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garik1379 [7]
3 years ago
15

A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to

the rod. The rod might be in translational equilibrium, rotational equilibrium, both, or neither.
(1) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net force on the rod is also zero.
(C) might not be in rotational equilibrium even if the net force on the rod is also zero.
(D) might be in rotational equilibrium even if the net force is not zero.
(2) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net torque about every axis through anyone point is found to be zero.
(C) might be in rotational equilibrium if the net torque about every axis through anyone point is found to be zero.
(D) might be in rotational equilibrium even if the net
torque about any axis through anyone point is not zero.
Physics
1 answer:
enot [183]3 years ago
6 0

Answer:

1. C

2.C

Explanation:

1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.

2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.

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Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a
nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

    p₀o = m₁ v₁ + m₂ v₂

    pf = 0

    m₁ v₁ + m₂ v₂ = 0

    m₁ / m₂ = -v₂ / v₁

    m₁ / m₂=  - (-6.2) / 4.7

     m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

    Xcm = 1/M    (m₁ x₁ + m₂ x₂)

We derive from time

   Vcm = 1/M   (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

    0 = 1/M   (m₁ v₁ + m₂v₂)

   m₁ v₁ + m₂v₂ = 0

    m₁ / m₂ = -v₂ / v₁

   m₁ / m₂ = 1.3

4 0
3 years ago
Carl went on a Euro Trip and was traveling from Spain to Russia. When he looked on the map it said Russia was 7035 km away from
bija089 [108]

Answer:

The average speed would be 279.167 meters per second.

Explanation:

Speed ​​is the magnitude that expresses the variation in position of an object and as a function of time. In other words, speed is defined as the relationship established between the distance traveled and the time elapsed in the movement of a mobile.

Speed can be calculated using the following expression:

Speed=\frac{distance}{time}

In this case:

  • distance= 7,035 km
  • time= 7 hours

Since the required unit of speed is meter per second, then you convert from km to meters for the case of distance, and from hours to seconds for the case of time. So:

  • distance= 7,035 km= 7,035,000 m (being 1 km= 1,000 m)
  • time= 7 hours= 420 minutes= 25,200 seconds (being 1 hour=60 minutes and 1 minute=60 seconds)

Replacing in the definition of speed:

Speed=\frac{7,035,000 meters}{25,200 seconds}

Solving:

Speed=279.167\frac{meters}{ seconds}

Then, <u><em>the average speed would be 279.167 meters per second.</em></u>

3 0
3 years ago
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on
ira [324]

Answer:

In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Explanation:

In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.

Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

4 0
3 years ago
A plane is flying at an elevation of 32000 feet. It is within sight of the airport and the pilot finds that the angle of depress
prohojiy [21]

Answer:x=79,202.77 ft

Explanation:

Given

plane is at a height of 32000 ft

angle of depression is 22^{\circ}

let x the distance between the point directly below the plane and airport be x

from diagram

\tan (22)=\frac{height}{x}

\tan (22)=\frac{32000}{x}

x=79,202.77 ft

distance between plane and airport

\sin (22)=\frac{32000}{h}

h=\frac{32000}{\sin (22)}

h=85,422.94 ft

6 0
4 years ago
A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an
elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)=(2.4-x^2)\hat{i}N

Initial position of the block=x=0

Initial velocity of block=v_i=0

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy=K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0

Work energy theorem:

K_f-K_i=W

Where K_f=Final kinetic energy

K_i=Initial kinetic energy

W=Total work done

Substitute the values then we get

K_f-0=\int_{0}^{2}F(x)dx

Because work done=Force\times displacement

K_f=\int_{0}^{2}(2.4-x^2)dx

K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}

K_f=2.4(2)-\frac{8}{3}=2.13 J

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy =K=2.4x-\frac{x^3}{3}

When the kinetic energy is maximum then \frac{dK}{dx}=0

\frac{d(2.4x-\frac{x^3}{3})}{dx}=0

2.4-x^2=0

x^2=2.4

x=\pm\sqrt{2.4}

\frac{d^2K}{dx^2}=-2x

Substitute x=\sqrt{2.4}

\frac{d^2K}{dx^2}=-2\sqrt{2.4}

Substitute x=-\sqrt{2.4}

\frac{d^2K}{dx^2}=2\sqrt{2.4}>0

Hence, the kinetic energy is maximum at x=\sqrt{2.4}

Again by work energy theorem , the  maximum kinetic energy of the block between x=0 and x=2.0 m is given by

K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx

k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}

K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0
4 years ago
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