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Nutka1998 [239]
3 years ago
10

How are some types of collisions different from others?

Physics
1 answer:
lyudmila [28]3 years ago
3 0
There are two general types of collisions, inelastic and elastic. 
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering) 
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2. Tony sets up an experiment inside his classroom. He spaces 5 plants at equal
brilliants [131]

Answer:

a

Explanation:

4 0
4 years ago
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A certain superconducting magnet in the form of a solenoid of length 0.56 m can generate a magnetic field of 6.5 T in its core w
Gnom [1K]

Answer:

The value  is N =36203 \  turns

Explanation:

From the question we are told that

   The length of the solenoid is  l = 0.56 \  m

   The magnetic field is B  =  6.5 \ T

    The current is I = 80 \ A

     The desired temperature is  T = 4.2 \ K

Generally the magnetic field is mathematically represented as

       B  = \frac{\mu_o * N * I }{L }

=>     N = \frac{B  * L }{\mu_o * I }

Here  \mu_o is the permeability of free space with value  

      \mu_o =  4\pi * 10^{-7} N/A^2

So

     N = \frac{6.5  * 0.56 }{ 4\pi * 10^{-7} *  80  }

=>   N =36203 \  turns

6 0
3 years ago
How many protons are in this atom if it has a balanced charge
Lerok [7]
It would be 20 protons left because 1-19= 20
7 0
3 years ago
What is an electric circuit?
agasfer [191]

Answer:

the answer to the question is a system of insulating element designed to control the path of electric current for a particular purpose

5 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
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