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Marysya12 [62]
2 years ago
5

On a distance vs time graph, what does the line of an object at constantspeed look like?​

Physics
1 answer:
9966 [12]2 years ago
4 0

Answer: Diagonal going up

Explanation:

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Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
postnew [5]

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

            c = speed of light

            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

Learn more about light here;

brainly.com/question/15200315

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7 0
1 year ago
What are groups 1,2 and 3 examples of on the periodic table
pishuonlain [190]
<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
Group 1 contains the  alkali metals ( lithium<span> (</span>Li<span>), </span>sodium<span> (</span>Na<span>), </span>potassium<span> (</span>K<span>), </span>rubidium<span> (</span>Rb<span>), </span>caesium<span> (</span>Cs<span>), and </span>francium(Fr).)<span>
Group 2 contains the alkaline earth metals (</span> beryllium<span> (</span>Be),magnesium<span> (</span>Mg<span>), </span>calcium<span> (</span>Ca<span>), </span>strontium<span> (</span>Sr<span>), </span>barium<span> (</span>Ba<span>) and </span>radium<span> (</span>Ra<span>) )
Group 3: </span><span> Scandium (Sc) and yttrium (Y) </span>
4 0
2 years ago
Photo shows all! Need help ASAP! Will mark brainlist
Goryan [66]

Answer:

The student is getting different info bc the students probable keeping track of the distance instead of the displacement.

Explanation:

5 0
2 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step
azamat

Answer:

Explanation:

v_f^2 = v_i^2-2a(x_f-x_i)

Subtract both sides by v_i^2:

- v_i^2+v_f^2 = -2a(x_f-x_i)

Divide both sides by -2*a:

\frac{v_i^2 - v_f^2}{2a} =x_f-x_i

Add both sides by x_i:

x_i+\frac{v_i^2 - v_f^2}{2a} =x_f

Subtract both sides by \frac{v_i^2 - v_f^2}{2a}:

x_i=x_f-\frac{v_i^2 - v_f^2}{2a}

8 0
3 years ago
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