Missing figure: find it in attachment.
Answer:
Force D
Explanation:
In order to answer the question, let's keep in mind that the force of gravity on an object on Earth is the attractive force exerted by the Earth on the object; its direction is always downward (towards the Earth's centre), and its magnitude is given by
F = mg
where m is the mass of the object and g is the acceleration of gravity.
It follows immediately that in the figure, the force of gravity is the only force acting downward: therefore, force D.
The other forces are called:
Force A: thrust (it is the forward force generated by the engines)
Force B: lift (it is the upward produced by the aerodynamics of the wings)
Force C: air resistance (it is the backward force due to the friction between the air and the surface of the plane)
Answer:
2.8512*10^8 inches
Explanation:
Seeing as the ratio for miles to inches is 1 mile = 63360 inches, set up an equation. 1 mile / 63360 inches = 4500 miles / x inches. Cross multiply and simplify to 2.8512*10^8 inches. Hope this helps! :)
Hello! Gravity is lost when you reach the outside of the Earth's orbit! So you would have to overcome gravity in order to officially be in space. Gravity is the first thing a rocket must overcome to reach space.
I hope this helped!
I am, yours most sincerely,
SuperHelperThingy
Answer:
Explanation:
Based on the wave model of light, physicists predicted that increasing light amplitude would increase the kinetic energy of emitted photoelectrons, while increasing the frequency would increase measured current.
Contrary to the predictions, experiments showed that increasing the light frequency increased the kinetic energy of the photoelectrons, and increasing the light amplitude increased the current.
Based on these findings, Einstein proposed that light behaved like a stream of particles called photons with an energy of \text{E}=h\nuE=hνstart text, E, end text, equals, h, \nu.
The work function, \PhiΦ\Phi, is the minimum amount of energy required to induce photoemission of electrons from a metal surface, and the value of \PhiΦ\Phi depends on the metal.
The energy of the incident photon must be equal to the sum of the metal's work function and the photoelectron kinetic energy:
This is more along the lines of "Does gravity affext potential energy"
Sort of. Potential energy is an odd one to imagine, sometimes.
It's the energy possessed by an object or system by dint of it's spatial
and mechanical configuration.
That definition alone is perhaps not so useful...and it's certainly not
official. But what it means is that an object can potentially have
energy due to where it is or what state the system is in.
Imagine we have a box and it's on the floor. The box, for all intents
and purposes, has no potential energy. It isn't going anywhere and it
just sits on the floor. It can't do any work in it's current position.
Now we hoist the box into the air. For any distance the box travels from
the floor, it gains potential energy. Now let's back track. We've
changed the box's spatial configuration by hoisting it into the air and
so have given it potential energy.
Why does it now have potential energy? Because we can now drop the box
(costing us no energy) and the box will fall. Maybe it falls onto a
passer-by and injures them.
Box on the floor = No energy.
We lift the box = We spend our energy and give the box potential energy
(as it wants to fall toward the ground).
We drop the box = Potential energy is converted to kinetic energy as the
box falls.
Box injures someone = The kinetic energy has done work upon the person.
So we can see how it all flows and connects. We have to put energy into
the box to fight against gravity, but you can't destroy or create
energy....so the energy we've spent is potentially stored 'inside' the
box.
Clearly, gravity effects a LOT of potential energies around us. In fact
to some small extent, it's probably impossible to entirely avoid it's
effects.
