What percent of the world’s earthquakes occur around the pacific rim/ring of fire?.

Answer:

a = 36 m/s

Explanation:

Given:

F = 9 N

m = 0.25 kg

a= ?

Use the equation:  

F = ma (divide by m on both sides to get a alone)

a = F/m= 9/0.25 = 36 m/s

The Schwarzschild Radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius or gravitational radius is a physical parameter in the Schwarzschild solution of Einstein's field equations that corresponds to the radius defining the event horizon of a Schwarzschild black hole.

Let's assume that the ring shaped disk is circular and thus, to find the mass of the object at the center of the milky way, we can use the circular orbit equation;

v = √(GM/r)

where;

G is the gravitational constant and has a value of 6.67 x 10^(-11) Nm²/kg²

R is the radius of the orbit,

M is the mass of the larger object

v = velocity = 190km/s = 190000 m/s

Also

1 light year = 9.4605 x 10^(15) m

Thus,7.5 light years = 7.5 x 9.4605 x 10^(15) m

so, M = v²r/G

Putting all the values in the equation,

M = [190000² x 7.5 x 9.4605 x 10^(15)]/6.67 x 10^(-11)

M = 3.84 x 10^(37) kg

Now, mass of the sun generally has a value of 1.9891 x 10^(30) kg

Thus, value of mass of object in solar masses = 3.84 x 10^(37)/1.9891 x 10^(30) = 1.931 x 10^(7) solar masses

The formula for Schwarzschild radius is given as;

Rs = 2GM/c²

Where c is speed of light = 3 x 10^(8) m/s

Thus,

Rs = 2 x 6.67 x 10^(-11) x 3.84 x 10^(37)/(3 x 10^(8))² = 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

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For accurate sampling, -44.1khz is minimum sampling rate for an a/d convertor.

The number of samples per second (or other unit) taken from a continuous signal to form a discrete or digital signal. The sample rate or sample rate specifies the number of samples per second (or other unit). of a continuous signal to form a discrete or digital signal..

The sampling rate can be thought of as the audio version of frames per second. This is the number of "clips" taken from an analogue sound wave to make a digital file. In addition, the sample rate also controls the highest frequency that can be accurately reproduced by a digital file.

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The distance the skier will go before they stop is 10193.68m.

Given ; Mass =100kg

coefficient of kinetic friction μ=0.05m

u=100m/s

v=0

μg=a

a= μg=0.05 x 9.81=0.4905 m/s²

v²-u²=2as

[v²-u²]/2a= s

-100²/2(0.4905)

s=10193.68m (neglecting the negative sign)

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System is set in motion with an amplitude of a and a period of T. If the amplitude of the motion is increased to 2a, the period of the motion changes.

A straightforward harmonic oscillator's period is independent of its amplitude. with the graphs of acceleration and velocity produced by the time derivatives. These oscillators also show how kinetic and potential energy can be transferred.

The acceleration of the mass increases proportionally with the increase in force, causing the mass to travel farther in the same amount of time. Therefore, increasing the amplitude has no overall impact on the oscillation's period.

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The vertical height of the building is determined as 44.1 m.

The height of the building is calculated by applying the following kinematic equation as shown below.

h = vyt + ¹/₂gt²

where;

h = (v sinθ)t + ¹/₂gt²

where

The given parameters include the following;

h = (0 x sin25)(3)  +  ¹/₂(9.8)(3²)

h = 44.1 m

Thus, the height of the building is equal to the vertical distance travelled by the brick thrown at the given initial velocity.

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The given statement is true. A lunar eclipse happens when the Moon passes into the shadow of the Earth. A solar eclipse happens when the Moon passes between Earth and the Sun.

A lunar eclipse happens when the Moon passes into the shadow of the Earth. This is only possible when the Sun, Earth, and Moon are precisely or extremely closely aligned with Earth between the other two, which can happen only on a full moon night when the Moon is close to either lunar node.

A solar eclipse happens when the Moon passes between Earth and the Sun, completely or partially blocking Earth's view of the Sun. A new moon corresponds with such an alignment, showing that the Moon is closest to the plane of the Earth's orbit. The Moon completely obscures the disk of the Sun during a total eclipse.

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The given statement is true. A lunar eclipse happens when the Moon passes into the shadow of the Earth. A solar eclipse happens when the Moon passes between Earth and the Sun.

