Answer:
so true though
Explanation:
Ultimate frisbee relies upon good sportsmanship since there are no referees and players must self-officiate the game. What is this known as?
Group of answer choices
Spirit of the Team
Spirit of the Frisbee
Spirit of Sportsmanship
Spirit of the Game
Answer:
spirit of the team / game
Evidence supporting the theory of continental drift includes:
a.
b.
c.
d .
Answer:
The apparent fit of the eastern coastline of South America and western coastline of Africa
Similarities of plants and animal fossils in South America and some parts of African continent which were separated by a vast ocean
Similarities in the sequence of rock layers of opposite sides of the Atlantic Ocean
A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satellite's orbital speed. 9.82278e7 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) Find the period of its revolution. h (c) Find the gravitational force acting on it.
Answer:
a) v = 5.59x10³ m/s
b) T = 4 h
c) F = 1.92x10³ N
Explanation:
a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:
[tex] F_{c} = F_{G} [/tex]
[tex]\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}[/tex]
[tex] v = \sqrt{\frac{gr^{2}}{r+h} [/tex]
Where:
g is the gravity = 9.81 m/s²
r: is the Earth's radius = 6371 km
h: is the satellite's height = r = 6371 km
[tex]v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s[/tex]
b) The period of its revolution is:
[tex] T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h [/tex]
c) The gravitational force acting on it is given by:
[tex] F = \frac{GMm}{(r + h)^{2}} [/tex]
Where:
M is the Earth's mass = 5.97x10²⁴ kg
m is the satellite's mass = 782 kg
G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²
[tex] F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N [/tex]
I hope it helps you!
explain and derive the equation for capillary action in the phenomenon of surface tension
Answer:
Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.
Look at this model of an atom. Where are the protons located and how many are there?
Answer:
protons are in the nucleus .
Explanation:
there are 6 protons
Katniss everdeen applies 20 n of force back on her bow what happens to the arrow when she lets go?
Objects in free fall are weightless.
Please select the best answer from the choices provided
T
F
Answer:
[tex]\boxed{True}[/tex]
Explanation:
Objects are weightless in free fall because their is no other force pushing them rather than the gravitational force.
Hope it helps!<3
Answer:
It's ture hope this answer help have a great day .!Explanation:
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleration α, the torque on the child is τ. The child moves to a position half way between the center and edge of the merry-go-round, and the angular acceleration increases to 2α. The torque on the child is now
Answer:
The torque on the child is now the same, τ.
Explanation:
It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:[tex]I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)[/tex]
So, τ = 3/2*m*r²*α (2)When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:[tex]I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)[/tex]
Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and
the coefficient of static friction between tires and dry road, Ms = 0.5. Calculate the maximum
speed the car can have and still make the turn successfully
The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have
• net vertical force:
∑ F = N - W = 0
• net horizontal force:
∑ F = Fs = m a
where
N = magnitude of normal force
W = car's weight
Fs = mag. of static friction
m = car's mass
a = v ²/R = mag. of the centripetal acceleration
v = car's speed
R = radius of curve
Now,
• compute the car's weight:
W = m g = (1500 kg) (9.8 m/s²) = 14,700 N
• solve for the mag. of the normal force:
N = 14,700 N
• solve for the mag. of the friction force, using the given friction coefficient:
Fs = 0.5 N = 7350 N
• solve for the (maximum) acceleration:
7350 N = (1500 kg) a → a = 4.9 m/s²
• solve for the (maximum) speed:
4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s
Based on the law of conservation of energy, which statement is correct?
A.
Energy is always being added to all parts of the Universe.
B.
Energy is often destroyed in some parts of the Universe.
C.
Energy in a closed system cannot change forms.
D.
Energy in an isolated system remains constan
Answer:
D
Explanation:
Nothing can enter or leave so it remains constant
2. A car accelerates at a rate of 1.4 m/s. Find the mass of the car if a 2250 N net force is
required to produce this acceleration.
