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3 years ago

6

Some people say that the “force of inertia” (and “force of momentum”) throws the passengers forward when a car brakes sharply. w

hat is wrong with this explanation

1 answer:

marishachu [46]3 years ago

5 0

the car stops, the inertia of people keeps them going. no force ... they were going before the car stopped

sudden stop = pain

gentle stop = ok

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A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

so

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

8 0

3 years ago

A cd player uses a ___ to convert info to an electrical signal

Alex_Xolod [135]

C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates

7 0

3 years ago

A cat is being chased by a dog. Both are running in a straight line at constant speeds. The cat has a head start of 3.8 m. The

Mars2501 [29]

In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s

7 0

3 years ago

I need to know the right answer to that question

ad-work [718]

The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation

v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.

8 0

3 years ago

When 1,250^3/4 is written in simplest radical form, which value remains under the radical?

GaryK [48]

Answer:

$125\sqrt[4]{8}$

Explanation:

A number of the form

$a^{\frac{m}{n}}$

can be re-written in the radical form as follows:

$\sqrt[n]{a^m}$

In this problem, we have:

a = 1,250

m = 3

n = 4

So, if we apply the formula, we get

$1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}$

Then, we can rewrite 1250 as

$1250 = 2\cdot 5^4$

So we can rewrite the expression as

$=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}$

7 0

3 years ago

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