Cuando Daniel hace oscilar un péndulo, este realiza 30,6 ciclos (completos) en 9 [s].

¿Cuál es la frecuencia del péndulo?

A )3,4 [Hz].

B )4,3 [Hz].

C )30 [Hz].

D )5 [Hz]

Answers

Answer 1
I believe the answer is c

Related Questions

what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc

Answers

Answer:

[tex]E=6.86\times 10^6\ N/C[/tex]

Explanation:

Given that,

Mass of the sphere, m = 2.1 g = 0.0021 kg

Charge, q = 3 nC

We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,

qE = mg

[tex]E=\dfrac{mg}{q}[/tex]

Put all the values,

[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]

So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].

What do interplanetary space missions study?


the moon


stars in other galaxies


planets in the solar system


the sun

Answers

The sun should be the answer










Answer:

C. Planets in the solar system

Explanation:

The one above is incorrect, and I know this is late. Even if it doesn't help you I hope it helps people in the future! YES I AM TALKING ABOUT YOU FUTURE PEOPLE!! I know this is the answer because I have taken 5.11 Quiz: Uncrewed Spacecraft in K12. There will only be the questions and correct answers below.

1. Which planetary body was Spirit designed to explore?

Mars.

2. What is the name of the most distant manmade object in space? (Credit: shathaadnan64/lak521)

Voyager 1.

3. Which group was designed to study Saturn? (Credit: Brainly User/snowballandtigoya1xa

Voyager 1, Huygens, and Cassini.

4. Why are scientists interested in exploring Mars?

Possible evidence of life.

5. What do interplanetary space missions study?

Planets in the solar system.

Have an amazing day!!

Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.

Answers

The answer is A because i know that it is and you will get the answer correct

What is the first quantum number of a 252 electron in phosphorus,
1322s22p3s23p3?
A. n=0
B. n= 3
O
C. n = 1
O D. n = 2

Answers

Answer:

the correct answer is B

Explanation:

The quantum numbers are the constants obtained when solving the Schrodinger equation, the first quantum number or principal quantum number (n), can take values ​​from zero to infinity.

This quantum number is placed as a coefficient in the quantum distribution.

In this case for phosphorus, the number is n = 3

the correct answer is B

The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge

Answers

Answer:

Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge

Explanation:

The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.

If λ is negative, E is negative so it points in the negative direction towards the line of charge.

Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.

Also as r increases, E decreases since E ∝ 1/r

So, the electric field decreases at larger distances from the line of charge.

So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.

A body of mass 5 kg is moved by a horizontal force of 0.5 N on a smooth frictionless table for 20 seconds. Calculate the change in kinetic energy.

A. 5 J

B. 20 J

C. 10 J

D. 30 J​

Answers

Answer: 10 J

Explanation:

Trust ;)

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 ss to speed up from rest to its top speed of 1 rotation every 1.30 ss . The astronaut is strapped into a seat 5.90 mm from the axis. What is the astronaut's tangential acceleration during the first 40.0 s?

How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.

Answers

Answer:

speed = 0.9 mm/s

Explanation:

time, t = 40 s

initial angular speed, wo = 0 rad/s

final frequency, f = 1/1.03 rps = 0.97 rps

final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s

time, t = 40 s

distance, r = 5.9 mm

The angular acceleration is given y the first equation of motion.

[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]

The linear velocity is

[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]

speed, v = 0.9 mm/s

You drive past a potential parking space in center city. Your new car is travelling at 85% the speed of light. If your car is 6.0 m long (which you measured the day you bought it) and you observe the space to be 3.0 m, should you try to park? Why is your friend on the sidewalk (who hasn't studied relativity) so sure that you can park? How does the situation appear to him?

Answers

Answer:

We should not try to park the car because its rest length is greater than the space available.

The car seems to be approximately equal to the friend (L = 3.16 m). Due to this reason he is sure to park.

Explanation:

We should not try to park the car because its rest length is greater than the space available.

The friend is sure about parking because the car appears short in length to him. For this, we will solve Einstein's length contraction formula from theory of relativity:

[tex]L = L_o\sqrt{1-\frac{v^2}{c^2}}[/tex]

where,

L = Relative length observed by friend = ?

