How is a light element like helium is formed in stars and how is a heavier element like gold formed by stars?

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

4 0

3 years ago

Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o

postnew [5]

The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
= (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
= 539.7 m²

Answer: 539.7 m²

8 0

3 years ago

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

beks73 [17]

Answer:

4.6 kHz

Explanation:

The formula for the Doppler effect allows us to find the frequency of the reflected wave:

$f'=(\frac{v}{v-v_s})f$

where

f is the original frequency of the sound

v is the speed of sound

vs is the speed of the wave source

In this problem, we have

f = 41.2 kHz

v = 330 m/s

vs = 33.0 m/s

Therefore, if we substitute in the equation we find the frequency of the reflected wave:

$f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz$

And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:

$f_B = |f'-f|=|45.8 kHz-41.2 kHz|=4.6 kHz$

6 0

3 years ago

Read 2 more answers

Alice suffered a minor electrical shock as she switched on her bathroom light. She is concerned and wants to prevent it from hap

snow_tiger [21]

Get a pull chord light switch installed in her bathroom by a qualified electrician asap. Meanwhile, keep hands as dry as possible, and try not to go near that switch until it's either been properly earthed, or whatever the problem actually is, and get qualified advice on what the problem is. Don't have wet feet either, and don't stand in puddles of water whilst operating - i she has to - the switch. 250V AC mains can be lethal, and at least painful.

7 0

2 years ago

Read 2 more answers

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

2 years ago