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natta225 [31]
3 years ago
15

In the picture below, the candle is heating the water in the tank. Which picture shows how the water will move as it gets hot?

Physics
2 answers:
Masteriza [31]3 years ago
5 0

The first picture, at the top, does.

adoni [48]3 years ago
5 0
First picture at the top it does
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g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo
mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D)  Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 ,  exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0
3 years ago
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially
mixer [17]

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0
3 years ago
A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?
Irina-Kira [14]
F = 400 Pa x 2.5 m2
F = 1 kN
4 0
3 years ago
An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

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W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
Temka [501]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

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If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

6 0
3 years ago
Read 2 more answers
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