Find the frequency, if the amplitude of a 3000g object in simple harmonic motion is 1000cm and the maximum speed of the object is 5.00 m/s
3.14 Hz
7.96 102 Hz
0 278 × 10 2 HZ
O 0.500 Hz
2.00 Hz

Answers

Answer 1

Answer:

A = 10 m     amplitude

m = 3 kg     mass of object

Vm = 5 m/s

w A = Vm      where w = omega

w = 2 * pi * f

2 * pi * f  10 = 5

f = 5 / (20 * pi)  = .0796 / sec


Related Questions

Chase grew up wanting to wear his sister's dresses over his brother's pants and button up shirts. When Chase turns 18, he decides to begin living as woman, though he's still only sexually attracted to women. He decides he doesn't want to undergo surgery. Chase is

Answers

Explanation:

she is a woman, i dont understand why youre still using he/him after she comes out

Can someone log into my acc FOR ME I will pay you to complete my physics assignments for money or points?!!

Answers

Answer: no sorry../

Explanation:


. A patient on the infectious disease floor takes 10 mL of levofloxacin syrup bid. If the product is only
available as a 5 mL unit-dose oral syringe, how many syringes will the technician prepare for a 24-
hour supply?

Answers

Answer:

The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.

Explanation:

brainliest plz . . .

Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.

Answers

Answer:

always runs slower than normal.

Explanation:

The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.

According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.

In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.

a youthful person run at 7km/h in a north- west direction across the derk of a ship in which is streaming due east at 40 km/h .ifind the velocity of the boy relative to the Sea
ii,And the velocity of the sea relative to the boy​

Answers

Answer:

velocity of ship wrt sea= 7i∧

velocity of women on deck= 40j∧

velocity of women relative to sea will be resultant of above two velocoties,

7i∧ + 40j∧ magnitude is

square root (7 x 7 + 40x40)

=√49+1600

=√1612

=40.14 m/s

A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of air is 1.29 kg/m^3. The skin and structure of the balloon has a mass of 910 kg. Neglect the buoyant force on the cargo volume itself.
Determine the largest mass of cargo the balloon can lift.

Answers

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

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A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​

Answers

image distance=-40cm=vObject distance=u=?Magnification=m=3

We know

[tex]\boxed{\sf m=-\dfrac{v}{u}}[/tex]

[tex]\\ \sf\longmapsto 3=-\dfrac{-40}{u}[/tex]

[tex]\\ \sf\longmapsto 3=\dfrac{40}{u}[/tex]

[tex]\\ \sf\longmapsto u=\dfrac{40}{3}[/tex]

[tex]\\ \sf\longmapsto u=13.3cm[/tex]

Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?

a.) a loudspeaker

b.) a headphone

c.) a light bulb

d.) a microphone

I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure

Answers

Answer:

I THINK C

Explanation:

BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.

Xác định ứng lực trong các thanh BC,
CF và FE của hệ giàn và chỉ rõ các thanh
chịu kéo hay nén. Cho P1=P2=600 lb,
P3=800 lb

Answers

Answer:

............................................

Explanation:

herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran

Answers

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

The speed of a sound wave

A. Depends on wavelength.
B. Depends on the medium.
C. Depends on amplitude.
D. None of the above.

Answers

Answer:

B) the medium

Explanation:

B. Depends on the medium.

A cowgirl ties one end of a 10.0 m long rope to a fence post (to the left of the cowgirl) and pulls on the other end so the rope is stretched horizontally with a tension of 140 N. The mass of the rope is

Answers

Answer:

Mass is zero if the whole rope is horizontal. Imaginary rope.

Explanation:

With any mass at all, only a small section of the rope will be truly horizontal. The rope curve will be a a catenary.

A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant speed. If the driver of the car hears the horn at a frequency of 246 Hz, then what is the speed of the car? Use 340 m/s for the speed of the sound

Answers

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a distance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?

Answers

Answer:

[tex]D_l=d[/tex]

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

[tex]V_{e1}=V_{e2}[/tex]

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

[tex]D=\frac{at^2}{2}[/tex]

Where

Acceleration is given as

[tex]a=\frac{V_o}{2d}[/tex]

And

Time

[tex]T=\frac{d}{v_0}[/tex]

Therefore horizontal displacement towards the left is

[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]

[tex]D_l=d[/tex]

A bus moving on a straight road increases its speed uniformly from rest to 20m's over a time period of 1 min. The distance travelled during the time is (a) 150 m (b) 300 m (c) 600 m (d) 900 m​

Answers

Explanation:

Given that,

Initial velocity (u) = 0 m/sFinal velocity (v) = 20 m/sTime taken (t) = 1 minute = 60 seconds

In order to find the distance travelled, firstly we need calculate the acceleration.

v = u + at

→ 20 = 0 + 60a

→ 20 = 60a

→ 20 ÷ 60 = a

→ ⅓ m/s² = a

Now, by using the 2nd equation of motion :

s = ut + ½at²

→ s = 0(60) + ½ × ⅓ × (60)²

→ s = ⅙ × 3600

→ s = 1 × 600

→ s = 600 m

Hence, the distance travelled is 600 m.

