All categories

3 years ago

A pendulum takes .5 seconds tp complete one cycle. what is the pendulum"s period and frequency?

1 answer:

5 0

You just told us how long the pendulum takes to complete one cycle. That's the definition of the period. Its 1/2 second.

The frequency is the reciprocal of the period. That's 2 per second. Or 2 Hz.

You might be interested in

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

Using the equation of heat:

Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

3 years ago

An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0

Diano4ka-milaya [45]

The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = µ x n x I where:
µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = µ o = 4 pi x 10^-7 Tm/A
I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT

5 0

3 years ago

The cheetahs in Kenya and the cheetahs in Tanzania make up separate _____.

Andreyy89

C. Populations.
Hope that's right.

6 0

3 years ago

Read 2 more answers

select all that apply the force of gravity a. changes slightly with the location on the earth b. decreases with height above sea

Reptile [31]

The answer to your question is C

7 0

3 years ago