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babymother [125]
3 years ago
7

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

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
Star X has an apparent magnitude of 1. Star Y has an apparent magnitude of 4. Both stars are in the same star cluster. Which sta
sashaice [31]

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})

3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})

Making \dfrac{L_1}{L_2} the subject of the formula:

\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

7 0
3 years ago
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