Which of the following is 8000 written in scientific notation?

Does wind have atoms and molecules in int

satela [25.4K]

All forms of matter are comprised of atoms. Wind in itself is the movement of particles(air). Wind does not have particles since it is an action and not a form of matter.

4 0

4 years ago

Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down

Lorico [155]

Answer:

sin 2θ = 1 θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

R = vo² Sin (2 15) /9.8

R = Vo² 0.051 m

In the table are all values in two ways

Angle (θ) distance R (x)

0 0 0

15 0.051 Vo² 0.5 Vo²/g

30 0.088 vo² 0.866 Vo²/g

45 0.102 Vo² 1 Vo²/g

60 0.088 Vo² 0.866 Vo²/g

75 0.051 vo² 0.5 Vo²/g

90 0 0

See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL

6 0

3 years ago

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstrat

Likurg_2 [28]

Answer:

Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

a = 2 lines 1/600

a = 2/600

a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

tan θ = y / L

as the measured angles are small

tan θ = sin θ / cos θ sin θ

sin θ = y / L

we substitute

a y/L = λ

y = λ L / a

for λ = 400 10-9 m

I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

i = 346.89 10⁻⁶ m

f

or λ = 700 nm

y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

y_f = 609.609 10⁻⁶ m

the separation of this spectrum

Δr = v_f - i

Dr = (609.609 - 346) 10 ⁻⁶

Dr = 263 10⁻⁶ m

4 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro

ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

4 0

3 years ago