Hellohellp fastttttttttt

Hellohellp Fastttttttttt

Answers

Answer 1

Answer:

Explanation:

[tex]w_f^2=w_o^2-2\alpha\theta\\\\0=w_o^2-2\alpha\theta\\\\\theta={w_o^2\over{2\alpha}}[/tex]

[tex]w_o=2\pif=2\pi*8Hz=16\pi{rad\over{s}}\\\\\alpha=2.2{rad\over{s^2}}\\\\\theta={(16\pi{rad\over{s}})^2\over{2*2.2{rad\over{s^2}}}}=574.23rad[/tex]

[tex]revolutions={\theta\over{2\pi}}={574.23\over{2\pi}}=91.39[/tex]


Related Questions

Which of the following accurately describes circuits?
options:

A)

In a parallel circuit, there's only one path for the current to travel.

B)

In a series circuit, the amount of current passing through each part of the circuit may vary.

C)

In a series circuit, the current can flow through only one path from start to finish.

D)

In a parallel circuit, the same amount of current flows through each part of the circuit.

Answers

' C ' is the only correct statement.

1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s , después cambia su trayectoria cuando es lanzada hacia arriba 25m durante 7 s. Calcular la velocidad y la rapidez al punto final (altura maxima) al que llegó la pelota.

2. Una mariposa vuela en línea recta hacia el sur recorriendo una distancia de 15 m durante 28 s, después cambia de dirección hacia el Oeste recorriendo una distancia de 50 m en un tempo de 80 s ¿cuál es la velocidad y rapidez de la mariposa?

3.- Una persona camina durante 21 minutos hacia el este de su casa una distancia de 1500 m y después cambia su dirección hacia el Norte recorriendo una distancia de 3350 m en un tiempo 32 minutos llegando al supermercado. ¿Calcula la velocidad y rapidez de la persona?

4.- Un automóvil se mueve al Oeste recorriendo una distancia de 80 km en 1.2 horas, posteriormente cambia su trayectoria hacia el Sur, recorriendo una distancia de 120 km en un tiempo 1.6 hora. ¿Calcula la velocidad y rapidez del automóvil?

Answers

Answer:

https://youtu.be/ymHHdoCGJOU

Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.

Answers

Answer:

An object in either state of equilibrium must have no net force acting on it.

Explanation:

Answer: An object in either state of equilibrium must have no net force acting on it.

Explanation:

Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction

Answers

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave

Ez Physics question will mark brainliest.

Answers

Answer:

The answer is B. longitude wave

What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making

Answers

Hi! I believe the answer is C. or The seasons because all the other options use springs when making cars, bikes, and clocks. I hope this helped. Goodluck :)

which of the following cannot be increased by using a machine of some kind? work, force, speed, torque

Answers

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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e. Your father bought you a pair of shoes. When you wore the shoes, you realized there was a problem. The shoes were too long Why might such a problem arise and how can it be mitigated?

Answers

The problem arose due to a difference in length. This was due to father not knowing the exact length of shoe used by the son. And this can be mitigated by the use of shoe fillers.

The length of an object implies how long the object is. And it is one of the fundamental unit of quantities measured in SI unit of meters.

Considering the given question, it can be observed that the father do not know the exact length of shoe that would fit the son appropriately. Thus the realized problem of the pair of shoes too long arose due to difference in length of the pair of shoes and the son's leg. This variation would not have occurred if the exact length of pair of shoes has been bought.

To mitigate this little problem, shoe fillers can be used.

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What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm

Answers

Hi! The answer is 15km I'm pretty sure

Answer:

The submarine's maximum safe depth in sea water is 801.678 m.

Explanation:

P=Po+(rho)*g*h

Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)

Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2

Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa

Initial Pressure= 1atm= 101.3kPa= 101300Pa

Water Density (rho) = 1000kg/m^3

Gravity= 9.8m/s^2

So rearranging for h= (P-Po)/((rho)*g)

h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m

Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.

Answers

D. I hope my answer helps you!

Which of the following groups is the largest ?

population
community
ecosystem
biome

Answers

Answer:

B. Community

Took science classes for 6 years now

the acceleration due to gravity jupiter is 25m/s Square . what does it mean

Answers

Answer:

The acceleration due to gravity of Jupiter is 25 m/s^2. means that, Any object dropped near Jupiter's Surface will accelerate downward (towards the Jupiter's surface) at the rate of 25 m/s^2 due to the gravity of Jupiter

Explanation:

hope it's help

Answer:

Explanation:

If any object is dropped from a height above the Jupiter's surface the object will fall towards jupiter's surface with a constant acceleration of 25m/s^2.

Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed​

Answers

Answer:

c.

magnitude of displacement

a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection

Answers

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).

Answers

Answer:

1.6031 miles

Explanation:

Given the following data;

Speed = 344 m/s

Time = 7.5 seconds

To find how many miles away was the lighting strike;

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 344 * 7.5

Distance = 2580 meters

Next, we would have to convert the value of the distance travelled in meters to miles;

Conversion:

1609.344 metres = 1 mile

2580 meters = X mile

Cross-multiplying, we have;

X * 1609.344 = 2580

X = 2580/1609.344

X = 1.6031 miles

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.

Answers

Answer:

I think D) less than 50 years

Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.

Who is an astronaut?

An astronaut observes and performs the experiments based on the universe.

A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.

Due to special relativity, between space and Earth, both moving with different speeds.

The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.

So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.

Thus, the correct option is d.

Learn more about astronaut.

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a baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn inclined at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline


Answers

Explanation:

The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g .

Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1

Answers

Answer:

x= 2*I1*d/(I1+I2) meter

15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?

Answers

Answer: (15 - 0)/1.8 = 8. 33m/s^2

Explanation:

The acceleration of the racehorse is 8.33 m/s²

The given parameters;

initial velocity of the racehorse, u = 0

final velocity of the racehorse, v = 15  m/s

time of motion of the horse, t = 1.8 s

The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]

Thus, the acceleration of the racehorse is 8.33 m/s²

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A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

Answers

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices

Answers

Answer:

check photo for solve

Explanation:

State the law of conservation of momentum

Answers

Explanation:

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant

A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.

a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?

b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?

Answers

Answer: it would be daddy

Explanation:

Because I’m daddy

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​

Answers

[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

In which situation should a parent be proactive and act to assume responsibility?

Answers

Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.

Explanation: i picked this and it is correct, you’re welcome:)

Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C

Answers

Answer:

The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.

Explanation:

charge, Q = 12.5 C

distance, d = 11 cm = 0.11 m

Let the electric field is E.

[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]

The direction of electric filed is away from the charge.

3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answers

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

Learn more about refractive index of light here;

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A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce

Answers

Answer:

The rms voltage of new generator is 3.4 V.

Explanation:

f = 3200 Hz

rms voltage, V = 2 V

rms current, i = 28 mA

Now

f' = 4700 Hz

rms current, i' = 70 mA

let the new rms voltage is V'.

[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars

Answers

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?

Answers

Answer:

The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.

Explanation:

Consider newtonian mechanics here.

Dynamic equation is

The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.

we'll consider newtonian mechanics here.

so the dynamic equations is S = ut + 0.5at^2

we know u=0.8

S=1.2+0.9

t=1.2/1.5

find a.

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