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2 years ago

11

The Earth's gravity keeps us from flying out to space. How can you explain that we have a stronger gravitational attraction to t

he Earth than to the Sun?
Question options:

The sun is bigger, but has a less mass than the Earth.

The sun has a larger mass so it has less gravitational pull.

We are closer to the sun than the Earth.

We are closer to the Earth than the sun.

2 answers:

julia-pushkina [17]2 years ago

6 0

It's the fourth choice.

This is because, since we are closer to the Earth, the Earth will have a stronger gravitational pull on us since again, we are closer.
That also explains tides, but that's just getting off topic. Hope I helped.

SOVA2 [1]2 years ago

6 0

The answer is we are closer to the earth than the sun i took test

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Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant

evablogger [386]

(a) $2.56\cdot 10^{-5} C$

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

$F=(0.021)(1.1)=0.0231 N$

The electrostatic force between the two balloons can be also written as

$F=k\frac{Q^2}{r^2}$

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

$Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C$

(b) $1.6\cdot 10^{14}$ electrons

The magnitude of the charge of one electron is

$e=1.6\cdot 10^{-19}C$

While the magnitude of the charge on one balloon is

$Q=2.56\cdot 10^{-5} C$

This charge can be written as

$Q=Ne$

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

$N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}$

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2 years ago

explain " you can not apply a force to an object with out that object applying the same force back to you"

stealth61 [152]

Answer:

Newton's Third Law Of Motion

Explanation:

3 0

2 years ago

Read 2 more answers

How many different values of l are possible for an electron with principal quantum number n=5?

lions [1.4K]

Answer:

different value to l if n=5 are 0,1,2,3,4

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2 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

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2 years ago

PLEASE HELP, PLEASE A CORRECT ANSWER!

LUCKY_DIMON [66]

Answer: I like your profile picture

Explanation:

6 0

2 years ago