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nikitadnepr [17]

3 years ago

12

What occurs during the Fall and Spring Equinoxes? *

1 answer:

Lostsunrise [7]3 years ago

5 0

At the time of the Fall equinox (always near September 21),
and the Spring equinox (always near March 21), the days and
nights are nearly the same duration.

The shortest days of the year are around the Winter solstice ...
always near December 21.

The longest days of the year are around the Summer solstice ...
always near June 21.

Eclipses can happen on any day or night of the year. They're
totally not connected with the equinoxes or solstices.

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Is this a plant or animal cell????

Lubov Fominskaja [6]

This is an Animal Cell.

4 0

3 years ago

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A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago

Analyze the effects on a passenger riding in a car traveling at

Mamont248 [21]

Without:

The passenger will get seriously injured
Colliding with the windshield at 50km/h
Could result in death

With safety:

The seatbelt holds tge passengers body back so they dont collide with anything
Their head is cushioned by the airbag
The passenger might get minor injuries but nothing serious will happen

5 0

2 years ago

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa

Strike441 [17]

Answer:

$1.24\times 10^{36}$

Explanation:

$q$ = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

$m$ = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

$F_{e} = \frac{kq^{2}}{r^{2}}$

$F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}$

$F_{e} = 230.4$ N

Gravitational force between the two protons is given as

$F_{g} = \frac{Gm^{2}}{r^{2}}$

$F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}$

$F_{g} = 1.86\times 10^{-34}$ N

Ratio is given as

$Ratio =\frac{F_{e}}{F_{g}}$

$Ratio =\frac{230.4}{1.86\times 10^{-34}}$

$Ratio = 1.24\times 10^{36}$

3 0

3 years ago

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What happens to the potential energy of a stationary charge when it begins to move freely from one point to another under the in

Ulleksa [173]

It decreases and the decrease in the energy level appear as a kinetic energy of the charge

8 0

3 years ago

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