Answer:
1607.14 kgExplanation:
The mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{2250}{1.4} \\ = 1607.14285...[/tex]
We have the final answer as
1607.14 kgHope this helps you
You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.
Answer:
Explanation:
When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed
At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.
F = ma
Acceleration is the change in the velocity of an object per change in time of motion.
At constant velocity, the acceleration of an object is zero.When acceleration of an object is zero, the force on the object is zero.A constant speed (magnitude only) and change in the direction of the object, implies a change in velocity of the object.at changing velocity, the acceleration on an object is positive, and hence net force acts on the object.Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
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Prob. 3: Manifestation of quantum phenomena (total 25 points) (a)-(e) 5 points each.
Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the following notion of quantum physics. In either experiments or observation, please attach a brief explanation and justification to your statement.
3-(A) Vibration of molecules can be treated by quantum mechanics using a parabolic potential. In classical cases, we can use mass(es) attached to a spring. In quantum mechanics, energy levels are quantized, unlike classical oscillators. How can you demonstrate that?
3-(B) The ground state energy of classical oscillator is zero, while that of quantum oscillators has finite (zero-point) energy. How can you demonstrate that (not by calculations but by some experiments or observations)?
3-(C) Electrons can have both orbital and spin angular momentum, and associated magnetic moment. There are two spin quantum numbers (+1/2 or -1/2) for an electron. Describe some quantum phenomena or observation which demonstrate(s) that an electron has a spin 1/2.
3-(D) Laser light is a coherent light. Which property of laser depends on this feature? What application is related to the coherence?
3-(E) Quantum particle(s) can tunnel through a potential barrier.
Answer:
3A. This phenomenon can be seen in the discrete emission of the molecules.
3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.
3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams
3D. This is due to the stimulated emission
3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)
Explanation:
This problem asks for some experimental explanations of various quantum phenomena.
3A. This phenomenon can be seen in the discrete emission of the molecules.
In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed
In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms
3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.
The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen
3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.
3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .
This is widely used for holographic and interference work
3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.
Answer:
Explanation:
Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the. Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature. Quantum mechanics arose gradually from theories to explain observations. By M Arndt · 2009 · Cited by 239 — Wave-particle duality of light: A paradigmatic example for this fact is the dual nature of light that manifests itself in Young's famous double-slit diffraction.
Two locomotives approach each other on parallel track. Each has speed of 95 km/h with respect to the ground.If they are initially 8.5 km apart. How long will it take before they reach each other?
Answer:
It will take 2.68 minutes for them to reach each other.
Explanation:
We use the two following kinematic equations, making the final position the same (for the moment they meet each other):
locomotive 1 --> [tex]x_f=v*t[/tex]
locomotive 2 --> [tex]x_f=8.5\,-\,v*t[/tex]
we make the two xf equal, and solve for the time (t) using v = 95 km/h:
[tex]95*t=8.5-95*t\\2*95*t=8.5\\t=8.5/(190)\\t = 0.0447\,\,hours[/tex]
converting the hours into minutes by multiplying this value times 60;
t = 2.68 minutes
Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the value of the resistance. Resonance occurs when omega squared space equals 1 divided by space L C At resonance, the power is maximum and the impedance is minimum. At resonance, R squared space equals space (X subscript L minus X subscript c )squared The impedance Z is always larger than the resistance R.
Answer:
True b and c
Explanation:
In an RLC circuit the impedance is
[tex]Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }[/tex]
examine the different phrases..
a) False. The maximum impedance is the value of the resistance
b) True. Resonance occurs when
(wL)² + (1 / wC)² = 0
w² = 1 / LC
c) True. In resonance the impedance is the resistive part and the power is maximum
d) False. In resonance the inductive and capacitive part cancel each other out
e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal
A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?
Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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A 1.13 kg skateboard is coasting along the pavement at a speed of 4.28 m/s when a 0.93
kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the
skateboard-cat combination?
