The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a pendulum (to nearest hundredth) if it makes one swing in 2.1 seconds? Use 3.14 for \pi .

Answers

Answer 1

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

or

T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

Hence, L(T) = 8T²/π²

But T = 2.1

Pi= 3.14

8(2.1)²/3.14²

35.28/9.85

= 3.6feet


Related Questions

Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you

Answers

Answer:

   f = 6.37 Hz,       T = 0.157 s

Explanation:

The expression you have is

       y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

      y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

         w = 2π f

         f = w / 2π

from the equation w = 40 rad / s

        f = 40 / 2π

        f = 6.37 Hz

frequency and period are related

       f = 1 / T

       T = 1 / f

       T = 1 / 6.37

       T = 0.157 s

which discribes what a velocity/time graph would look like with no accelaration

Answers

It would be a straight line. No acceleration means constant velocity.

Which may result from an increase in friction?

A: decreased traction
B: increased speed
C: reduced wear and tear
D: generation of heat​

Answers

Answer:

Its Generation of Heat. or "D"

Explanation:

Friction causes generation of heat and causes increased wear and tear.

The friction is a resistive force and is related to heat energy, so an increase in friction results in the generation of heat, so option D is correct.

What is friction?

Two solid objects cannot roll or slide over one another due to the force of friction. Although frictional forces can be useful, such as the traction needed to walk without slipping, they can also present a large degree of resistance to motion. Automobile engines need about 20% of their power to overcome frictional forces in moving parts.

The fundamental source of friction between metals appears to be the adhesion forces between the contact zones of the surfaces, which are always microscopically uneven. Friction is produced by shearing, these "welded" seams, and the rubbing action of the rougher, tougher surface against the softer, smoother surface.

The friction is a resistive force trying to oppose the force applied As friction is related to heat when we increase the heat, the friction increases, and vice versa.

To know more about Friction:

https://brainly.com/question/28356847

#SPJ6

Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 9 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. If the squid expels 2 kg of water out of its body with a speed of 8 m/s, what would be its own escape speed

Answers

Answer:

v_squid = - 2,286 m / s

Explanation:

This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.

Initial moment. Before expelling the water

          p₀ = 0

the squid is at rest

Final moment. After expelling the water

         [tex]p_{f}[/tex] = M V_squid + m v_water

         p₀ = p_{f}

          0 = M V_squid + m v_water

           c_squid = -m v_water / M

The mass of the squid without water is

            M = 9 -2 = 7 kg

let's calculate

           v_squid = 2 8/7

           v_squid = - 2,286 m / s

The negative sign indicates that the squid is moving in the opposite direction of the water

pls help me with this question​

Answers

Answer:

16 ms2 is the answer for this question

One star has a temperature of 30,000 K and another star has a temperature of 6,000 K. Compared to the cooler star, how much more energy per second will the hotter star radiate from each square meter of its surface?

Answers

Answer:

The hotter star radiates 625 times more energy per second from each square meter of its surface

Explanation:

Temperature of the hotter star is 30000 K

temperature of the cooler star = 6000 K

From Stefan-Boltzmann radiation laws, for a non black body

P = εσA[tex]T^{4}[/tex]

where

P is the energy per second or power of radiation

ε is the emissivity of the body

σ is the Stefan-Boltzmann constant of proportionality

A is the area of the sun

T is the temperature of the sun

The sun can be approximated as a black body, and the equation reduces to

P = σA[tex]T^{4}[/tex]

For the hotter body,

P = σA([tex]30000^{4}[/tex]) = 8.1 x 10^17σA  J/s

For the cooler body,

P = σA([tex]6000^{4}[/tex]) = 1.296 x 10^15σA   J/s

comparing the two stars energy

==> (8.1 x 10^17)/(1.296 x 10^15) = 625

This means that the hotter star radiates 625 times more energy per second from each square meter of its surface

The transfer of charge from clouds to the earth or cloud to cloud is called

Answers

That's called "lightning".

Answer:

The lightning itself is the transfer of charge from one region of a cloud to another or between the cloud and Earth. The narrow channel within which the flash of lightning occurs is heated suddenly to ~ 30,000 K, with essentially no time to expand.

Explanation:

Hope it helps

a 15kg television sits on a shelf at a height of 0.3 m how much gravitational potential energy is added to the television when it is lifted to a shelf of height 1.0m?

Answers

Answer:

103 Joules

Explanation:

In this problem we are required to find the potential energy possessed by the television

Given data

mass of television m = 15 kg

height  added above the ground, h= 1-0.3 = 0.7 m

acceleration due to gravity g = 9.81 m/s^2

apply the formula for potential energy we have

P.E= m*g*h

P.E = 15*9.81*0.7 = 103 Joules

Please help asap. A soccer player can kick a 0.370 kg football at 55 km/h. How much work does the soccer player have to do on the ball in order to give it that much kinetic energy?

