A 1kg dictionary is resting on a table that is 1.5m high. How much potential energy
does it have?

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

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

Answer:

protons are in the nucleus .

Explanation:

there are 6 protons

Answer:

[tex]\boxed{\boxed{\sf all \ of \ the \ above }}[/tex]

Good conductors have: current moves easily,  low resistance,   conserves energy-easy for electrons to move. Hence, Option (C) is correct.

Materials that easily permit the flow of electricity are referred to be electrical conductors. Conductivity is the quality of conductors that enables them to conduct electricity.

Electric current is the name given to the movement of electrons through conductor. Voltage is the amount of power necessary to cause that current to flow through the conductor.

Such an element receives a charge that is dispersed along its entire surface, causing the electrons inside the element to migrate. Charges are transferred to an electrical conductor, and they disperse until the minimal force of repulsion between electrons in locations of excess electrons. Such an item transfers its charge to another conductor when it comes into contact with it, reducing the overall repulsion caused by charge in the process.

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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

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.

Answer:

Volume =  24 cm^3

Explanation:

We recall that the volume of the box is calculated via the formula:

Volume = length * height * width

and that in a product, the number of significant figures for the result should coincide with the number of significant figures of the factor that has the least of them.

In this case, all measures have the same number of significant figures: two. so we calculate the product, and then limit the answer value to exactly two significant figures:

Volume = 4.1 cm * 2.8 cm * 2.1 cm = 24.108 cm^3, which must be rounded to two significant figures as: 24 cm^3

Answer:

Volume = 24.108cm³

Explanation:

Volume = length × breadth × height

Volume = 4.1 × 2.8 × 2.1

Volume = 24.108

Answer:

a) The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.

b) The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.

c) The total energy of the system is 1.802 joules.

d) The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].

Explanation:

a) If all non-conservative forces can be neglected and spring has no mass, then the mass-spring system exhibits a simple harmonic motion (SHM). The kinematic formula for the position of the system ([tex]x(t)[/tex]), measured in meters, is:

[tex]x(t) = A\cdot \sin(\omega \cdot t +\phi)[/tex] (1)

Where:

[tex]A[/tex] - Amplitude, measured in meters.

[tex]\omega[/tex] - Angular frequency, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\phi[/tex] - Phase, measured in radians.

The kinematic equation for the velocity formula of the system ([tex]v(t)[/tex]), measured in meters per second, is derived from (1) by deriving it in time:

[tex]v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)[/tex] (2)

The velocity when it passes the equilibrium point occurs when the cosine function is equal to 1 or -1. Then, that velocity is determined by following formula:

[tex]v = \pm \omega\cdot A[/tex] (3)

The angular frequency is calculated by this expression:

[tex]\omega = 2\pi\cdot f[/tex] (4)

Where [tex]f[/tex] is the frequency, measured in hertz.

If we know that [tex]f = 3\,hz[/tex] and [tex]A = 0.13\,m[/tex], then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:

[tex]\omega = 2\pi\cdot (3\,hz)[/tex]

[tex]\omega \approx 18.850\,\frac{rad}{s}[/tex]

[tex]v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)[/tex]

[tex]v = \pm 2.451\,\frac{m}{s}[/tex]

The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.

b) First, we need to determine the spring constant of the system ([tex]k[/tex]), measured in newtons per meter, in terms of the angular frequency ([tex]\omega[/tex]), measured in radians per second, and mass ([tex]m[/tex]), measured in kilograms. That is:

[tex]k = \omega^{2}\cdot m[/tex] (5)

If we know that [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]m = 0.60\,kg[/tex], then the spring constant is:

[tex]k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)[/tex]

[tex]k = 213.194\,\frac{N}{m}[/tex]

Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:

[tex]U_{k} + K = K_{max}[/tex] (6)

[tex]\frac{1}{2}\cdot k\cdot x^{2} + \frac{1}{2}\cdot m\cdot v^{2} = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (7)

Where:

[tex]U_{k}[/tex] - Current elastic potential energy, measured in joules.

