/tex]
[tex]W_2 = 8\:\text{N}[/tex]
Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.
Forces involving W1:
[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]
[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]
Forces involving W2:
[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]
[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]
Substitute (2) into (3) and we get
[tex]T_1\sin 53 - W_1 = W_2[/tex]
Solving for [tex]T_1[/tex],
[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.
Answer:
A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT
Explanation:
This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.
Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s
subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s
Initial instant. Before the crash
p₀ = m₁ v₀₁ + m₂ v₀₂
Final moment. After the crash
p_f = m₁ v_{f1} + m₂ v_{f2}
p₀ = p_f
m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}
as the shock is elastic, energy is conserved
K₀ = K_f
½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]
m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)
let's make the relationship
(a + b) (a-b) = a² -b²
m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
let's write our two equations
m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)
m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
we solve
v₀₁ + v_{f2} = v_{f2} + v₀₂
we substitute in equation 1 and obtain
M = m₁ + m₂
[tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]
[tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2
we calculate the values
m₁ + m₂ = 0.160 +0.3000 = 0.46 kg
v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]
v_{f1} = -0,250 - 2,961
v_{f1} = - 3,211 m / s
v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]
v_{f2} = 0.570 - 0.6909
v_{f2} = -0.12 m / s
now we can answer the different questions
A) v_{f1} = -3.2 m / s
B) the negative sign indicates that it moves to the left
C) v_{f2} = -0.12 m / s
D) the negative sign indicates that it moves to the LEFT
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTwhat is the time taken by moving body with acceleration 0.1m/s2 if the initial or finak velocities are 20m/s and 30m/s respectively?
Answer:
t= 100s
Explanation:
use v=v0+at
plug in givens and solve for t
30=20+0.1*t
t= 100s
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?
Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are
• A : x = (19.5 m/s) cos(30°) t
• B : x = 62.2 m + (19.5 m/s) cos(60°) t
These positions are the same at the moment one projectile is directly above the other, which happens for time t such that
(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t
Solve for t :
(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m
t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))
t ≈ 8.71 s
The Cleveland City Cable Railway had a 14-foot-diameter pulley to drive the cable. In order to keep the cable cars moving at a linear velocity of 14 miles per hour, how fast would the pulley need to turn (in revolutions per minute)
Answer:
13.94 rpm
Explanation:
Given that,
The diameter of the pulley, d = 14 foot
Radius, r = 7 foot
The linear velocity of the pulley, v = 14 mph = 20.53 ft/s
We need to find the angular velocity in rpm.
We know that, the relation between the linear velocity and the angular velocity is as follows :
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]
or
[tex]\omega=13.94\ rpm[/tex]
So, the angular velocity of the pulley is 13.94 rpm.
