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3 years ago

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A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

ce of 1.30 m in 1.20 s. What was the initial speed of the bottle cap, just after leaving the rubber band, in m/s

1 answer:

zavuch27 [327]3 years ago

3 0

Answer:

Initial velocity will be 1.356 m/sec

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity $u_x=ucos\Theta =ucos37^{\circ}=0.7986u$

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time $t=\frac{distance}{speed}$

$1.20=\frac{1.3}{0.7986u}$

$u=1.356m/sec$

So initial velocity will be 1.356 m/sec

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A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

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Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

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A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

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Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

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Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

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Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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