Answers

Answer: 43.2 J

Explanation:

Work = change in KE

initial KE = 0

final KE = 1/2mv^2 = 1/2(0.370 kg)(15.2778 m/s)^2 = 43.2 J

i'm not sure about sig figs though

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that the ball reaches the bottom of the cliff 8 s after the shoot? (Take g = 10 m/s^2 and the height of the robot is negligible.)
A 25 m
B 45 m
C 80 m
D 145 m​

Answers

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

A 15 kg object is pulled by a force of 68 N. If the surface exerts a friction force of 23 N, with what acceleration does the object move?

Answers

Answer:

3 m/s²

Explanation:

Sum of forces in the x direction:

∑F = ma

68 N − 23 N = (15 kg) a

a = 3 m/s²

A plane drops a package for delivery. The plane is flying horizontally at a speed of 120m/s,and the package travels 255 m horizontally during the drop. We can ignore air resistance.What is the package's vertical displacement during the drop?

Answers

Answer:

Package's vertical displacement(s) = 22.12 meter

Explanation:

Given:

Speed of plane = 120 m/s

Total distance = 255 m

Find:

Package's vertical displacement(s)

Computation:

Time taken = Distance / Speed

Time taken = Total distance / Speed of plane

Time taken = 255 / 120

Time taken = 2.125 s

Acceleration due to gravity(g) = 9.8 m/s²

Initial velocity (u) = 0

So,

Package's vertical displacement(s) = ut + (1/2)gt²

Package's vertical displacement(s) = (0)(2.125) + (1/2)(9.8)(2.125)²

Package's vertical displacement(s) = 22.12 meter

Answer: -22.1

Explanation:

I just did the Khan Academy and that was the answer, not the one provided by that one person. :)))

If y=5sin (3x -40)
Calculate the frequency and period​

Answers

Answer:

0.477 Hz

2.09 s

Explanation:

y = A sin(ωx − φ)

A is the amplitude, ω is the angular frequency, and φ is the phase shift.

ω = 3 rad/s

f = ω / 2π ≈ 0.477 Hz

T = 1/f ≈ 2.09 s

PLEASEEE HELP!!!!! I HAVE BEEN STRUGGLING FOR 2 DAYS
If i workout 90 minutes on earth, if I am on a rocket traveling 0.80c, according to the timer on the rocket, how long should I exercise?

Answers

Answer:

You should still workout 90 min.

The proper time is measured by a single clock in a single place.

The proper time on earth is 90 min.

The clock on the rocket is also in a single place in the frame of the rocket so you still need to workout for 90 min.

A 140-Hz sound travels through pure carbon dioxide. The wavelength of the sound is measured to be 1.92 m. What is the speed of sound in carbon dioxide?

Answers

Answer:

V = 268.8 m/s

Explanation:

The speed of a wave in general is given by the following formula:

V = fλ

where,

V = Speed of that wave

f = Frequency of the wave

λ = wavelength of the wave

In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:

V = Speed of Sound in Carbon dioxide = ?

f = frequency of sound wave = 140 Hz

λ = wavelength of sound wave = 1.92 m

Therefore,

V = (140 Hz)(1.92 m)

V = 268.8 m/s

A disk-shaped merry-go-round of radius 3.03 mand mass 145 kg rotates freely with an angular speed of 0.681 rev/s . A 65.4 kg person running tangential to the rim of the merry-go-round at 3.41 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round?

Answers

Answer:

[tex]\omega_2=0.891\ rev/s[/tex]

Explanation:

Given that

Radius , r= 3.03 m

Mass of disk , M= 145 kg

Initial angular velocity

ω=0.681 rev/s

Mass of person , m= 65.4 kg

Velocity of person , V= 3.41 m/s

Initial mass moment of inertia

[tex]I_1= \dfrac{M\times R^2}{2}[/tex]

[tex]I_1= \dfrac{145\times 3.03^2}{2}=665.61\ kg.m^2[/tex]

Final mass moment of inertia

[tex]I_2= \dfrac{M\times R^2}{2}+m\times R^2[/tex]

[tex]I_2= \dfrac{145\times 3.03^2}{2}+65.4\times 3.03^2=1266.04\ kg.m^2[/tex]

[tex]Final\ angular\ velocity =\omega_2[/tex]

By using angular momentum equation

[tex]I_1\times \omega+m\times V\times R=I_2\times \omega_2[/tex]

[tex]665.61\times 0.681+65.4\times 3.41\times 3.03=1266.04\times \omega_2[/tex]

[tex]1129.01= 1266.04\times \omega_2[/tex]

[tex]\omega_2=\dfrac{1129.01}{1266.04}[/tex]

[tex]\omega_2=0.891\ rev/s[/tex]

Thus the angular velocity will be 0.891 rev/s

Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona los frenos y se detiene 4 s después. Si la vaca estaba a 25 m de la tractomulacuando el conductor pisó el freno. ¿atropelló la vaca? Justifique su respuesta.