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Answer:

I = 1/2 M R^2          moment of inertia of solid disk

I = 1/2 * 41.0 kg * (.3 m)^2

I = 20.5 kg * .09 m^2 = 1.85 kg-m^2

Using the concepts of motion, we got that Ma is the force exerted on the  passengers inside the car when a  car traveling with a constant velocity quickly comes to rest.

We know very well that according to newton second law of motion,

if a moving object is moving with some velocity v at any particular time t, then there is force acting on that object and that force is given by

F= dp /dt,

where dp is defines as change in momentum of the object. In other words,

dp=M×([tex]v_f_i_n_a_l[/tex] - [tex]v_i_n_i_t_i_a_l[/tex])

where [tex]v_f_i_n_a_l[/tex] and v[tex]_i_n_i_t_i_a_l[/tex] are the velocities of the object in different time.

So suppose car has some mass M kg

It is given that final velocity of the object is =0 as the car comes to rest,

suppose initial velocity is v

Therefore, F=[M×(0-v)]/t

=>              F= -M×(v/t)

=>              F= -Ma because acceleration =velocity/time

Here negative sign shows that force acts on the person in opposite direction of frictional force applied by the road.

Hence, when a car traveling with a constant velocity quickly comes to rest. while slowing down, the passengers inside the car are pushed forward by the force - Ma.

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The speed of the ball is 192m/s. The long of the ball is 2.85s.The height of the ball is 2.13m.

A)The speeds at these points are equal since the starting position and the point at which it touches the ground have the same elevation. Solving the product of the horizontal and vertical velocities yields the final speed. Which is

vf=√(V[tex]ox[/tex])²+(V[tex]oy[/tex])²

vf=√(23m/s)²+(14m/s)²

vf=192m/s

B)Take into account the starting point and the highest point, respectively.

We are aware that the vertical velocity is zero at the highest point. Additionally, we are aware that the duration of the flight from start to finish is twice as long.

So, we'll apply the formula t=v[tex]f[/tex]-v[tex]o[/tex]/a

   t=2(v[tex]top[/tex]-v[tex]o[/tex])/a

    t=2(0m/s-14m/s)/-9.81m/s

t=2.85s

C)Using the formula, the highest height that can be reached can be determined.(v[tex]f[/tex])²=(v[tex]0[/tex])²+2ay

This is how the maximum height is determined:

                    (v[tex]f)[/tex]²=(v[tex]0)[/tex]²+2ay

                    y[tex]max[/tex]=(v[tex]top[/tex])²-(v[tex]o[/tex])²÷2a

    y[tex]max[/tex]=(0m/s)²-(14m/s)²÷2(-9.81m/s)²

            =2.13m

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The speed of the plane will have a factor of 0.668 with the satellite speed at the same centripetal acceleration.

We need to know about centripetal acceleration to solve this problem. The centripetal acceleration depends on the angular velocity and the radius. It can be written as

a = v² / R

where a is centripetal acceleration, v is velocity and R is radius.

From the question above, the given parameters are

R1 = 6.7 x 10⁶ m

R2 = 15 m

a1 = a2

By substituting the given parameters, we can calculate the speed of plane

a1 = a2

v1² / R1 = v2² / R2

v2² / v1² = R1 / R2

v2² / v1² = 6.7 x 10⁶ / 15

v2² / v1² = 6.7 x 10⁶ / 15

v2/v1 = 0.668

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Angular velocity (in rev/s) after a 15.0 kg child gets onto it by grabbing its outer edge is 1.575 rad/s.

The angular velocity of the merry-go-round after a child gets on it is equal to the angular velocity of the child plus the angular velocity of the merry-go-round.

The angular velocity of a rotating object can be calculated using a formula:

ω = ω + (m/r)

where ω is the original angular velocity, m is the mass of an object that's been added to it, r is its radius, and (m/r) is the change in mass divided by its radius. When we plug in our values for this problem, we get:

ω = 0.350 rad/s + ((15 kg)/(1.8 m)) rad/s

= 0.350 rad/s + 1.25 rad/s

= 1.575 rad/s

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When  the core of a star reaches a temperature of about 100 million degrees (k), the new event named  triple-alpha process happens where the helium will begin to fuse into carbon.

The triple-alpha mechanism, which turns three helium nuclei into one carbon atom, will start to fuse helium into carbon when the temperature in the core reaches roughly 100 million degrees. There is a lot of heat produced by this.

The triple-alpha process is unusually sensitive to temperature; when the reaction's temperature is doubled, it proceeds nearly a trillion times faster! Therefore, while the fusing helium heats the core, which cannot expand to cool down, the increasing temperature prompts the helium fusion to abruptly proceed millions of times faster, which very quickly heats the core even more, which in turn prompts the helium to fuse way, way faster.

