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ANEK [815]
3 years ago
9

Here is the situation: A puck is resting on the floor of a large moving van. Assume that the floor of the van is frictionless. T

he van and the puck are initially at rest, then the van undergoes a positive uniform acceleration. Determine whether each statement below is:a. True b. False.
Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

Statement 1: Someone outside the van sees the puck remain stationary with respect to the ground.

TRUE. This is because the puck relative to the ground initially was stationary. When the van starts to move forward or accelerate, the puck will tend to maintain its stationary state with the ground by moving in the opposite direction. This movement is only relative to the van and not the ground because the puck and the ground are at rest as the puck has no net force acting on it.

Someone in the van sees the puck move toward the front of the van.

FALSE. The puck has no net force acting in it so by newton's first law it should have no relative movement to the ground. In order to achieve this, the puck by newton's third law of motion must move in the opposite direction (backwards) as the van to maintain its stationary state with the ground.

The acceleration of the van with respect to the ground is > the acceleration of the puck with respect to the ground.

If the van turns to the right, the puck will hit the left wall.

WHAT DOES THIS HAVE TO DO WITH THE STATEMENTS? I'll assume the truck continues in a straight line and this statement if irrelevant. It does say "If"

The magnitude of the acceleration of the van with respect to the ground is = the magnitude of the puck's acceleration with respect to the van.

The puck is stationary with respect to the van during this acceleration.

FALSE

Explanation:

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Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

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6 0
3 years ago
When driving at night, only use your high-beam headlights___
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ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav
Montano1993 [528]

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

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