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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.

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What portion (division) of a meter stick is a centimeter?
Anastasy [175]
A meter is 100 meters. So a hundredth of a meter stick is a centimeter.<span />
8 0
4 years ago
Read 2 more answers
If a person walks 3 blocks East, turns and walks 5 blocks north. The person then turns and walks 3 blocks west and then 5 blocks
torisob [31]
The person still walked 16 blocks but ended up in the same spot he started from
3 0
3 years ago
Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field
ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)²  and  Ex = 2008 N / C

Explanation:

a)   The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

         E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1.  Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

          r² = x² + y²

          E1 = k q / (x² + y²)

Particle 2.   located at x = -24.9 m

          r² = x² + y²

          E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

     cos θ = CA / H = x / (x2 + y2) ½

     E1x = E1 cos θ

      E2x = = E1 cos θ

The resulting field

      Ey = 0

      Ex = E1x + E2x 2 E1x

      Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

      Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

      Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

      Ex = 497.15   x / (x² + 0.062)  ³/₂  

Point where can the value of the electric field x = 38.1 cm = 0.381 m

       Ex = 497.15 0.381 / (0.381² + 0.062)  ³/₂  

       Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

       Ex= 189.41 /0.0943

       Ex = 2008 N / C

c)  E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

       Ex = 497.15 x / (x² + 0.062)  ³/₂  

Let's use some arithmetic

       Ex / 497.15 = x / (x² + 0.062)  ³/₂  

       [Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

       ∛[Ex / 497.15]² = (∛x²) / (x² + 0.062)                 (1)

The roots of this equation are the solution to the problem,

     

For Ex = 1.00 kN / C = 1000 N / C

 

      [Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

       1.312 = (∛x² ) / (x² + 0.062)

       1.312 (x² + 0.062) = ∛x²

       1.312 X² - ∛x² + 1.312 0.062 = 0

       1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0
3 years ago
A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

7 0
3 years ago
A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to
Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass m_1 just before the collision

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(3)}

v_1 = 7.67m/s

Therefore the momentum just before collision would be

p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s

Momentum after the collision

p_1 = 20*u_1+40u_1\\p_1 = 60u_1

Since the momentum is conserved we have that

153.36= 60u_1

u_1 = \frac{153.36}{60}

u_1 = 2.56m/s

The velocity of mass m_2 after the collision is given by

v_2 = \frac{2m_1}{m_1+m_2} u_1

v_2 = \frac{2(20)}{20+40}(2.56)

v_2 = 1.71m/s

Therefore the change in momentum of mass 2 is

p_2 = m_2v_2

p_2 = 40*1.71

p_2 = 68.4kg\cdot m/s

Therefore the impulse acting on m2 during the collision between the two boxes is p = 68.4kg\cdot m/s

8 0
4 years ago
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