A circular parallel-plate capacitor whose plates have a radius of 25 cm is being charged with a current of 1.3 A. What is the magnetic field 11 cm from the center of the plates

Answers

The magnetic field at 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.

Given;

radius of the circular plate, d = 25 cm = 0.25 m

current in the plate, I = 1.3 A

distance from the center of the circular plate, r = 11 cm = 0.11 m

To find:

magnetic field (B)

The magnetic field from the given distance is calculated as from Biot Savart equation:

[tex]B = \frac{\mu_o I}{2\pi r} \\\\where;\\\\\mu_o \ is \ permeability \ of \ free \ space \ 4\pi \times 10^{-7} \ T.m/A\\\\B = \frac{(4\pi \times 10^{-7} ) \times (1.3)}{2\pi \times 0.11} \\\\B = 2.364 \ \times 10^{-6} \ T[/tex]

Therefore, the magnetic field 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.

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You spaceship has a snazzy lounge called Ten-Forward that needs a new spacecouch. Sadly the couch you bought is too long and won't fit into your car when it is stationary. The couch has a length Lc = 14.0 m and mass m = 49 kg, and your shuttlecraft has a length of L = 9.0 m How fast would you need to run, in terms of the speed of light c, in order to get the couch to fit inside the length of the shuttlecraft?

Answers

Answer:

v = speed

c = speed of light

using the equation

L = Lc Sqrt( 1 - (v/c)2)

9 = 14 Sqrt( 1 - (v/c)2)

v/c = 0.765

v = 0.765 c

HELP ME ASAP PLSSS!!​

Answers

I hope this helped !

Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor​

Answers

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

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A circular loop of wire 10 cm in radius carries a current of 20 A. The axial magnetic field 15 cm from the center of the loop is approximately:

a. 37 mu-T
b. 13 mu-T
c. 21 mu-T
d. 41 mu-T
e. 18 mu-T]

Answers

Answer:

ans is c

Explanation:

chk photo

The axial magnetic field 15 cm from the center of the loop is approximately 21 μT. Hence, option (c) is correct.

What is magnetic field?

In the vicinity of a magnet, an electric current, or a shifting electric field, there is a vector field called a magnetic field where magnetic forces can be seen. Electric charges in motion and the intrinsic magnetic moments of elementary particles connected to the fundamental quantum characteristic known as spin create a magnetic field.

Given parameters:

Radius of the circular loop: r = 10 cm = 0.10 m.

Current passing through the loop: I = 20 A.

Axial distance of the point: z = 15 cm = 0.15 m.

Hence, the axial magnetic field at that point:

B = (μ₀/4π) 2πR²I/(z²+R²)^(3/2)

By putting these values, we get B = 21 × 10⁻⁶ T = 21 μT.

Learn more about magnetic field here:

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A solid metal sphere of radius 3 m carries a total charge of -5.5 uc. What is the magnitude of the
electric field at a distance from the sphere's center of (a) 2.9 m and (b) 8 m? How would the answers
differ if the sphere was (c) a thin shell.
IN​

Answers

Answer:

2.9::: 5.87*10*3 N/C

8: 7.73 × 10 ^2  N/C

Explanation: https://study.com/academy/answer/a-solid-metal-sphere-of-radius-3-00-m-carries-a-total-charge-of-5-50-muc-what-is-the-magnitude-of-the-electric-field-at-each-of-the-following-distances-from-the-sphere-s-center-a-3-10-m-b-8-00-m.html

A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor

Answers

Explanation:

Given:

[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]

[tex]V = 12 V[/tex]

[tex]I = 1.2 A[/tex]

Recall that power P is given by

[tex]P = VI[/tex]

so the amount of energy dissipated [tex]\Delta E[/tex] is given by

[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]

[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]

1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia. ​

Answers

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Describe sound and record​

Answers

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long after the asteroid hit the Moon, which is 3.84 x 10^5 km away, would the light first arrive on Earth?

Answers

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?

Answers

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

                       

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

How many types of physics?

Answers

Answer:

Two Main Branches of Physics

it is Classical Physics and Modern Physics.

Explanation:

Further sub Physics branches are Mechanics, Electromagnetism, Thermodynamics, Optics, etc. The rapid progress in science during recent years has become possible due to discoveries and inventions in the field of physics.

hope it helped

What do scientists use to determine the temperature of a star?

Answers

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?

Answers

Answer:

2.5 km

Explanation:

Answer:

2.5 km

Explanation:

Distance = speed x time

So =5 x 0.5

A person runs up the stairs elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.
Determine the work done by the person climbing the stair case.

Answers

Answer:

Work done = 2289.084 Joules

Explanation:

Given the following data;

Mass = 102 Kg

Height = 2.29

Time = 1.32 seconds

We know that acceleration due to gravity, g = 9.8 m/s²

a. To find the work done by the person;

Here, work would be done in the form of gravitational potential energy.

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

Work done = 102 * 2.29 * 9.8

Work done = 2289.084 Joules

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