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1 year ago

where are the turning points of the motion? express your answers in meters. enter your answers separated by a comma.

1 answer:

8 0

When K = 0, the motion's turning points take place. Think about frictionless one-dimensional motion in the x-direction. So whenever Etotal = U, K(x) = 0 and v = 0. (x).

<h3>How to describe friction</h3>

friction is the resistance to a thing moving or rolling over another solid object. Although frictional forces can be advantageous, such as the traction required to walk without slipping, they also provide a significant amount of resistance to motion.

<h3>What leads to friction?</h3>

Because of the imperfections on the two surfaces in contact, friction is produced. Small imperfections can be found on even the smoothest surfaces, and when two surfaces interact with one another, they cause friction.

To know more about friction visit:

brainly.com/question/13000653

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A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

3 years ago

Two formats of sources are

Scilla [17]

Answer:

APA and MLA are the two format sources.

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3 0

3 years ago

A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside

vitfil [10]

Answer:

The value is $B = 0.0452 \ T$

Explanation:

From the question we are told that

The number of turns is N = 1000

The length is L = 50 cm = 0.50 m

The radius is r = 2.0 cm = 0.02 m

The current is I = 18.0 A

Generally the magnetic field is mathematically represented as

$B = \mu_o * \frac{N }{L} * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0$

=> $B = 0.0452 \ T$

3 0

3 years ago

What does more damage; a slow semi truck, or a fast sports car

Goshia [24]

The fast sports car does more damage then the slow semi truck

7 0

3 years ago

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.

scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

Pressure at sea level = 1 atm = 101300 Pa

we know

P = ρ g h

$h = \dfrac{P}{\rho\ g}$

$h = \dfrac{101300}{1.3\times 9.8}$

h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

at x = 0 air density = 0

at x= h ρ_l = ρ_sl

assuming density is zero at x - distance

$\rho_x = \dfrac{\rho_{sl}}{h}\times x$

now, Pressure at depth x

$dP = \rho_x g dx$

$dP = \dfrac{\rho_{sl}}{h}\times x g dx$

integrating both side

$P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx$

$P =\dfrac{\rho_{sl}\times g h}{2}$

now,

$h=\dfrac{2P}{\rho_{sl}\times g}$

$h=\dfrac{2\times 101300}{1.3\times 9.8}$

h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0

3 years ago