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natta225 [31]
3 years ago
15

Which statement is true about the atom?

Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

Its d

atome contain

negative electrons,

positive protons and uncharged neutrons.

Explanation:

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For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
3 years ago
Read 2 more answers
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
Gennadij [26K]

Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

3 0
3 years ago
Read 2 more answers
A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit
JulsSmile [24]
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =\frac{y*d}{D} =  \frac{3.55*10^{-3}*1.25*10^{-4}  }{2.25} = 1.97*10^{-7} m
λ ~ 1.97 ×10⁻⁷m
7 0
3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

F_n = 306 N

Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15 \times 9.81}

v = 12.1 m/s

So this speed is independent of the mass of the rider

5 0
3 years ago
Why is it important foe scientists to know how many valence electrons a given element has
Vikentia [17]
Because the number of valence electrons of an element determines the properties and in particular the reactivity of that element.

In fact, elements of the first group (i.e. only one valence electron) have high reactivity, because they can easily give away their valence electron to atoms of other elements forming bonds. On the contrary, elements of the 8th group (noble gases) have their outermost shell completely filled with electrons, so they do not have valence electrons, and they have little or no reactivity at all.
3 0
3 years ago
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