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Inga [223]
3 years ago
7

As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the ot

herforce is conservative and does +50 J of work. Between A andB,
a. the kineticenergy of object increases, mechanical energydecreases.
b. the kineticenergy of object decreases, mechanical energydecreases.
c. the kineticenergy of object decreases, mechanical energyincreases.
d. the kineticenergy of object increases, mechanical energyincreases.
e. None of theabove.
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.

You might be interested in
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
3 years ago
What effect does a catalyst have on a system in equilibrium?
Jet001 [13]

ANSWER:

What effect does a catalyst have on a system in equilibrium?

The system is unaffected.

~batmans wife dun dun dun....

8 0
3 years ago
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

           P_1 V_1 ^r  =  P_2  V_2  ^r

Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

8 0
3 years ago
Science cannot be used to answer questions about _____.
aksik [14]
Science cannot be used to answer questions about Moral judgements
8 0
4 years ago
Read 2 more answers
What is the acceleration of a 31 kg object pushed with a force of 42N
Ray Of Light [21]
I belive it could be 6.5 but I could be wrong
3 0
3 years ago
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