What is the final temperature after 80.0 j?

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

4 years ago

in a hydraulic garage the small piston has a radius of 5 cm and the large piston has radius of 15 cm what force must be applied

viktelen [127]

The force applied to small piston = 2.2 x 10³ N

<h3>Further explanation</h3>

Given

a radius of 5 cm and 15 cm

weight 20000 N

Required

Force applied

Solution

Pascal Law :

F₁/A₁=F₂/A₂

A₁ = π.5²

A₂ = π.15²

F₁/ π.5² cm² = 20000/π.15² cm²

F₁ = 2222.22 N⇒2.2 x 10³ N

5 0

3 years ago

Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b

Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is $n_2 = 1.333$ , and for air $n_1 = 1.00$ : the angle of light with the normal is $90^o-60^o = 30^o$ ; therefore Snell's law gives

$n_1sin(\theta_1)= n_2sin(\theta_2)$

$1.00*sin(\theta_1) = 1.33 sin(30^o)$

$sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}$

$sin (\theta_1) = 0.665$

$\theta _1 = sin^{-1}(0.665)$

$\boxed{\theta_1 = 41.68^o}$

4 0

3 years ago

If a gasoline engine has an efficiency of 21 percent and losses 780 J to the cooling system and exhaust during each cycle, how m

NISA [10]

As we know that efficiency of engine is given as

$efficiency = \frac{W}{Q_i}$

also we know that

$Q_i - Q_o = W$

given that

$Q_o = 780 J$

efficiency = 21%

now we have

$0.21 = \frac{Q_i - Q_o}{Q_i}$

$0.21 Q_i = Q_i - 780$

$0.79Q_i = 780$

$Q_i =987.3J$

now we have

$W = 987.3 - 780 = 207.3 J$

7 0

3 years ago

Two spherical balloons are filled with water. The first balloon has a radius of 3 cm, and the second has a radius of 6 cm. How m

Marianna [84]

The volume of water in the larger balloon is 8 times greater than in the smaller balloon

We'll begin by calculating the volume of each balloon.

<h3>For smaller balloon:</h3>

Radius (r) = 3 cm
Pi (π) = 3.14
Volume (V) =?

V = 4/3 πr³

V = 4/3 × 3.14 × 3³

<h3>For larger balloon:</h3>

Radius (r) = 6 cm
Pi (π) = 3.14
Volume (V) =?

V = 4/3 πr³

V = 4/3 × 3.14 × 6³

Finally, we shall determine how much greater the larger balloon is to the smaller balloon

Volume of smaller balloon = 113.04 cm³
Volume of larger balloon = 904.32 cm³
Greatness =?

Greatness => large / small

Large / small = 904.32 / 113.04

Large / small = 8

Cross multiply

<h3>Large = 8 × small </h3>

Therefore, the larger balloon is 8 times greater than the smaller balloon.

Learn more on volume of sphere: brainly.com/question/9178703

4 0

3 years ago

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