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docker41 [41]
4 years ago
11

The net force on a car is zero in both the horizontal and vertical directions which two situations could be true about the motio

n of the car
Physics
2 answers:
Alina [70]4 years ago
8 0

Two possible situations:

- The car is at rest. In this case, there are only 2 forces acting on the car, both in the vertical direction: the weigth of the car (downward) and the normal reaction of the asphalt on the car (upward). The two forces are equal in magnitude but in opposite directions, so the net force is zero.

- The car is traveling at constant velocity. In this case, there are 4 forces: in the vertical direction, as before, we have the weigth of the car (downward) and the normal reaction of the asphalt on the car (upward). The two forces are equal in magnitude but in opposite directions, so the net force in the vertical direction is zero. However, this time the car is also moving, so there are forces acting along the horizontal direction: the driving force, which pushes the car forward, and the frictional force, which goes in the opposite direction. Since the car is moving at constant velocity, its acceleration is zero, so the net force must be zero: therefore, the magnitude of these two forces is equal.

galben [10]4 years ago
8 0
Since both the net force of the horizontal and vertical directions on the car is zero, we can imply that it is in equilibrium in two cases: either it is at rest, or it is moving on a constant velocity. I hope my answer has come to your help. God bless and have a nice day ahead!
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Noah stands 170 meters away from a steep canyon wall. He shouts and hears the echo of his voice one second later. What is the sp
xxMikexx [17]

answer: 340 m/s

explanation: in this instance, the sound wave travels 340 meters in 1 second, so the speed of the wave is 340 m/s. remember, when there is a reflection, the wave doubles its distance. in other words, the distance traveled by the sound wave in 1 second is equivalent to the 170 meters down to the canyon wall plus the 170 meters back from the canyon wall.

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4 years ago
4. Brian stands at the edge of a cliff and throws a stone horizontally over the edge with a
Verdich [7]

Answer:A).3.16 s B).56.92

A).12.5m/s

B).7.81m

Explanation:

5 0
4 years ago
Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi
jeka94

Answer:

     I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

       d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

       d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

        I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

         d = (d₁ + d₂) / 2

         d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

         r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

        M = M_man - 0.2 M_man = 80 (1-0.2)

        M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis:         Ic = ½ M r²

A rod that rotates at the end:                            I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

           I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

          D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

          I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

         I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

         I₁ = 8.258 + 5.33 + 4.129

         I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

          I₂ = ½ M r² + 2 m D²

          I₂ = ½ 64 0.508² + 2 8 0.508²

          I₂ = 8,258 + 4,129

          I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

          I₁ / I₂ = 17,717 /12,387

          I₁ / I₂ = 1.43

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Answer:

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