By conservation of momentum,
Pinitial = Pfinal
m1v1 + m2v2 = (m1 + m2)*vf
m1 = mass of skateboard = 1.13 kg
m2 = mass of cat = 0.93 kg
v1 = initial velocity of skateboard = 4.28 m/s
v2 = initial velocity of cat = 0 m/s
vf = final velocity of skateboard-cat combo
So plug in the values and solve for vf,
1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf
vf = 2.35 m/s
Can I get help on this question please I don’t understand
Answer:
A
Explanation:
Weight = 55 * 9.8 = 539 N
Scale = 686 N
Since the scale reading is larger than the lady’s weight, the elevator must be accelerating as it moves upward.
Net upward force = 686 – 539 = 147 N
147 = 55 * a
a = 147 ÷ 55
The acceleration is approximately 2.67 m/s^2.
ASAP PLS HELP
What is chemical potential energy?
Answer:
chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition.
Explanation:
In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m and 0.6 m, respectively. The wavelength is 630 nm. Suppose that the radius of the aperture can be increased from 0.5 mm, determine: (a) The first two radii when the center intensity at the reception screen is maximum. (b) The first two radii when the center intensity is minimum.
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
a 2 kg block is attached to a horizontal ideal spring with a spring constant 200N/m. when the
Some metals have a molecular structure that makes them good conductors. Explain how understanding this relationship can help engineers make more powerful batteries.
Answer:
Explained below.
Explanation:
Conductors can be defined as materials that permit electricity to flow through them easily.
Now, metals have a molecular structure that makes them good conductors because electrons in the atoms of these conductors tend to move freely from one atom to the other. So a majority of metals make good conductors because these metals tend to hold their electrons loosely. In short, it can help engineers make powerful batteries because then it means that they are capable of giving much more electrical energy since nowadays, advanced batteries make use of ion charges for the batteries.
Which lists three organic biological molecules?
O carbohydrates, salts, metals
O salts, proteins, minerals,
O proteins, lipids, carbohydrates
O lipids, metals, minerals
Answer:
B
Explanation:
I'm learning it in science.
Answer:
its not b i just took the test and b was wrong
Explanation:
What is the magnitude of the Box's Acceleration?
The Box's Acceleration : g sin θ
Further explanationNewton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object
∑F = m. a
F = force, N
m = mass = kg
a = acceleration due to gravity, m / s²
We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.
Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then
[tex]\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta[/tex]
Using the law of conservation of energy, what is the kinetic energy at e?
Answer:
Send the pic so I can see
Many curves have banked turns, which allow the cars to travel faster around the curves than if the road were flat. Acutally, cars could also make turns on these banked curves if there were no friction at all. Explain this statement using the free-body diagram shown in the figure.
Answer:
the normal force that is perpendicular to the surface has a component towards the center of the curve.
Explanation:
When curves with superelevation or inclination, the normal force that is perpendicular to the surface has a component towards the center of the curve. This component is what maintains the vehicle and gives it centripetal acceleration.
Consequently, the vehicle does not need the friction force since it does not have a tendency to slide and this is zero.
The phrase "ten meters per second squared" describes the
Answer:
Acceleration
Explanation;
change in velocity over timeTherefore metre per second over second, which is meter per second squaredGive reason:
a) In 'coin on card' experiment a smooth card is used.
Answer:
Please mark as brainliest!!
Explanation:
In coin card experiment smooth card is used so that the card can slide easily from glass.
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass.
Calculate the potential energy of a coffee cup that is resting on a 24.5 meter ledge, and
weighs 5.4 Newtons
Answer:
E_{pot} = 132.3 [J]
Explanation:
In order to solve this problem, we must use the definition of the potential energy which can be calculated by means of the following formula.
[tex]E_{pot}=W*h[/tex]
where:
Epot = potential energy [J]
W = weight = 5.4 [N]
h = elevation = 24.5 [m]
[tex]E_{pot}=5.4*24.5\\E_{pot}=132.3[J][/tex]
ASAP DUE TO DAY PLS HELP MEEEEEE
1. Kinetic energy is the energy of:
A. potential
B. objects at rest
C. objects in motion
D. accelerating objects
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