Answer:

**Complete question:**

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

**Answer:**

The exit velocity is** 629.41 m/s**

**Explanation:**

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

k = 1.4

Work done is given as;

inlet velocity is negligible;

Therefore, the exit velocity is** 629.41 m/s**

The tension of a string is found to be 4050 N and the mass of 1 m of the string is 0.5 kg. What could be the velocity of a wave traveling in that string?

The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

A ball having a mass of 500 grams is dropped from a height of 9.00 meters what is its kinetic energy when it hits the ground

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.

The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

A ball having a mass of 500 grams is dropped from a height of 9.00 meters what is its kinetic energy when it hits the ground

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

Where,

= Mass of each object

= Initial Velocity of each object

= Final velocity

Replacing we have that,

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

Since there is no initial speed, then

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

Therefore the Kinetic friction coefficient is 0.7105

**Answer: 132.02 J**

**Explanation:**

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = **132.02 J**

To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by

Where,

Original Length of the bar

= Change in temperature

= Coefficient of thermal expansion

On the other hand our values are given as,

Replacing we have,

The width of the expansion of the cracks between the slabs is 0.5832cm

The **width** of the **expansion cracks** between the **slabs** to prevent buckling should be **0.5832cm**.

According to this question, the following **information** are given:

- Lo = Original
**length**of the bar - ∆T = Change in
**temperature** - α = Coefficient of thermal energy

The values are given as follows:

- Lo = 18m
- T1 = 25°C, T2 = 52°C
- α = 12 × 10-⁶/°C

∆L = Loα (T2 - T1)

∆L = 18 × 12 × 10-⁶ (27)

∆L = 3.24 × 10-⁴ × 18

∆L = 5.832 × 10-³m

Therefore, the **width** of the **expansion** of the cracks between the slabs is **0.5832cm**.

Learn more about **width** at: brainly.com/question/26168065

a water molecule,

A. the electronegative atom becomes strongly positive

B. the hydrogen atom becomes partially positive

O C. the oxygen atom becomes partially negative

If answer is right WILL GIVE BRAINLIEST

D. the hydrogen atom becomes partially negative

**Answer:**

I'm leaning twards **A**

**Explanation:**

I think B but I’m not sure

**Answer:**

** The add mass = 5.465 kg**

**Explanation:**

**Note: Since the spring is the same, the length and Tension are constant.**

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

**Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.**

*Substituting into equation 4*

*m₂ = (2.07)²(0.5)/(1.18)²*

*m₂ = 4.285(1.392)*

*m₂ = 5.965 kg.*

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

**Thus the add mass = 5.465 kg**

Answer:

The gel that is applied before ultrasonic imaging is a **conducting material.** It acts as a medium between transducer and skin. **The ultrasonic waves easily transmit from the probe to the tissues because of gel.** A tight bond is created between the probe and skin layer and the gel acts as a **coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves. A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. **Hence, the technician apply ultrasound gel to the patient before beginning the examination

The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.