What's unique about the philosophical approach to the question of truth?

A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol

ELEN [110]

Answer:

v = 719.2 m / s and a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

v - v₀ = $v_{e}$ ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

m_fuel = 75 10

m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

M = 3000 -750 = 2250 kg

let's calculate

v - 0 = 2500 ln (3000/2250)

v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

Push = v_{e} dM / dt

let's calculate

Push = 2500 75

Push = 187500 N

If we use Newton's second law

F = m a

a = F / m

let's calculate

a = 187500/2250

a = 83.33 m / s²

7 0

2 years ago

A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then trav

bija089 [108]

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

$t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h$

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

$d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km$

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

$v_{average}=\frac{155km}{2.25h}=68.9km/h$

6 0

2 years ago

Suppose you want to make an interferometer with two slits to show the wave nature of light to your friends. The most convenient

nevsk [136]

Answer:

a) the pattern is bright, b) y = 8.25 mm, c) the distance increases with the wavelength increases

Explanation:

This is a team building exercise to perform interference experiments, where the separation of the slits is d = 1 m = 1 10⁻³ m and the distance to the screen L = 1.5 m

Constructive interference occurs when the path difference is equal to an integer number of wavelengths

d sin θ = m λ

a) The center of the two-ray pattern has the same length so the intensity is MAXIMUM, that is, the pattern is bright

b) to find the distance let's use trigonometry

tan θ= y / L

tan θ =sin θ/cos θ= sin θ

because the angles are small

we substitute

y = $\frac{m \lambda L}{d}$

let's calculate

y = 10 550 10⁻⁹ 1.5 / 1 10⁻³

y = 8.25 10⁻³ m

c) for this part the incident light must be white or polychrome light

let's find the distance (y) for some colors, in the first order

Violet λ = 400

y = 1 400 10⁻⁹ 1.5 / 1 10⁻³

y = 6 10⁻⁴ m

green λ = 500 nm

y = 1 500 10⁻⁹ 1.5 / 1 10⁻³

y = 7.5 10⁻⁴ m

yellow λ = 600 nm

y = 1 600 10⁻⁹ 1.5 / 1.10⁻³

y = 9 10⁻⁴ m

lam λ = 700 nm

y = 1 700 10⁻⁹ 1.5 / 1. 10⁻³

y = 1.05 10⁻³ m

violet, green, orange, red

the distance increases with the wavelength increases

d) To find the wavelength we must measure the distance y from the center of the slit for various interference orders and using the equation we can find the wavelength of rationality

4 0

2 years ago

When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these

Zigmanuir [339]

Answer:

a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

v = $\sqrt{ \frac{T}{ \mu } }$

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

m = 2m₀

let's find the new linear density

μ = m / l

iinitial density

μ₀ = m₀ / l

final density

μ = 2m₀ / lo

μ = 2 μ₀

we substitute in the equation for the velocity

initial v₀ = $\sqrt{ \frac{T_o}{ \mu_o} }$

with the new dough

v = $\sqrt{ \frac{T_o}{ 2 \mu_o} }$

v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }

v = 1 /√2 v₀

v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

μ = μ₀

in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

μ = 3m₀ / l₀

μ = 3 m₀/l₀

μ = 3 μ₀

v = $\sqrt{ \frac{T_o}{ 3 \mu_o} }$

v = 1 /√3 v₀

v = 0.577 v₀

d) l = 2l₀

μ = m₀ / 2l₀

μ = μ₀/ 2

v = $\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }$

v = √2 v₀

v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

μ = 10 m₀/2l₀

μ = 5 μ₀

we look for speed

v = $\sqrt{ \frac{T_o}{5 \mu_o} }$

v = 1 /√5 v₀

v = 0.447 v₀

5 0

2 years ago

Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different

kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

P = P_a* ( 1 - βh/T_a)^(g / R*β)

(g / R*β) = 32.174 / 1716*0.0035 = 5.252

P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

P = 1240 lb/ft^2

- The air density method which is expressed as:

P = P_a - γ*h

P = 2116.2 - 0.07647*14,110

P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

P = P_a* e^ ( -g*h / R*T_a )

P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

P = 1270 lb/ft^2

6 0

2 years ago

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