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3 years ago

Which type of force can be in the same it opposite direction?

Physics

1 answer:

LuckyWell [14K]3 years ago

4 0

Answer:

C) unbalanced

Explanation:

Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion. When you add equal forces in opposite direction, the net force is zero.

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Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

vodomira [7]

Answer:

A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs

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3 years ago

What are two unique characteristics of a social psychologists?

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3 years ago

In an experiment, a torque of a known magnitude is exerted along the edge of a rotating disk. The disk rotates about its center.

lisabon 2012 [21]

Answer:

B) The amount of time the torque is applied to the disk, because the time interval is related to the angular impulse of the disk.

Explanation:

Angular impulse = Torque x time

= change in angular momentum

So,

Torque x time = change in angular momentum

change in angular momentum = Torque x time

Torque is already known .

Hence to know the change in angular momentum what is needed to know is time duration of torque acting on the body .

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3 years ago

run at the same speed and in the same direction, and they both run for the same amount of time, what can you say about the dista

LenaWriter [7]

Answer:

the are equivalent

Explanation:

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3 years ago

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$