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
Alenkasestr [34]
3 years ago
5

Calculation on simple pendulum

Physics
1 answer:
kiruha [24]3 years ago
4 0

Answer:State the question so I can answer

Explanation:

You might be interested in
John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
3 years ago
Meteor Infrasound A meteor that explodes in the atmosphere creates infrasound waves that can travel multiple times around the gl
kherson [118]

Answer:

0.04455 Hz

Explanation:

Parameters given:

Wavelength, λ = 6.5km = 6500m

Distance travelled by the wave, x = 8830km = 8830000m

Time taken, t = 8.47hours = 8.47 * 3600 = 30492 secs

First, we find the speed of the wave:

Speed, v = distance/time = x/t

v = 8830000/30492 = 289.58 m/s

Frequency, f, is given as velocity divided by wavelength:

f = v/λ

f = 289.58/6500

f = 0.04455 Hz

8 0
3 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
The outside diameter of your teacher's rear bicycle tire is 16 inches. How far will he travel if the rear wheel makes 1200 revol
FinnZ [79.3K]

Answer:

241,274.32 inches

Explanation:

How far will he travel if the rear wheel makes 1200 revolutions on the road?

Since the rear wheel makes one revolution in the distance of a circumference of a circle, C with diameter, d = 16 inches

C = πd²/4

So, the distance, travelled in 1200 revolutions is D = 1200 × C = 1200πd²/4

Substituting d = 16 into D, we have

D = 1200πd²/4

D = 1200π(16)²/4

D = 76800π

D = 241,274.32 inches

3 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0
Grace [21]

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2    where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0
3 years ago
Other questions:
  • Someone please help with the picture up top??
    10·1 answer
  • How do you convert mass to weight and visa versa
    7·2 answers
  • What is the speed of a 6-kg ball with a momentum of 72 kg∙m/s?
    14·1 answer
  • if air resistance is negligible what maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/
    14·1 answer
  • The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
    6·1 answer
  • An easy way to draw a circle is to hold a string fixed at one end (for instance, with a pin) and attach a pen to the other end.
    7·1 answer
  • Please answer help me
    12·1 answer
  • In typical game play situation, (with no overtime), when is a game over?
    12·1 answer
  • write the difference between the fall of a parachute on the earth and its fall on the moon? On what condition does a body have f
    10·1 answer
  • A house is advertised as having 1560 square feet under roof. What is the area of this house in square meters?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!