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The momentum of car B immediately after the collision if the momentum of car A is 21 000 kg m/s after collision is 15000 kg m / s

p = m v

p = Momentum

m = Mass

v = Velocity

m1 = 2000 kg

u1 = 18 m / s

pi1 = 2000 * 18

pi1 = 36000 kg m / s

m2 = 1200 kg

u2 = 0

pi2 = 0

pf1 = 21000

According to law of conservation of momentum

pi1 + pi2 = pf1 + pf2

pi1 = Initial momentum of car A

pi2 = Initial momentum of car B

pf1 = Final momentum of car A

pf2 = Final momentum of car B

pf2 = pi1 + pi2 - pf1

pf2 = 36000 + 0 - 21000

pf2 = 15000 kg m / s

Therefore, the momentum of car B immediately after the collision is 15000 kg m / s

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Force is to be known for this problem.

What is force?

In physics, force is the push or pull that an object feels when its mass causes it to change its velocity. An external force has the ability to change a body's resting or moving state. It moves and has a size.

What constitutes speed?

the speed at which the position of an object is shifting, in any direction. Speed is defined as the product of the distance traveled and the time needed to cover that distance. Since speed simply has a focus and no magnitude, it is a scalar number.

2T cos∅ =μg

⇒ if weight (μg)

rises

⇒∅ increases

⇒ cos∅ decreases

⇒T increases

⇒F will also increases

But ∴ Cos 90° = u

⇒ not possible.

Therefore, the force is to be needed.

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The magnitudes and directions of the velocities of the two marbles after the collision is: v1f is left and v2f is right.

The initial marble weighs m1 = 0.015 kg.

Second marble's mass is m2=0.045kg.

The first marble's initial velocity is v1i=0.40m/8 before to contact.

The second marble's initial velocity is v2i=0m/8 prior to impact.

After the impact, the first marble's final speed is v1f.

Following the collision, the initial marble's final velocity is v2f.

If the collision is elastic, the kinetic energy is preserved.

We'll write 1/2 m1 for KE = 1/2 mv2 (v1i)

2 + 1/2 m2(vi) (vi)

2 = 1/2 m1(v1f)2 + 1/2 m2 (v2f) (v2f)

2.

Using this equation, we can establish a new relationship between velocities and determine each marble's post-collision velocity:

v1f=(m1-m2)/(m1+m2)

v1\sv1f=(0.015kg-0.045kg)/(0.015kg+0.045kg)

0.40m/s\s=-0.20m/s

First marble is moving leftward at 0.20 m/s, as indicated by the negative sign.

Find the second marble's final velocity.

​v2f=2m1v1/(m1+m2)\s

v2f=2⋅0.015kg⋅0.40m/s/(0.015kg+0.045kg)\s

v2f=0.20

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(a)  The diver's velocity when he hits the water is 13.795 m/s.

(b) The diver's net force while in the water was -3092.7N.

Newton's second law states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration an of its center of mass, F = ma.

The three equations of motion are: v = u + at, v2 = u2 + 2as, and s = ut + 12at2.

Given

A 65-kg diver jumps off a 9.7-m tower

(a) Using newton's equation of motion,

v² = u² + 2gs     ......... (1)

Where v = final velocity,

u = initial velocity,

s = height of the tower,

g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s (height jump), g = 9.81 m/s2.

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

As a result, the driver's velocity when he hits the water is 13.795 m/s.

(b)

F = ma.    ......... (2)

Where F is the force applied to the diver, m is the diver's mass, and an is the diver's acceleration below the water's surface.

Also using

v² = u² + 2as .........  (3)

When the diver enters the water, u = 13.795 m/s, and when the diver comes to a complete stop, v = 0 m/s.

Given are the following values: s = 2.0 m, u = 13.795 m/s, and v = 0 m/s.

Substitute in equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 65 kg,

Substitute into equation 3

F = (-47.58)(65)

F = -3092.7

The Force is negative because it acts against the diver's motion.

As a result, the net force exerted on the diver while in the water equals -3092.7N.

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Within a uniform electric field of magnitude 28,000 n/c produced by two parallel plates, an electron is expelled at a speed of 3.51 * 10^15 m/s^2. The distance between the plates is 4 cm.