L₀ = rest length = 6 m

v = relative speed = 85% of speed of light = 0.85c

Therefore,

[tex]L = (6\ m)\sqrt{1-\frac{(0.85c)^2}{c^2}}[/tex]

L = 3.16 m

Hence, the car seems to be approximately equal to the friend. Due to this reason he is sure to park.

an object moves clockwise around a circle centered at the origin with radius m beginning at the point ​(0,​). a. find a position function r that describes the motion of the object moves with a constant​ speed, completing 1 lap every s. b. find a position function r that describes the motion if it occurs with speed .

Answers

Answer:

Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.

Explanation:

help me with following question.

Answers

Answer:

The answer for this question is 50k N.m

compare the time period of two pendulums of length 4m and 9m​

Answers

area= length × length

area = 4m × 9m

ans 36

The answer is 45-54

A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant

Answers

Answer:

(a) 4.20 J

(b) 16.74 J

Explanation:

For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;

C = A∈₀ / d              --------------------(i)

Where;

∈₀ = constant called permittivity of vacuum.

The energy U stored in such capacitor is given by;

U = [tex]\frac{1}{2}[/tex]CV²             ----------------------(ii)

or

U =  [tex]\frac{1}{2}[/tex](Q²/C)        -------------------(**)

Where;

V = potential difference or voltage across the plates.

Q = charge on the plates.

(a) If the charge is held constant

Combine equations (i) and (**) to give;

U =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)     -----------------------(iii)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iii)

8.40 =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)

8.40 =  [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)

Multiply through by 2

2 x 8.40 = Q² x (0.023 / A∈₀)

16.80 = Q² x (0.023 / A∈₀)

Divide through by 0.023

16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023

730.4 = Q² / (A∈₀)

Make Q² subject of the formula

Q² = 730.4(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

Q = constant [this means that Q² still remains 730.4(A∈₀) ]

The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;

U =   [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)

U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)

U = 4.20J

Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J

(b) If the voltage is held constant

Combine equations (i) and (ii) to give;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / d)V²     -----------------------(iv)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iv)

8.40 =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²

Multiply through by 2 x 0.023

2 x 0.023 x 8.40 = (A∈₀)V²

2 x 0.023 x 8.40 = (A∈₀)V²

0.385 = (A∈₀)V²

Make V² subject of the formula

V² = 0.385/(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

V = constant [this means that V² still remains 0.385/(A∈₀) ]

The energy stored is found by substituting these values of d and V² into equation (iv) as follows;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]  

U = [tex]\frac{1}{2}[/tex](0.385/0.0115)

U = 16.74

Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J

Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is

Answers

a particle of violet light has exactly the same energy as a particle of red light

With explanation pls

Answers

Answer:

[tex](x \times 1) = (400 \times 2.5) \\ x = 1000 \: newtons[/tex]

[tex]y = 0[/tex]

what is the light synthesis ?​

Answers

Answer:

Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. ... During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.

Explanation:

thank me later

Photosynthesis is a process through which plants and other organisms transform light energy into chemical energy, which is then released to power the organism's metabolic activities through cellular respiration.

You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?

Answers

Answer:

(a) I = 1650000 A

(b) 4.125 T

Explanation:

Magnetic field, B = 5.5 T

distance, r = 0.06 m

(a) Let the current is I.

The magnetic field due to a long wire is given by

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]

(b) Let the magnetic field is B' at distance r = 0.08 m.

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]

How are the Northern Lights are formed.

Answers

Answer:

Bottom line: When charged particles from the sun strike atoms in Earth's atmosphere, they cause electrons in the atoms to move to a higher-energy state. When the electrons drop back to a lower energy state, they release a photon: light. This process creates the beautiful aurora, or northern lights.

Explanation:

^-^I hope it's help u

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?

Answers

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

HELP ME PLEASEEEEEEEEEEEEEE

Answers

Answer: The correct statements are:

The atoms are very attracted to one another.The atoms are held tightly together.

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

The atoms are very attracted to one another.The atoms are not moving freely.It will not spread out evenly to fill any container.The atoms are held tightly together.The forces of attraction are strong to bring molecules together.The atoms are close and in fixed positions.

