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3 years ago

A 6.0 kg object undergoes an acceleration of 2.0 m/s2 .

1 answer:

6 0

A. Fnet=ma
6*2=12N of force acting on the object in the direction it is accelerating

B. Fnet=ma
4*2=8N of force action on the object in the direction it is accelerating

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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

andre [41]

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

6 0

2 years ago

A 65 kilogram astronaut weighs 638 newtons at the surface of earth what is the mass of the astronaut at the surface of the moon,

Montano1993 [528]

The mass of the astronaut is still 65 kilograms. Mass is constant or doesn't change no matter where you are.

3 0

3 years ago

Read 2 more answers

What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

Arlecino [84]

The go-kart's velocity $v$ after time $t$ is given by

$v=v_0+at$

where $v_0$ is its initial velocity and $a$ is its acceleration. After $t=5\,\mathrm s$ , the go-kart stops completely, so

$0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}$

where $a$ because we know the go-kart is slowing down.

7 0

3 years ago

If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 52.0 m/s in the opposite directi

stich3 [128]

Explanation:

Below is an attachment containing the solution.

7 0

2 years ago

g A wheel of radius 1.2 meters initially rotates clockwise around its center with an angular speed of 10 rad/s, and it steadily

Lemur [1.5K]

Answer:

α = 5 rad / s²

Explanation:

This is a rotational kinematics exercise.

They indicate the initial velocity wo = 10 rad / s

w = w₀ + α t

α = $\frac{w-w_o }{t}$

let's calculate

α = $\frac{30-10}{4}$

α = 5 rad / s²

The velocity, the angular relation are the same in all the points of the wheel, the velocities and linear accelerations are the ones that change

a = α r

v = w r

5 0

2 years ago