1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MrRa [10]
3 years ago
7

The length (L) of a simple pendulum is directly proportional to the square of the time period (T). Which type of graph can best

describe this relationship?
Physics
2 answers:
iris [78.8K]3 years ago
8 0
L=T^2 so that would be a parabola
forsale [732]3 years ago
6 0

Answer:

L=T^2 so its parabola

Explanation:

You might be interested in
Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in
sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0
3 years ago
(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa
Vedmedyk [2.9K]

Missing data: the wave number

k=60 cm^{-1}

(a)  z = 0.003 sin (6000y-31.4 t)

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

z = A sin (ky-\omega t)

where

A is the amplitude of the wave

k is the wave number

\omega is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

k = 60 cm^{-1} = 6000 m^{-1} is the wave number

T = 0.20 s is the period, so the angular frequency is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s

So, the wave equation (in meters) is

z = 0.003 sin (6000y-31.4 t)

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

v_t = z' = -A \omega cos (ky-\omega t)

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

v_{t}_{max} =\omega A

where

\omega = 31.4 rad/s\\A = 0.003 m

Substituting,

v_{t}_{max}=(31.4)(0.003)=0.094 m/s

(c) 5.24 mm/s

The wave speed is given by

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

The frequency can be found from the angular frequency:

f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz

While the wavelength can be found from the wave number:

\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m

Therefore, the wave speed is

v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s

7 0
3 years ago
What happens to parallel light rays that strike a concave lens?
zhenek [66]
Due to the shape of the lens , parallel rays will be deviated
5 0
3 years ago
Read 2 more answers
Write a question you would like to see on a quiz to calculate NET FORCE.
Pie

Answer:

How do you find the net force?

Explanation:

i think this is want you asked for?

5 0
2 years ago
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
Other questions:
  • what is the best term to describe information sent as patterns and codes in the controlled flow of electrons through a circuit
    6·1 answer
  • What is the claim evidence and reasoning, for the top one
    9·1 answer
  • (b) An example of an inverse square field is the gravitational field F = −(mMG)r/|r|3. Use part (a) to find the work done by the
    6·1 answer
  • Fossil fuels are..
    12·1 answer
  • A 4.6-kg block of ice originally at 263 K is placed in thermal contact with a 15.7-kg block of silver (cAg = 233 J/kg-K) which i
    7·1 answer
  • An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens
    6·1 answer
  • Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th
    7·1 answer
  • The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner
    15·1 answer
  • If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?
    5·1 answer
  • an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!