Answers

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s)  = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) =  -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s  

v(t) =  (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

The following passage has not been edited. There is an error in each line. Write the

incorrect word and the correction in your answer sheet against the correct question

number. The first one has been done as an example. ( 1 x 4 = 4 )

Community service sensitize people to Error: sensitize ; Correction: sensitizes

other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______

development to the underprivileged (b) Error: _______ ; Correction: ______

sections with society. Courses about social (c) Error: ______ ; Correction: _______

work prepares frontline workers to (d) Error: _______ ; Correction: ______​

Answers

Answer:

(a) Error: Other's ; Correction: Others'

(b) Error: to ; Correction: for

(c) Error: with ; Correction: of

(d) Error: prepares ; Correction: prepare

Explanation:

a)  The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.

b)  The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.

c)  The error is in the word "with"  and the correct word will be "of" as of indicates relationships between other words including things that made of other things.

d)  The error is in the word "preapres" and the correct word will be "prepare".

Which of the following object is in dynamic equilibrium?

Answers

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

The specific latent heat of vaporisation of water is 2200 kJ kg, and the specific
heat capacity of water is 4200 J/ kg.K. A heating element is immersed in an
insulated cup of water. It takes five minutes to boil the water completely from an
initial temperature of 30°C. Assuming no heat is lost to the surroundings,
determine the mass of the water if the heating element is operating at 1000 W.​

Answers

Answer:

0.12 kg

Explanation:

The amount of energy added is:

1000 W × (5 min × 60 s/min) = 300,000 J = 300 kJ

Heat to boil the water is:

q = mCΔT + mL

q = m (CΔT + L)

300 kJ = m (4.2 kJ/kg/K × (100°C − 30°C) + 2200 kJ/kg)

300 kJ = m (2494 kJ/kg)

m = 0.12 kg

The density of water is 1000 kg m^3. What is the value expressed in gcm^-3 units? please help me..
(1) 1000 (2) 100 (3) 1 (4) 0.1 (5) 0.01​

Answers

The answer is 1 which is option 3

Here's the neat, cool way to convert units like this:

-- 1 kilogram  =  1,000 grams

-- 1 meter  =  100 centimeters

So . . . . .

(1000 kg/m³) x (1000 g/kg) x (1 m/100 cm)³ =

(1,000 kg/m³) x (1,000 g/kg) x (1 m³/1,000,000 cm³) =

(1,000 x 1,000 x 1 / 1,000,000) (kg-g-m³ / m³-kg-cm³)  =  1 g/cm³

Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. From what height above the ground did Mariana drop the ball?

Answers

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

A ball is thrown vertically upwards. It returns 6s later. Calculate : (1) the greatest height reached by the ball, and (2) the initial velocity of the ball. (Take g=9.8m/s2) ​

Answers

Answer:

greatest displacement = 44.1m

initial velocity= 29.4m/s

Explanation:

Greatest displacement

s=1/2at^2

= (9.8/2 ×9)m

= 44.1m

initial velocity

s=ut-1/2at^2

44.1= 3u -(1/2×9.8×9)

44.1=3u-44.1

3u=88.2

u=29.4m/s

the resistor of values 6 ohm,6 ohm are connected in series and 12 ohm are connected in parallel. the equivalent resistance of the circuit is

Answers

Answer:

The equivalent or total resistance of the circuit is 6

Explanation:

6 &6 are in series

6+6=r

r= 12

1/Rtotal= 1/12+1/2

1/Rt=2/12=1/6

Rt=6

Ratan took a thin, solid piece of material. When he put it in water, it rose and floated. He took it out of the water and cut out holes in it as shown below. When Ratan puts the material with holes back in the water, what will happen?

Answers

Answer:

The material will still remain afloat.

Explanation:

The fact that the material floated on water as a solid piece means that the material is less dense than water. When a material is less dense than another liquid material, the material floats in the denser material. Cutting holes in the material will not reduce the buoyancy of the material. If the material had not floated as a single solid piece, then it will be said to be denser than the liquid material, and in such a case, if a strategic hole is cut on it to reduce its weight while maintaining its immediate volume, then it can be made to float. This is the principle behind metal floating ships.

The region of magnetic influence around either pole of a magnet is called the magnetic field. The magnetic field line points out from the south magnetic pole and in from the north magnetic pole. This statement is:

Answers

Answer:

This statement is not true

Explanation:

Because The normal magnetic field line points out from the north magnetic pole and in from the south magnetic pole.

Answer:

false

Explanation:

hope this helps :)

Two students are trying to measure how high a ball bounces when it is dropped from different heights. They dropped a ball from P and it bounced up till Q. They now have to record two measurements as shown below. Which measurements should they take?A.A B.B C.C D.D

Answers

Answer:

A.A

Explanation:

The balls usually bounce 60% of the original height because it stores 60% of the energy it had before the bounce. When a ball is dropped from a great height it has kinetic energy before it hits the ground which is the result of the bounce of ball. The size of ball does matter in this case, Large balls will bounce higher.

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)

Answers

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

how much heat is required to raise the temperature of 5kg of iron from 50°c to250°c​

Answers

Answer:

462000J

Explanation:

Quantity of heat= mass x specific heat capacity of iron x change in temp

specific heat capacity of iron is 462J/Kg/K

change in temp = 250-50= 200°C

200°C is equivalent to 200K since 1°C is 1K

Q= mct

= 5x462x200

= 462000J

formula of minimmum pressure​

Answers

Answer:

pressure=force/area

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