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3 years ago

Which of these statements suggests that the orbital period of Venus is shorter than that of Saturn?

Physics

1 answer:

Snowcat [4.5K]3 years ago

5 0

Answer:

C) Venus is closer to the Sun than Saturn is.

Explanation:

This is the right answer.

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A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light λ is 580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D = 1.8 m.

Width of one fringe = $\frac{\lambda\times D}{d}$

Putting the values we get fringe width

= $\frac{580\times10^{-9}\times1.8}{.000315}$

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be

= 4.86 mm.

3 0

3 years ago

Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given

sergiy2304 [10]

Answer:

The angular velocity is

5.64rad/s

Explanation:

This problem bothers on curvilinear motion

The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s

We know that the velocity v is given as

v= ωr

Where ω is the angular velocity

r is 300mm to meter = 0.3m

the radius of the circle

described by the level

v=1.64m/s

Making ω subject of the formula and solving we have

ω=v/r

ω=1.64/0.3

ω=5.46 rad/s

3 0

3 years ago

What is the function of eye lens of human eye

Arte-miy333 [17]

Answer:

the lens is located in the eye. by changing its shape, the lens changes the focal distance of the eye. in other words,it focuses the light rays that pass through it (and onto the retina) in order to create clear images of objects that are positioned

at various distances.

8 0

3 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$