Why is it important to collect forensic evidence and witness accounts as soon as possible after an accident?

Answer:

The process by which something absorbs or is absorbed by something else.

Answer:

a. 9.5838kg

b. 340.321kj

Explanation:

See attachment please

Answer:

The ability to do work or cause change

Explanation:

Answer: satellite is a artificial sub planet and every sub planet must around to it's main planet for their connect.

Explanation:

Satellite mainly 3types.

1. GEO (22,000mi away)

2. MEO (10,000mi away)

3. LEO (1,000-3,000mi away)

If any satellite will stabilised more then 22,000mi then it's can't around as a ring to world. And takes long time more then 24hours. And it's periodic time may disturbance.

The velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above are 1.18m/s and  2.51 * 10^5 Pa

The given parameters are:

Start by calculating the radius using:

[tex]r = 0.5d[/tex]

So, we have:

[tex]r_1 = 0.5 \times 4cm[/tex]

[tex]r_1 = 2cm[/tex]

Express as meters

[tex]r_1 = 0.02m[/tex]

Next, calculate the area using:

[tex]A =\pi r^2[/tex]

So, we have:

[tex]A_1 = \pi \times 0.02^2[/tex]

[tex]A_1 = \pi \times 0.004[/tex]

[tex]A_1 = 0.004\pi[/tex]

Also from the question, we have:

[tex]d_2 = 2.6cm[/tex]

[tex]h = 5m[/tex]

Calculate the radius using:

[tex]r = 0.5d[/tex]

So, we have:

[tex]r_2 = 0.5 \times 2.6cm[/tex]

[tex]r_2 = 1.3cm[/tex]

Express as meters

[tex]r_2 = 0.013m[/tex]

The area is then calculated as:

[tex]A =\pi r^2[/tex]

So, we have:

[tex]A_2 = \pi \times 0.013^2[/tex]

[tex]A_2 = \pi \times 0.000169[/tex]

[tex]A_2 = 0.000169\pi[/tex]

The velocity is then calculated using:

[tex]A_1v_1 = A_2v_2[/tex]

Make v2 the subject

[tex]v_2 = \frac{A_1v_1}{A_2}[/tex]

So, we have:

[tex]v_2 = \frac{0.0004\pi \times 0.5}{0.000169\pi}[/tex]

[tex]v_2 = \frac{0.0004\times 0.5}{0.000169}[/tex]

[tex]v_2 = 1.18343195266[/tex]

Approximate

[tex]v_2 = 1.18[/tex]

The pressure is then calculated as follows:

[tex]P_1 + 0.5 \times density \times v_1^2 + density \times g \times h_1 = P_2 + 0.5 \times density \times v_2^2 + density \times g \times h_2[/tex]

Where:

[tex]g = 9.81ms^{-1}[/tex]

[tex]density = 1000kgm^{-3[/tex]

[tex]h_1 = 0[/tex]

So, we have:

[tex]3 \times 10^5 + 0.5 \times 1000 \times 0.5^2 + 1000 \times 9.8 \times 0 = P_2 + 0.5 \times 1000 \times 1.18^2 + 1000 \times 9.8 \times 5[/tex]

[tex]300000 + 125 + 0 = P_2 + 696.2 + 49000[/tex]

Collect like terms

[tex]300000 + 125 - 696.2 - 49000 = P_2[/tex]

[tex]250428.8 = P_2[/tex]

Rewrite as:

[tex]P_2 =250428.8[/tex]

Rewrite as:

[tex]P_2 = 2.51 \times 10^5\ Pa[/tex]

Hence, the velocity and pressure are 1.18m/s and  2.51 * 10^5 Pa

Read more about pressures and velocities at:

https://brainly.com/question/3362972

The velocity in the 2.6-cm diameter pipe is 2.0 m/s and the pressure in the 2.6-cm diameter pipe is 2.73x10^5 Pa.

To solve this problem, we can use the principle of continuity and the Bernoulli equation.

Continuity equation

The continuity equation states that the mass flow rate through a pipe is constant. This means that the product of the density of the fluid, the cross-sectional area of the pipe, and the velocity of the fluid is the same at all points in the pipe.

ρ * A * v = constant

where:

ρ is the density of the fluid in kg/m³

A is the cross-sectional area of the pipe in m²

v is the velocity of the fluid in m/s

Bernoulli equation

The Bernoulli equation states that the total energy of a fluid is constant along a streamline. This total energy is made up of the fluid's pressure energy, kinetic energy, and potential energy.

P + 1/2ρv² + ρgh = constant

where:

P is the pressure of the fluid in Pa

ρ is the density of the fluid in kg/m³

v is the velocity of the fluid in m/s

g is the acceleration due to gravity in m/s²

h is the height of the fluid above a reference point in m

Solving for the velocity and pressure in the 2.6-cm diameter pipe

We can use the continuity equation to solve for the velocity in the 2.6-cm diameter pipe.

ρ * π(0.02 m)² * v_2 = ρ * π(0.04 m)² * v_1

where:

v_1 is the velocity in the 4.0-cm diameter pipe

v_2 is the velocity in the 2.6-cm diameter pipe

Substituting in the known values, we get:

v_2 = (π(0.04 m)² * v_1) / π(0.02 m)²

v_2 = 4v_1

Therefore, the velocity in the 2.6-cm diameter pipe is four times greater than the velocity in the 4.0-cm diameter pipe. Since the velocity in the 4.0-cm diameter pipe is 0.50 m/s, the velocity in the 2.6-cm diameter pipe is 2.0 m/s.

We can use the Bernoulli equation to solve for the pressure in the 2.6-cm diameter pipe.

P_1 + 1/2ρv_1² + ρgh_1 = P_2 + 1/2ρv_2² + ρgh_2

where:

P_1 is the pressure in the 4.0-cm diameter pipe

P_2 is the pressure in the 2.6-cm diameter pipe

h_1 is the height of the water in the 4.0-cm diameter pipe above a reference point

h_2 is the height of the water in the 2.6-cm diameter pipe above a reference point

Assuming that the height of the water in the two pipes is the same, we can cancel out the ρgh terms.

P_1 + 1/2ρv_1² = P_2 + 1/2ρv_2²

Substituting in the known values, we get:

3.03x10^5 Pa + 1/2ρ(0.50 m/s)² = P_2 + 1/2ρ(2.0 m/s)²

P_2 = 3.03x10^5 Pa + 1/2ρ(0.50 m/s)² - 1/2ρ(2.0 m/s)²

P_2 = 2.73x10^5 Pa

Therefore, the pressure in the 2.6-cm diameter pipe is 2.73x10^5 Pa.

For such more question on velocity

https://brainly.com/question/30505958

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