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2 years ago

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g 2. The _ spans the distance from the _ to the location of the applied force. moment arm; pivot point moment of inertia

; center of mass torque; center of mass None of the above

1 answer:

Alik [6]2 years ago

7 0

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

6 0

2 years ago

Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti

Ket [755]

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

4 0

2 years ago

Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

vichka [17]

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

7 0

2 years ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0

2 years ago