The electron is a subatomic particle with the symbol e or with an electric field of one elementary negative charge. Due to their lack of known components or substructure, electrons, which are part of the first generation of the lepton particle family, are typically considered to be elementary particles. When the electric field strength at the electron's location, E, is taken into account, the force acting on it has the size F=eE. Newton's second law provides the electron's acceleration as follows:

a= F/m = eE/m

= (1.60*10^-19)(2.00*10^4)/9.11*10^-31

= 3.51 *10^15 m/s^2

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(a) At time, t = 0.25 s, the y position of the driver is 0.31 m.

(b) At time, t = 0.50 s, the y position of the driver is 1.2 m

(c) At time, t = 0.75 s, the y position of the driver is 2.76 m

The position of the driver at the various time is determined by applying the following kinematic equation.

y = vt + ¹/₂gt²

where;

Since the driver runs horizontally, the initial vertical velocity = 0

y = 0 + ¹/₂gt²

y = ¹/₂gt²

at time, t = 0.25 s, the y position of the driver is calculated as;

y = ¹/₂gt²

y = ¹/₂(9.8)(0.25)²

y = 0.31 m

at time, t = 0.5 s, the y position of the driver is calculated as;

y = ¹/₂gt²

y = ¹/₂(9.8)(0.5)²

y = 1.2 m

at time, t = 0.75 s, the y position of the driver is calculated as;

y = ¹/₂gt²

y = ¹/₂(9.8)(0.75)²

y = 2.76 m

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The difference between linear speed and angular speed is linear speed calculates the distance to time whereas angular speed calculates the angle to time.

The difference between the linear speed and the angular speed is that linear speed is just a measurement determined by calculating the distance of an arc versus time taken, and angular velocity is also a measurement determined by calculating the angle of an arc versus time taken.

The formula of the linear speed is

                          s = d / t

where s is the speed

          d is the distance

          t is the time taken

The formula of the angular speed is

                       ω = θ / t

where ω is the angular speed

           θ is the angular rotation

           t is the time taken.

The relation between the linear speed and angular speed in a uniform circular motion is

                           s = rω

where s is the linear speed

           r is the radius

          ω is the angular speed.

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The work done by the person operating the jack is 4,800 J

Efficiency is described as the potential to perform something with the least quantity of wasted time, cash, and attempt or competency in performance. Effectiveness is described because the diploma to which some thing is successful in producing a preferred result; achievement.

calculation:-

work input = (1200 J)/(0.25)

= 4800 J.

There are numerous styles of performance, consisting of allocative and effective performance, technical efficiency, 'X' efficiency, dynamic performance and social efficiency.

The prime focus of performance is for customers to get the goods and offerings they want at the bottom possible resource price. Inside the have a look at of economics, the dreams of efficiency and equity are frequently in warfare with each other.

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Robin started at point a. she walked 4m south, then 8m east, then 4m north, then 8m west so that she was back at point a. if her walk took 5 minutes her velocity in meters per minute is 0.96.

Displacement = distance / time = 24m/5 = 4.8

Velocity = displacement / Time = 4.8/5 = 0.96m/min

Its scientific definition and velocity are presumably similar. You are aware that a significant displacement over a short period of time indicates a large velocity, and that velocity has measures of distance / time, like miles/hour or kilometers/hour.

The change in position divided by the distance travelled is the definition of average velocity. However, the average speed of an object tells us nothing about what happens to it between the starting and finishing points. For instance, average velocity does not allow us to determine whether a passenger in an airplane briefly stops or reverses before relocating to the rear of the aircraft. We must consider shorter distances travelled over shorter times in order to obtain more information.

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The magnification of an astronomical telescope is  -27.24    .

The magnification of a telescope is actually a relationship between two independent optical systems: the telescope itself and the eyepiece that is being used.

Focal length of the telescope = T = 79 cm = 790mm

Focal length of the eyepiece = E = 2.9 cm = 29mm

M = -[tex]\frac{T}{E}[/tex]

M = -[tex]\frac{79}{2.9}[/tex]

M = -27.24    

Therefore, the magnification of an astronomical telescope is  -27.24.

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The wavelength of an electron is 215.5nm.

A description of the de Broglie wavelength is as follows:

The formula for wavelength is = h/mv, where h stands for Planck's constant, m for mass, and v for velocity.

The equation p = h/λ, where h is the Planck constant and is the wavelength, expresses the De Broglie relation, which states that there is a correlation between an electron's wavelength and momentum.

