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3 years ago

What aspect of motion can you conclude is common among freely falling objects ?

Physics

1 answer:

Liula [17]3 years ago

6 0

Answer:

Free-fall is defined as the movement where the only force acting on an object is the gravitational force.

By the second Newton's law, we have that:

F = m*a

Where F = Force, m = mass, a = acceleration.

We can write this as:

a = F/m

And the gravitational force can be written as:

F = (G*M/r^2)*m

Where G is the gravitational constant, M is the mass of the Earth in this case, and r is the distance between both objects (the center of the Earth and the free-falling object)

As the radius of the Earth is really big, the term inside the parentheses is almost constant in the region of interest, then we can write:

G*M/r^2 ≈ g

And the gravitational force is:

F = g*m

And by the second Newton's law we had:

a = F/m = (g*m)/m = g

a = g

Then the acceleration does not depend on the mass of the object.

Then the thing that is common among the free-falling objects is the vertical acceleration.

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Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines."

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

How does geosphere interact with biosphere

Marina86 [1]

With a partner, describe interactions in this scene, tracing the movement of materials or energy through all four of Earth's spheres if possible. Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.

6 0

3 years ago

Need help with this!!! 15 points!!!!

irinina [24]

Answer:

// EDU HACK REDDIT EDITON //

if (running == null) {

var running = null; console.log("Program Loading...");

var iframe = document.getElementById("stageFrame");

for (i = 0; i < 1; i++) {

setInterval(function() {

if (running == null) { running = true; console.log('Program Loaded! Enjoy!') }

if (iframe.contentWindow.document.getElementsByClassName('FrameRight') != null) { iframe.contentWindow.document.getElementsByClassName('FrameRight')[0].onclick(); }

if (iframe.contentWindow.document.getElementById("invis-o-div") != null) { iframe.contentWindow.document.getElementById("invis-o-div").remove(); }

},1000);

}

} else {

console.log("Program already running!")

}

Explanation:

5 0

3 years ago

7. A 3.0 kg object travels vertically at a constant

vladimir2022 [97]

Answer:

220.7

Explanation:

distance traveled is 1.5*5=7.5m

PE gain is MGH=3*9.81*7.5=220.7

7 0

3 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules

6 0

4 years ago