All categories

2 years ago

If the Earth got closer to the Sun, how would the gravity between the Sun and Earth change?

Physics

2 answers:

SSSSS [86.1K]2 years ago

8 0

It is 2 it would decrease

Neko [114]2 years ago

4 0

Answer:

I think its 2. It would decrease

Explanation:

Hope this helps

You might be interested in

Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a

irina1246 [14]

Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

I₁ = I₀ / 2

I₁ = 22.4 / 2

I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

I₂ = I₁ cos² θ

I₂ = 11.2 cos² 69

I₂ = 1.44 Lux

8 0

3 years ago

Evaporating water produces gaseous water. What has occurred during this process?

Katyanochek1 [597]

Answer:

Heat has accelerated water atoms enough to break the surface tension which leads the liquid to turn into a gas

Explanation:

The state of a substance depends on the distribution of its atoms, therefore any atmosphere change (in this case heat) enough to change the atoms Distribution results in a change of state.

brainliest please ;)

3 0

3 years ago

How long does it take a man to travel 6 km if his speed is 3km/h?

Ymorist [56]

why did my answer get deleted??

oh yeah i put a link on there- oopsies.

I wont this time!

I got 30!

3 0

2 years ago

Imagine imagining an imagination.

ozzi

I’m imagining imagining imagining an imagination...

4 0

3 years ago

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

Mass of the pendulum, = M
Length of the pendulum, = L
Initial angular speed, $\omega _i$ = 1 rad/s

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0

2 years ago