The wavelength of a wave is the distance between its two subsequent crests or troughs.

h = 6.62607015 × 10⁻³⁴ joule second.

m =9.11 x 10 ⁻³¹Kg

v = 2.963 x 10⁶m/s

λ = h / ( mv)

λ = 6.62607015 × 10⁻³⁴/9.11 x 10 ⁻³¹×2.963 x 10⁶

λ = 2.155×10⁻⁷

λ = 215.5nm.

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a) magnitude of the magnetic field = 8.02 X [tex]10^{-3}[/tex] T

b)Energy Density = 5.03 J/[tex]m^{3}[/tex]

c) Total magnetic energy inside the solenoid = 2.01 X [tex]10^{-5}[/tex] J

Calculation

a) B = uNI/L

B = (4π X [tex]10^{-7}[/tex])(37)(1.15)/(.015)

B = 8.02 X [tex]10^{-3}[/tex] T

b) Energy Density = [tex]B^{2}[/tex]/2u

Energy Density = [tex](8.02 *10^{-3} )^{2}[/tex]/(2)(4π X [tex]10^{-7}[/tex])

Energy Density = 5.03 J/[tex]m^{3}[/tex]

c) Total magnetic energy inside the solenoid

Multiply by volume for total energy.

Total Energy = 5.03(π[tex]r^{2}[/tex])(L)

Total Energy = (5.03)(π)[tex](8.5*10^{-3})^{2}[/tex])(.015)

Total Energy = 2.01 X [tex]10^{-5}[/tex] J

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a)The  magnitude of the magnetic field [tex]$=\mathbf{8 . 0 2} \times 10^{-3} \mathbf{T}$[/tex]

b)The  Energy Density [tex]$=\mathbf{5 . 0 3} \mathrm{J} / \mathrm{m}^3$[/tex]

c) Total magnetic energy inside the solenoid [tex]$=\mathbf{2} . \mathbf{0 1} \mathbf{X} 10^{-5} \mathbf{J}$[/tex]

a) [tex]$B=u N I / L$[/tex]

[tex]&B=\left(4 \pi \times 10^{-7}\right)(37)(1.15) /(.015) \\[/tex]

[tex]&B=8.02 \times 10^{-3} \mathrm{~T}[/tex]

b) Energy Density[tex]$=B^2 / 2 \mathrm{u}$[/tex]

Energy Density[tex]$=\left(8.02 * 10^{-3}\right)^2 /(2)\left(4 \Pi \times 10^{-7}\right)$[/tex]

Energy Density [tex]$=5.03 \mathrm{~J} / \mathrm{m}^3$[/tex]

c) Total magnetic energy inside the solenoid

Multiply by volume for total energy.

Total Energy [tex]$=5.03\left(\pi r^2\right)(\mathrm{L})$[/tex]

Total Energy [tex]$\left.=(5.03)(\pi)\left(8.5 * 10^{-3}\right)^2\right)(.015)$[/tex]

Total Energy [tex]$=2.01 \times 10^{-5} \mathrm{~J}$[/tex]

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This can be calculated by using motion equations, The distance covered on the runway before coming to rest

Initial speed is 180 km/hr = 50 m/s

time took to come to rest is = 30 sec

so we can write, here we are assuming that there is no frictional force is applied externally or internally to the system.

u = 50 m/s

v = 0 m/s

t = 30 s

we know that distance travelled = x

so, x =

putting the appropriate values, we get that

x = [tex](\frac{50+0}{2} ) 30[/tex]

x = 750 m

Hence the distance travelled before coming to rest is 750 m.

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The efficiency of a machine that requires 20 J of input work to do 10 J of output work. The efficiency of the machine η = 50 %

The engine efficiency  is defined as ratio of the useful work done to the heat provided.

By using the formula for efficiency of a machine and substituting the values in it, we get :

η = (Wo/Wi × 100)%

η = (10/20 × 100)%

η = 50 %

The efficiency of the machine will be η = 50 %

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The energy that powers radio galaxies, quasars, and other active galactic nuclei is thought to be produced by accreting matter onto a supermassive black hole. This process is called active galactic nucleus (AGN) fueling.

The gas that surrounds the black hole is heated to millions of degrees as it falls into the black hole. Because the gas is so hot, it emits a great deal of radiation in many different wavelengths.

This radiation can be seen across the electromagnetic spectrum, from low-energy gamma rays to high-energy X-rays. In some cases, jets of charged particles are ejected at nearly the speed of light along magnetic field lines; these jets can create shock waves in the surrounding medium that emit radio waves.

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0.2×51000)

Explanation:

change 5ms into cms before multiplying by 0/2 kgs