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finlep [7]
2 years ago
9

For the following substances: air, water and iron. order of speed of sound waves from low to high

Physics
1 answer:
pantera1 [17]2 years ago
5 0

D)

the slowest through gases, faster through liquids, and fastest through solids.

is that Right?

pls let me know

You might be interested in
Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna
Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

I=\frac{V}{R}

Where R is the equivalent resistance of the resistors in series

R=0.0510+0.0090=0.0600[ohm]

I=\frac{12.0}{0.0600}=200A

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V

The power dissipated supplied to the motor is given by:

P=I^2*R_m\\P=(200)^2*0.0510=2.04kW

now solving adding a 0.0900 ohm resistor:

I=\frac{12.0}{0.15}=80A

V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V

P=I^2*R_m\\P=(200)^2*0.0510=0.326kW

5 0
3 years ago
What is the magnitude and direction (right or left) of the
Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

Fy_{net}=F_{n}+F_{g} (1)

Where F_{n}=12 N is the Normal force, directed upwards and F_{g}=-12 N is the weight of the box (gravity force), directed downwards.

Fy_{net}=12 N-12 N (2)

Fy_{net}=0 N (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

Fx_{net}=F_{left}+F_{right} (4)

Where F_{left}=-3 N and F_{right}= 15 N

Fx_{net}=-3 N + 15 N (5)

Fx_{net}=12 N (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0
3 years ago
Why are the element from period 2 grouped together
valentinak56 [21]
All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.
4 0
3 years ago
The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.
Assoli18 [71]
A magnetic flux would be the correct answer

8 0
3 years ago
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