What are two most important processes in the cycling of oxygen in and out of the atmosphere?

; (b) A uniform beam 150cm long weighs 3.5kg and

noname [10]

Explanation:

(b) A uniform beam 150cm long weighs 3.5kg and

supported on knife-edges at its ends. The beam

supports a weight 7kg at a distance 30cm from

one end. Find the reactions of the supports.

4 0

3 years ago

The pulley is a uniform disk with a mass of 0.75 kg and a radius of 6.5

larisa [96]

Answer:

please find the answer in the attached images , in 2 parts, btw thanks for so many points. i hope my answers are correct.

5 0

3 years ago

Read 2 more answers

In Lesson 4.1, you learned that the machine used to fuel missions to Titan is not making enough liquid oxygen. We are waiting fo

LenaWriter [7]

Answer: The ideas that I have is how useful the Universal Space Agency or USA, is how great the fuel missions are

Explanation:

8 0

2 years ago

A compound wall consists of parallel layers of two different materials, 10 cm of brick and 2 cm of wood. If the difference of te

RSB [31]

The temperature difference between the wooden wall is 16⁰C, and the heat current per square metre of the wall is 7,325 W/m².

<h3>Temperature difference between the wooden wall</h3>

The temperature difference between the wooden wall is calculated as follows;

Let the brick wall = wall A
Let the wooden wall = wall B
Let the area of the walls = A

$\frac{dQ_A}{dt } = \frac{dQ_B}{dt} \\\\\frac{K_A \Delta T_A \times A}{L_A} = \frac{K_B \Delta T _B \times A}{L_B}\\\\\frac{K_A \Delta T_A }{L_A} = \frac{K_B \Delta T _B}{L_B}\\\\\frac{0.5 \times 20}{0.1} = \frac{0.125 \times \Delta T _B }{0.02} \\\\100 = 6.25\Delta T _B\\\\\Delta T _B = \frac{100}{6.25} \\\\\Delta T _B = 16 \ ^oC$

<h3 /><h3>Heat flowing in the walls</h3>

Q = KL(ΔT)

Q = 0.5 x 0.1 x (20 + 273)

Q = 14.65 W

<h3>Heat current per square meter of the Walls</h3>

QA = W/A

QA = (14.65)/A

Let the area of the wall = 10 cm x 2 cm = 0.1 m x 0.02 m = 0.002 m².

QA = 14.65/0.002

QA = 7,325 W/m²

Learn more about thermal conductivity here: brainly.com/question/11213835

4 0

3 years ago

Suppose a conducting rod is 52 cm long and slides on a pair of rails at 2.75 m/s. What is the strength of the magnetic field in

Eva8 [605]

Answer:

5.6 Tesla

Explanation:

L = 52 cm = 0.52 m

V = 2.75 m/s

e = 8 V

Let B be tha magnitude of magnetic field. Use the formula for the motional emf

e = B × V × L

B = e / V L

B = 8 / (2.75 × 0.52)

B = 5.6 Tesla

6 0

4 years ago