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ml2/3. Write your answer with one decimal place.

Answers

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point =  mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A>s. (a) What is the mutual inductance of these solenoids

Answers

Answer: The mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].

Explanation:

Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m

[tex]N_{1}[/tex] = 6750

[tex]N_{2}[/tex] = 15

Radius = [tex]\frac{0.120 cm}{2} = 0.6 cm = 6 \times 10^{-4} m[/tex]

As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.

[tex]Area = \pi \times r^{2} = \pi \times (6 \times 10^{-4})^{2}[/tex]

Formula used to calculate mutual conductance of two solenoids is as follows.

[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}[/tex]

where,

M = mutual conductance

A = area

[tex]\mu_{o}[/tex] = relative permeability = [tex]4 \pi \times 10^{-7} Tm/A[/tex]

[tex]N_{1}[/tex] = no. of coils in outer solenoid

[tex]N_{2}[/tex] = no. of coils in inner solenoid

l = length

Substitute the values into above formula as follows.

[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}\\= \frac{4 \pi \times 10^{-7} Tm/A \times \pi (6 \times 10^{-4})^{2} \times 6750 \times 15}{0.5 m}\\= 2.88 \times 10^{-7} H[/tex]

Thus, we can conclude that the mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].

Which of these is NOT an effect of humor?


strengthened immune system

reduced stress levels

reduced feelings of anxiety

feelings of jealousy and envy

Answers

feelings of jealousy and envy

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.390 mm wide. The diffraction pattern is observed on a screen 3.10 m away. Define the width of a bright fringe as the distance between the minima on either side.

Answers

Answer:

Y = 5.03 x 10⁻³ m = 5.03 mm

Explanation:

Using Young's Double-slit formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = Fringe Spacing = Width of bright fringe = ?

λ =  wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = Screen distance = 3.1 m

d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m

Therefore,

[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]

Y = 5.03 x 10⁻³ m = 5.03 mm

As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.5 T over a square area with a length of 10.2 cm on a side when the magnets are separated by a distance of 12.8 cm . The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 400 loops.
What is the minimum rotation rate of the coil so it will produce the required voltage? Express your answer using two significant figures.

Answers

Answer:

The rotation rate is 15.3 rad/s.

Explanation:

maximum voltage, V = 120 V

Magnetic field, B = 1.5 T

length, L = 10.2 cm

width, W = 12.8 cm

Number of loops, N = 400

Let the rate of rotation is w.

The maximum voltage is given by

V = N B A w

120 = 400 x 1.5 x 0.102 x 0.128 x w

w = 15.3 rad/s

If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N

Answers

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

What is Force?

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

Learn more about force.

https://brainly.com/question/13191643

#SPJ2

how can you prove that acceleration is a derived unit​

Answers

a = (dx / dt)²

Explanation:  Unit of distance is m (metres) and unit of time is s (seconds) speed v is  first derivative of distance x versus time:

  v = dx / dt, unit is m/s.    Acceleration is second derivative of

 speed versus time  a = (dx / dt)² = (dv/dt) , unit is m/s²

Answer:

Explanation:

Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.

5N
5 N
19 N
19 N

Pls help look at the pic

Answers

Answer:

b. is the correct answer ....

two 100 ohm resistors are connected inparallel and one identical resister in series. The maximum power that can be delivered to any one resistor is 25W. What is the maximum voltage that can be applied between the terminals A and B ?
A. 50V
B. 75V
C. 100V
D. 125V

Answers

SOLVED DOWN BELOW

Explanation:

In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law

I = 25/200

I= .125 amps or 125 ma

__________

R= r1 * r2 / r1 +r2

R= 100 * 100 / 100 + 100

R= 10000 / 200

R= 50 ohms

A. 50v volunteered

Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion with a negative velocity and a negative acceleration.

Answers

Answer:

An example of positive velocity is throwing a ball upwards

An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height

Explanation:

In a vertical movement the acceleration is always downwards, therefore negative since it is created by the attraction of the Earth on the body.

An example of positive velocity is throwing a ball upwards

An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height

magnetism/ magnetic field ana magnetic forces

Answers

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

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