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worty [1.4K]
4 years ago
6

Who studies this branch of physics? What types of careers are there in this branch of physics? What particular subjects should s

omeone study if they want a job in this branch of physics?
Physics
1 answer:
kolezko [41]4 years ago
8 0

The answers to all three questions depend on WHICH branch

of Physics you're talking about . . . a piece of information you've

neglected to specify.

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The D.A.R.E. program (“Drug Abuse Resistance Education”) began in 1983. In this program, police officers taught special 45-minut
melisa1 [442]

Answer:

the correct answer is practicing refusal strategies can help students stay sober

Explanation:

7 0
3 years ago
Do quasars reside within or without side of galaxies?
sveticcg [70]

They almost entirely reside within galaxies because quasars are a subset of blackholes with a large and fast enough accretion disk to generate a beam of interstellar material perpendicular to itself. This typically only occurs in the largest black holes at the center of galaxies (supermassive blackholes) or at least stellar black holes---which still occur within galaxies because the material is necessary to form them.

6 0
3 years ago
A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel
Svetradugi [14.3K]

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force  = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

8 0
3 years ago
Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.
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