All categories

2 years ago

Where are most mountains formed?

1 answer:

4 0

Hey there,

Most mountains are formed mostly in Mount Vesuvius, Himalayas, and <span>Mount Kilimanjaro.

Hope this helps!</span>

You might be interested in

Need answer fast plz

vazorg [7]

Answer:

a = 1.20m\ $s^{2}$

Explanation:

225 x 9.8

= 441N

710 - 441

= 269N

$\frac{269}{225}$ = 1.19

a = 1.20

8 0

2 years ago

Energy from the Sun that bounces off the surface of the Earth and returns to space is called...

Reika [66]

D is the answer trust me have faith it’s for the glory of humanity

5 0

3 years ago

What is the value of work done on an object when a

tino4ka555 [31]

W = F * s
Here, F = 50 N
s = 15 m

Substitute their values,
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!

7 0

2 years ago

IN THIS FORMULA FOR WATER WHAT DOES THE SUBSCRIPT 2 INDICATE

masha68 [24]

The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.

7 0

3 years ago

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

nikitadnepr [17]

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;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago