I need help solving this,

Two parallel rods are each 0.69 m in length. They are attached at their centers to a spring that is initially neither stretched

diamong [38]

Answer:

Explanation:

Let the separation required be d .

Force between rod = 10⁻⁷ x 2 I₁ I₂ L / d

where I₁ and I₂ are current in them , d is distance of separation and L is length of wire .

Force between rod = 10⁻⁷ x 2 x 1200 x 1200 x .69 / d

= .1987 /d

Restoring Force by spring = k x where k is force constant and x is compression .

= 130 x .03

= 3.9 N

For balancing

Restoring Force by spring = Force between rod

.1987 /d = 3.9

d = .1987 /3.9

= .0509 m

= 5.09 cm .

5 0

3 years ago

Describe the relationship between force and mass when the acceleration of two objects remain constant

LekaFEV [45]

Answer:

If they both remain constant they ain't moving, It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation

Explanation:

6 0

3 years ago

P.E

Vladimir79 [104]

Answer:

pagtalon?

Explanation:

because that had to be the answer

5 0

3 years ago

Pls try to understand my doubt and clear it <br>.

jeka94

I can’t read the last few numbers, is that an 8 ??

7 0

3 years ago

A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$ , $v=2.85\frac{\rm m}{\rm s}$ , and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$ , so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$

Plugging in the known quantities, we end up with

$\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27$

The second case has no real solution, since $-1\le\cos(\theta)\le1$ for all $\theta$ . This leaves us with

$\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}$

7 0

2 years ago