A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside?

Answers

Answer 1

Answer:
   Potential energy = (mass) x (gravity) x (height)

   1,568 J = (40 kg) x (9.8 m/s²) x (height)

Divide each side
by 392 Newtons:    Height = (1,568 N-m) / 392 N

= 4 meters .

Answer 2

Answer:

Answer:

B) 4.0m

Explanation:

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Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE

Answers

The work done to launch the spacecraft in low earth orbit is $(GM_emh)/(R_e(R_e+h))$

Work-energy theorem:

The gravitational force is a conservative force. So, the total energy of the system must be conserved.

According to the work-energy theorem:

work done = - change in potential energy of the system.

W = -ΔPE

Initially, the potential energy of the satellite on the surface is:

PE = $-(GM_em)/(R_e)$

where m is the mass of the satellite

Let the orbit be at a height of h from the surface, so the potential energy in the orbit is :

PE' = $-(GM_em)/(R_e+h)$

ΔPE = PE'-PE

ΔPE = $-GM_em((1)/(R_e+h)-(1)/(R_e))$

$\Delta PE=-(GM_emh)/(R_e(R_e+h))$

Now work done:

W = - ΔPE

Thus,

$W=(GM_emh)/(R_e(R_e+h))$

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Answer:

Work done to shift the spacecraft from Earth surface to low Earth radius is given as

$W = (GM_e m)/(2R_e)$

Explanation:

As we know that spacecraft is at surface of Earth initially

So we will have

$U_i = -(GM_e m)/(R_e)$

now when it is at low radius Earth Orbit then we have

$U_f = -(GM_e m)/(2(R_e + h))$

now we know that work done to shift the spacecraft from Earth Surface to Low earth orbit is change in total energy

$W = -(GM_e m)/(2(R_e + h)) + (GM_e m)/(R_e)$

so we have

$W = (GM_e m)/(R_e) (-(1)/(2(1 + h/R_e)) + 1)$

$W = (GM_e m)/(R_e) (-(1)/(2)(1 - (h)/(R_e)) + 1)$

$W = (GM_e m)/(2R_e)(1 + (h)/(R_e))$

since we know h << Re

so work done is given as

$W = (GM_e m)/(2R_e)$

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.66 s, and the top-to-bottom height of the window is 1.80 m. How high above the window top does the flowerpot go? (Assume that in both sailing upward and then falling downward, the flowerpot is seen to span the full vertical height of the window.)

Answers

Answer:

Height covered after crossing window top = 0.75 m

Explanation:

In the question,

The height of the window pane = 1.8 m

Time for which the flowerpot is in view = 0.66 s

So,

The time for which it was in view while going up is = 0.33 s

Time for which it was in view while going down = 0.33 s

So,

Let us say,

The initial velocity of the flowerpot = u m/s

So,

Using the equation of the motion,

$s=ut+(1)/(2)at^(2)\n1.8=u(0.33)-(1)/(2)(9.8)(0.33)^(2)\n1.8=0.33u-0.5336\n0.33u=2.33\nu=7.07\,m/s$

So,

Velocity at the top of the window pane is given by,

$v=u+at\nv=7.07-(9.8)(0.33)\nv=3.836\,m/s$

Now,

Let us say the height to which the flowerpot goes after crossing the window pane is = h

So,

Using the equation of motion,

$v^(2)-u^(2)=2as\n(0)^(2)-(3.836)^(2)=2(-9.8)h\nh=(14.714)/(19.6)\nh=0.75\,m$

Therefore, the height covered by the flowerpot after window is = 0.75 m

How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?a. The bright lines are at the same energies as the dark lines.
b. The dark lines are at higher energies than the bright lines.
c. The bright lines are at higher energies than the dark lines.
d. You cannot relate the two types of spectra.

Answers

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon ! If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove. It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels. Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up. We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

Final answer:

The dark absorption lines of an atom's spectrum correspond to the same energies as the bright emission lines. They both reflect energy changes in electron states.

Explanation:

The dark lines of an atom's absorption spectrum are at the same energies as the bright lines of its emission spectrum, therefore the correct answer is a. The bright lines are at the same energies as the dark lines. Absorption spectra are produced when electrons absorb energy and move to a higher energy level, while emission spectra are observed when electrons lose energy and return to a lower energy level.

The dark lines (absorption) and bright lines (emission) coincide because the energy required to move an electron from a lower to higher energy level matches the energy released when an electron drops from a higher to lower state.

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In a closed system, glider A with a mass of 0.40 kg and a speed of 2.00 m/s collides with glider B at rest with a mass of 1.20 kg. The two interlock and move off. What speed are they moving at?

Answers

The speed are they moving at will be 0.5 m/sec.Law of conservation of momentum is applied.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of 1st gilder= 0.40 kg

(u₁) is the initial velocity = 2 m/s

(m₂) is the mass of 2nd gilder = 1.20 kg

(u₂) is the initial velocity of 2nd gilder = 0 m/s

(v) is the velocity after collision =.?

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

$\rm m_1u_1 + m_2u_2 = v(m_1 + m_2)\n\n(0.40* 0.25) + (1.2 * 0) = v * (0.40+1.20) \n\n 125 + 0 = v * (1250) * 125 \n\n \rm v= (0.8)/(1.6) \n\n \rm v= 0.5 \ m/sec$

Hence, the speed are they moving at will be 0.5 m/sec

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Conservation of momentum.
m1*u1 + m2*u2 =m1*v1 + m2*v2
here u2=0 body at rest, v1=v2=v both interlock and move together
0..4*2+1.2*0 = 0.4*v+1.2*v
v= 0.8/1.6 = 0.5 m/sec

Walking across a carpet is an example of charge being transferred bya. contact.
b. induction.
c. static electricity.
d. friction.

Answers

Answer : The correct answer is- D) Friction.

There are three modes of charging, which are Conduction, induction, and friction.

The static electricity is produced due to the rubbing of one substance with another. This is called charing by friction.

Hence, the correct answer is friction as there is friction between carpet and the feet of person walking over it.

Walking across a carpet is an example of charge being transferred by friction. The answer is letter D. The carpet, which is at rest, will resist the motion cause by your walking. This resistive motion d the friction the carpet obtained from you and you from the carpet.

If you mix 40.0 ml of a 0.200 m solution of k2cro4 with an aqueous solution of agno3, what mass of solid forms? (hint: most chromates are insoluble.)