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1 year ago

How does position depend on time on a free falling motion, for short distance, near the surface of the earth?

Physics

1 answer:

Korvikt [17]1 year ago

7 0

Answer:

if an object is in free fall for a longer time its position will increase at an exponential rate because the object is accelerating

You might be interested in

Can somebody please help?

DedPeter [7]

Answer:

Im not sure

Explanation:

I don't take physics cuz im in 9th grade. so. idk but I will find out and come back with an answer.

7 0

3 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0

3 years ago

This equation is used to calculate the properties of a gas under nonideal conditions.

goblinko [34]

Answer:

Van der Waal's equation

Explanation:

The Van der Waal's equation is use to calculate the properties of a gas under nonideal or real gases conditions.

$(P + \frac{an^2}{V^2})(V-nb) = nRT$ .

Here P, V ,T ,n and R have usual meaning as in the ideal gas equation

that is PV=nRT

with the difference of constant a and b. a and b are constants representing magnitude of intermolecular attraction and excluded volume respectively respectively.

5 0

4 years ago

A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 8.0 times 10^1 N is being applied to the

Yanka [14]

Answer:

48 Newtons

Explanation:

There is no acceleration and net forces.
$8*10x^{-1} N cos(53) = friction$
Answer: 48.15 Newtons

7 0

3 years ago

The air pressure on the bottom of an airplane’s two wings is greater than the pressure on the top, providing an upward lift. Cal

dangina [55]

P=F/A

A=F/P

P = 0.1 atm = 10132.5 Pa

A=4 x 10⁵/10132.5

A=39.477 m²

for one wing=39.477 : 2 = 19.739 m²

5 0

2 years ago