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allochka39001 [22]
3 years ago
12

A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational

Physics
1 answer:
velikii [3]3 years ago
3 0

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

           sin  35 = Wₓ / W

            cos 35 = W_y / W

            Wₓ = W sin 35

            W_y = W cos 35

            Wₓ = 2500 9.8 sin 35

             Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis  

       N-W_y = 0

       N = W_y

X axis  

       F -Wₓ = m a

       F = Wₓ + m a = mg sin 35 + m a

       F = m (a + g sin 35)

   

let's calculate

       F = 2500 (5 + 9.8 sin 35)

       F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

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Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

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What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored
Musya8 [376]

The total energy  stored in the capacitors is determined as  2.41 x 10⁻⁴ J.

<h3>What is the potential difference of the circuit?</h3>

The potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

where;

  • C is capacitance of the capacitor
  • V is the potential difference

For a parallel circuit the voltage in the circuit is always the same.

The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

2U = CV²

V = √2U/C

V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)

V = 12 V

The equivalent capacitance of C1 and C2 is calculated as follows;

1/C = 1/C₁ + 1/C₂

1/C = (1)/(0.9 x 10⁻⁶)  +  (1)/(16 x 10⁻⁶)

1/C = 1,173,611.11

C = 1/1,173,611.11

C = 8.52 x 10⁻⁷ C

The total capacitance of the circuit is calculated as follows;

Ct = 8.52 x 10⁻⁷ C   +   2.5 x 10⁻⁶ C

Ct = 3.35 x 10⁻⁶ C

The total energy of the circuit is calculated as follows;

U =  ¹/₂CtV²

U =  ¹/₂(3.35 x 10⁻⁶ )(12)²

U = 2.41 x 10⁻⁴ J

Learn more about energy stored in a capacitor here: brainly.com/question/14811408

#SPJ1

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