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Artyom0805 [142]
3 years ago
7

Which one is not physical quantity question is mistake

Physics
2 answers:
Murljashka [212]3 years ago
6 0

Answer:

length

hope it helps you

Anastaziya [24]3 years ago
5 0

Answer:

Length

Explanation:

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two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation
ad-work [718]

The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

  • k is the spring constant
  • m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

3 0
2 years ago
The velocity of sound in air saturated with water vapour at 30°C
Luba_88 [7]

Explanation:

The velocity of sound depends on the density of the medium.  So we need to find the density of air at each set of conditions.  The density of air is:

ρ = (Pd / (Rd T)) + (Pv / (Rv T))

where Pd and Pv are the partial pressures of dry air and water vapor,

Rd and Rv are the specific gas constants of dry air and water vapor,

and T is the absolute temperature.

At the first condition:

Pv = 31.7 mmHg = 4226.3 Pa

Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 30°C = 303.15°C

ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)

ρ = 0.94746 + 0.03021

ρ = 0.97767 kg/m³

At the second condition:

Pv = 0 Pa

Pd = 650 mmHg = 86660 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 0°C = 273.15°C

ρ = (86660 / 287.00 / 273.15) + (0 / 461.52 / 273.15)

ρ = 1.1054 + 0

ρ = 1.1054 kg/m³

The square of the velocity of sound is proportional to the ratio between pressure and density:

v² = k P / ρ

Since the atmospheric pressure is constant, we can say it's proportional to just the density:

v² = k / ρ

Using the first condition to find the coefficient:

(340)² = k / 0.97767

k = 113018.652

Now finding the velocity of sound at the second condition:

v² = 113018.652 / 1.1054

v = 319.75

6 0
3 years ago
Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equ
sdas [7]

Answer:

x_{3}=2.35

Explanation:

Given x^2-2x-1=0,x_1=2

From Newton's method

x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

f(x)=x^2-2x-1

f'(x)=2x-2

Now

x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}

f(x)=x^2-2x-1

f(2)=2^2-2\times 2-1

f(x)=-1

f'(2)=2x-2

f'(1)=2\times 2-2

f'(1)=2

x_{2}=2+\dfrac{1}{2}

x_{2}=2.5

x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}

f(2.5)=2.5^2-2\times 2.5-1

f(2.5)=0.45

f'(2.5)=2\times 2.5-2

f'(2.5)=3

x_{3}=2.5-\dfrac{0.45}{3}

So

x_{3}=2.35

8 0
3 years ago
11 kg is a familiar weight for a bag of flour. You are baking cookies for a Save The Rain Forest fund drive. It takes 500 g of f
arlik [135]

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

8 0
3 years ago
Metals are used in many products because of the characteristics properties that most metals have. Which product requires the hig
nadezda [96]
Alright sounds good buddy
5 0
2 years ago
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