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Sergio039 [100]
3 years ago
15

7) T F If two forces of equal magnitude act on an object that is hinged at a pivot, the force acting farther from the pivot must

produce the greater torque about the pivot.
Physics
1 answer:
saul85 [17]3 years ago
4 0

Answer:

False

Explanation:

The torque exerted by a force is given by:

\tau=Fd sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the pivot

\theta is the angle between the directions of F and d

We see that the magnitude of the torque depends on 3 factors. In this problem, we have 2 forces of equal magnitude (so, equal F). Moreover, one of the forces (let's call it force 1) acts farther from the pivot than force 2, so we have

d_1 > d_2

However, this does not mean that force 1 produces a greater torque. In fact, it also depends on the angle at which the force is applied. For instance, if the first force is applied parallel to d, then we have

\theta_1 =0\\sin \theta=0

and the torque produced by this force would be zero.

So, the statement is false.

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Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
4 years ago
A sound wave traveling downward with a speed of about 4,000 m/s suddenly slows to 1,500 m/s not far below the Earth’s surface. W
mezya [45]
<h2>Answer: an underground lake</h2>

Explanation:

In general, sound (mechanical waves) travels faster in solids than in liquids, and faster in liquids than in gases. This is because <u>the speed of the mechanical waves is determined by a relationship between the elastic properties of the medium </u>in which they are propagated and the mass per unit volume of the medium (that is:<u>density</u>).

In other words: The speed of sound varies depending on the medium through which the sound waves travel.

So, if we are told the sound wave initially had a speed of 4,000 m/s and it suddenly decreases to 1,500 m/s, this means the sound waves passed from a solid medium to a liquid medium.

Hence, the correct option is: an underground lake.

8 0
4 years ago
Read 2 more answers
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

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3 years ago
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Marat540 [252]
B. fungi

I hope this helps!
5 0
4 years ago
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