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3 years ago

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Which one of the following does not accurately describe the universal gravitational law? Question 10 options: A) Gravitational f

orce is proportional to the square of the distance between the centers of both bodies. B) Every object in the universe that has mass attracts every other object in the universe that has mass. C) Gravitational force is proportional to the product of the two masses. D) The object with the smaller mass does most of the moving because the object with the larger mass has too much inertia to move any noticeable amount.

2 answers:

just olya [345]3 years ago

5 0

Answer:

Choice C seems to be the right answer.

Explanation:

Kaylis [27]3 years ago

4 0

Answer:

D is right answer

Explanation:

$f = \frac{Gm1m2}{ {r}^{2} }$

•Gravitational force directly proportional to product of their masses

•And is inversely proportional to square of distance between them

•Every object in the universe that has mass attracts every other object in the universe that has mass.

THEREFORE OPTION 4 WILL BE THE ANSWER

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

3 years ago

Help with #3 please

Marizza181 [45]

Answer:

The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²

Explanation:

Data

Ta = 0.125 s

Tb = 0.08 s

Δtab = 0.5 s

distance = 1 cm

Process

1.- Calculate va

va = 1/0.125 = 8 cm/s

vb = 1/0.08 = 12.5 cm/s

2.- Calculate Δv

Δv = 12.5 - 8

Δv = 4.5 cm/s

3.- Calculate acceleration

a = Δv / Δt

a = 4.5/0.5

a = 9 cm/s²

4 0

3 years ago

When two notes are played simultaneously, creating a discordant sound, it is called _____.

mart [117]

Answer:

The correct answer is option D.

Explanation:

Acoustic : A branch of physics which study the properties of sound.

Consonance: Combination of notes occurring simultaneously due to relationship between their respective frequencies.

Timbre: A characteristic of a musical note which makes it distinct from another wave which also have same pitch and intensity.

Dissonance :When combination of two notes are played simultaneously with lack of harmony in between them.

Hence, the correct answer is option D.

5 0

3 years ago

A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui

GalinKa [24]

Answer:

$n_l = 1.97$

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

$\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

focal length of lens in liquid is

$\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}$

rearrange fro $n_l$

$n_l = \frac{n_g f_l}{f_l+f(n_g-1)}$

$n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}$

$n_l = 1.97$

7 0

3 years ago