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natita [175]
2 years ago
7

TRUE OR FALSE: some rays of light carry radiation.

Physics
2 answers:
vaieri [72.5K]2 years ago
8 0

Answer:

true

Explanation:

Daniel [21]2 years ago
6 0

Answer:

True

Explanation:

You might be interested in
A car changes chemical energy from fuel into thermal energy and ________ energy.
Shkiper50 [21]

Answer:

Mechanical energy

Explanation:

A car changes chemical energy from fuel into thermal energy and mechanical energy.

Mechanical energy can be defined as the type of energy that is possessed by an object due to its motion or position. Mechanical energy is the sum of potential energy and kinetic energy, that is, the sum of energy in motion and stored energy. Examples of mechanical energy includes driving a car, riding a bicycle, listening to music etc.

Types of mechanical energy

1. Motion energy (kinetic energy)

2. Stored energy(potential energy)

Mechanical energy = Kinetic energy + Potential energy

4 0
3 years ago
NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25
poizon [28]
22. reduction
25. Le Chatelier's principle
8 0
3 years ago
What is the SI (metric) unit of FORCE?<br><br> A. meter<br> B. newton
Tresset [83]

What is the SI (metric) unit of FORCE?

  • B. newton

with symbol ( N )

All the best !

4 0
2 years ago
Read 2 more answers
The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26
Delvig [45]

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3

(b) The energy stored in the magnetic field within the solenoid

U_B = u_B V\\\\U_B = u_B AL

U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\

6 0
3 years ago
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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