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3 years ago

A 100 g wire is held under a tension of 250 n with one

Physics

1 answer:

sashaice [31]3 years ago

6 0

So??
what is the question??

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One glass microscope slide is placed on top of another with their left edges in con- tact and a human hair under the right edge

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3 years ago

The resultant of 2 forces at right angles is 100 lbs. If one of the forces makes an angle of 30 degress with the resultant, comp

Ipatiy [6.2K]

Answer:

86.6 lbs

Explanation:

Let the force is X.

Resultant force, R = 100 lbs

Other force is Y. Angle between resultant force and force X is 30°.

According to the diagram

$Cos30=\frac{X}{R}$

$0.866=\frac{X}{100}$

X = 86.6 lbs

Other force Y

$Sin30=\frac{X}{R}$

$0.5=\frac{Y}{100}$

Y = 50 lbs

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4 years ago

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3 years ago

How does a rubber rod become negativley charged trough friction/

natali 33 [55]

Answer: Answer below hope it helps

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3 years ago

Read 2 more answers

The cannon on a battleship can fire a shell a maximum distance of 26.0 km.

Alla [95]

It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is $\theta = 45^{\circ}$ .
In fact, the laws of motions on both x- and y- directions are
$S_x(t)= v_0 cos \theta t$
$S_y(t)= v_0 \sin \theta t - \frac{1}{2} gt^2$
From the second equation, we get the time t at which the projectile hits the ground, by requiring $S_y(t)=0$ , and we get:
$t= \frac{2 v_0 \sin \theta}{g}$
And inserting this value into Sx(t), we find
$S_x(t) = 2 \frac{v_0^2}{g} \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)$
And this value is maximum when $\theta=45^{\circ}$ , so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
$S_x(t)= \frac{v_0^2}{g} \sin (2\theta)$
And by using $S_x = 26 km=26000 m$ , we find the value of the initial velocity v0:
$v_0 = \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s$

8 0

3 years ago