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Anestetic [448]
3 years ago
15

Application - You have two objects that you want to push the same distance (5 meters) across the floor. The objects have differe

nt masses. The first object is a dresser (40 kg) and the second object is a kitchen chair (11.5 kg). Which object will take more force to push the the 5 meters across the floor? Why?
Physics
1 answer:
Mamont248 [21]3 years ago
8 0

I see you're in Middle School, so I've got a hunch that they want you
to say "the dresser because it has more mass".  But that's a poor
answer, to a poor question.

The fact is that there's no way to tell.

The force it takes to move either object across the floor does NOT really
depend on just its mass.  It depends on both the object's mass AND the
friction between the object and the floor.  And THAT depends on the shape
of the feet where they touch the floor, and what kind of material the feet and
the floor are made of.

So it seems to me that we really don't have enough information to answer
the question with.

But again, I suspect that the answer they want is "the dresser because
it has more mass".

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A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t
Maurinko [17]

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height <em>h</em> and has a speed <em>v, </em>Its energy is given by,

E=mgh+\frac{1}{2} mv^2

Substitute 2.0 kg for <em>m</em>, the mass of the pendulum, 9.81 m/s² for <em>g</em>, the acceleration due to gravity, 0.10 m for <em>h and 4.0 m/s for </em>v<em>.</em>

E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J

The pendulum has an initial energy of 20 J. the energy lost is given by,

\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J

Thus, the energy lost by the pendulum is (A) 2.0 J

4 0
3 years ago
An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel
Delicious77 [7]

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels.  The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

7 0
3 years ago
A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta
My name is Ann [436]

Answer:

a).K=528.92 \frac{N}{m}

b).m=4.84kg

Explanation:

a).

The work of the spring is find by the formula:

w_s =\frac{1}{2}*k*x^2

So knowing the work can find the constant K'

3.2J =\frac{1}{2}*k*(0.11m)^2

Solve for K'

K=\frac{2*W_s}{x^2}=\frac{2*3.2J}{0.11m^2}

K=528.92 \frac{N}{m}

b).

The force of the spring realice a motion so using the force and knowing the accelerations can find the mass

F=m*a

m=\frac{F}{a}=\frac{K*x}{a}

m=\frac{528.9*0.11m}{12m/s^2}

m=4.84kg

8 0
3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
In the year 2081 in a shipping port on the moon, workers for Ore-Space, Inc., hoist a 500.0 kg hunk of anorthosite moon rock by
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Answer:

2,800 n

Explanation:

hope this helps, have a nice day/night! :D

7 0
3 years ago
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