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2 years ago

In the context of studying major bodies of our solar system, what category of object does our moon best fit?.

Physics

1 answer:

Mumz [18]2 years ago

4 0

The simplest, obvious category is " <em>Satellites</em> " .

But there's a lot of good reason to call it "Component of Binary Planet" .

As a fraction of the size and mass of its planet, our moon is the largest in the solar system.

Several writers have pointed out that an alien astronomer studying our solar system might describe the 3rd body from the Sun as a "Binary planet".

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What is the moon phase today

Elis [28]

Waxxing Gibbous Phase

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2 years ago

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What is found in both plant and animal cells but is much larger in plant cells

djverab [1.8K]

The vacuoles because the plant's central vacuole is bigger.

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3 years ago

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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

2 years ago

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Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

romanna [79]

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

7 0

3 years ago

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$