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3 years ago

IF a rock has a mass of 92,224 kg, what is it's weight?

Physics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

904 717.44 N

Explanation:

The mass of an object is a fundamental property of the object.

The weight of an object is the force of gravity on the object.

This is defined as follows

Weight (W)= mass(m) × weight of a unit mass(g)

( gravitational field intensity = g)

assume g =9.81 m/s2

weight = 92 224 * 9.81

= 904 717.44 N

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Examine the scenario. Two neutral objects, a balloon and a sweater, are rubbed against each other. Which choice most accurately

Ne4ueva [31]

Electrons move from the sweater to the balloon. The sweater becomes positively charged, while the balloon becomes negatively charged.

Explanation:

Sweater is a conductive material, which means it readily gives away its electrons.

Consequently, when you rub a balloon on Sweater, this causes the electrons to move from the Sweater to the balloon's surface.
The rubbed part of the balloon now acquired a negative charge. Objects made of rubber, such as the balloon, are basically electrical insulators, meaning that they resist electric charges flowing through them.
This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral after electrostatic process.

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3 years ago

How is the Sun similar to the stars that can be seen at night?

prisoha [69]

Answer:

Explanation:

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3 years ago

Read 2 more answers

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du

natulia [17]

Given: Normal pull of gravity g = 9.8 m/s²;

g = 0.855 m/s² (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or

r = 21,610 Km

.

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3 years ago

Hugs 5 things your dog might do when hugged

Nezavi [6.7K]

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4. play: they will want to play with you once they see you love them and want to be around them
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3 years ago