Answer:
Gravity and
Air resistance
Explanation:
The two forces acting on a skydiver are gravitational force and air resistance.
Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.
Air resistance is the drag of air on a body as it passes to it. It is resisting force.
- When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
- It soon balances both force and attain terminal velocity.
- Air resistance is a frictional force that opposes motion.
- This frictional force pushes in the opposite direction of motion
- Motion direction is downward due to the celerity caused by gravity.
Answer: The bug will remain motionless
Explanation:
According to Newton's first Law of Motion (sometimes called Law of Inertia):
<em>An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion.
</em>
In other words:
An object or body will keep its state of motion until an external force changes its state
This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.
In addition, according to his law, an object in rest can be in equilibrium (net force equals to zero), and a moving object can also be in equilibrium, as long as it keeps a constant velocity.
<h2>
This is why the bug, which is at rest will remain at rest, although the ants are simultaneously pulling it in different directions, since the resultant of all these forces is zero.</h2>
Answer: 3
Explanation:
Given
One cloud is traveling at rate of 
combined velocity of the two is 
Suppose the masses of the clouds be 
Conserving momentum

The answer is: " 208 g " .
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Explanation:
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The formula/ equation for density is:
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D = m / V ; That is, "mass divided by volume" ;
Density is expressed as:
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"mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
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{Note the exact equivalent: 1 cm³ = 1 mL }.
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→ The formula is: " D = m / V " ;
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in which:
"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);
"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
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We want to find the mass, "m" ; so we take the original equation/formula for the density:
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D = m / V ;
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And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
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Multiply each side of the equation by "V" ;
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V * { D = m / V } ; to get:
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V * D = m ; ↔ m = V * D ;
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Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
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m = V * D ;
m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;
→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
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