[tex]K[/tex] - Current translational kinetic energy, measured in joules.

[tex]K_{max}[/tex] - Maximum translational kinetic energy, measured in joules.

[tex]v[/tex] - Current velocity of the system, measured in meters per second.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v_{max}[/tex] - Maximum velocity of the system, measured in meters per second.

If we know that [tex]k = 213.194\,\frac{N}{m}[/tex], [tex]x = 0.10\,m[/tex], [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the velocity of the mass-spring system is:

[tex]\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}[/tex]

[tex]v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }[/tex] (8)

[tex]v = \sqrt{\left(\pm 2.451\,\frac{m}{s} \right)^{2}-\frac{\left(213.194\,\frac{N}{m} \right)\cdot (0.10\,m)^{2}}{0.60\,kg} }[/tex]

[tex]v \approx \pm 1.567\,\frac{m}{s}[/tex]

The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.

c) The total energy of the system ([tex]E[/tex]), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:

[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (9)

If we know that [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the total energy of the system is:

[tex]E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}[/tex]

[tex]E = 1.802\,J[/tex]

The total energy of the system is 1.802 joules.

d) Given that initial position of the mass-spring system is a maximum, then we conclude that the equation of motion has the following parameters: ([tex]A = 0.13\,m[/tex], [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]\phi = 0.5\pi\,rad[/tex])

From (1) we obtain the resulting formula:

[tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex] (10)

The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].

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

Answer:

Explanation:

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

Hope this helps you

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!

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.

Answer:

spirit of the team / game

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]

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:

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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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.

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.

Answer:

C. Some of the mass of the four hydrogen nuclei was converted into

energy

Explanation:

a p e x , just took the quiz

Some of the mass of the four hydrogen nuclei was converted into energy.

When four hydrogen nuclei fuse to form one helium nuclei, mass defect, some mass is converted into energy, that is the reason of energy of the sun.

"When two nuclei form a big nuclei, the phenomenon is known as fusion."

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Answer:

B

Explanation:

I'm learning it in science.

Answer:

its not b i just took the test and b was wrong

Explanation:

Answer:

A stunt person jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following best describes why no injury occurs?

The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.

ANSWER: A

Answer:

A. The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.

During stunts, people usually put an-air filled bag on the ground for safe landing and to prevent injuries.

Momentum is calculated by m* v where m is for mass and v is for velocity.

Force is directly proportional to momentum

F∝M

The bag in this scenario helps to increases the amount of time the force acts on the person. This implication means that there is a reduction in the change of momentum. Since Force and momentum have a directly proportional relationship then the Force taken to hit the floor is greatly reduced.

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Answer:

It enabled them to run faster from predators.

Explanation:

The evolution of hooves from a five-digit feet enabled horses to run faster from predators as well as support their larger weights and longer legs. Hope this helped and have an awesome day! :)

It enabled them to run faster from predators. ]

Explanation: The evolution of hooves from a five-digit feet enabled horses to run faster from predators as well as support their larger weights and longer legs.

Answer:

[tex]976562.5\ Joules[/tex]

Explanation:

[tex]We\ are\ given:\\The\ mass\ of\ the\ missile=125\ kg\\The\ velocity\ of\ the\ missile=125\ m/s\\Hence,\\As\ we\ know\ that,\\Kinetic\ energy\ possessed\ by\ an\ object\ with\ mass\ m\ moving\\ with\ a\ velocity\ v:\\E_k=\frac{1}{2}mv^2\\Here,\\The\ Kinetic\ Energy\ possessed\ by\ the\ missile=\frac{1}{2}*125*(125)^2=976562.5\ Joules[/tex]

Answer:

The torque on the child is now the same, τ.

Explanation:

       [tex]I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)[/tex]

       [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]

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.

Answer:

Send the pic so I can see

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:

Explanation:

The Box's Acceleration : g sin θ

Newton'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]