3 years ago

What is the function of physics education

1 answer:

8 0

Answer:

Physics is the natural science that studies matter its fundamental constituents, its motion and behavior through space and time, and the related entities of energy and force

Explanation:

You might be interested in

Nearsightedness can be corrected by using eyeglass lenses that are

Nat2105 [25]

Concaved lenses...........

3 0

3 years ago

Read 2 more answers

18. Which would be the most reliable source of information to use for a history report? (2 points)

Paladinen [302]

Answer:

encyclopedia most reliable I think

3 0

3 years ago

Read 2 more answers

Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.

ser-zykov [4K]

Answer:

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.

6 0

3 years ago

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

3 years ago

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in

dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}$

Put all the values,

$t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s$

The velocity of the ball is :

$v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s$

The kinetic energy of the ball is :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J$

So, the required kinetic energy is 0.076 J.

6 0

3 years ago