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Slav-nsk [51]
3 years ago
8

An object dropped from Cliff falls with a constant acceleration of 10 metre per second square find its Speed of two seconds afte

r it was dropped
Physics
2 answers:
Nastasia [14]3 years ago
8 0

Answer:

20 meters per second

Explanation:

All you have to do is multiply 10 meter by the speed of two seconds which is 20 METERS PER SECOND

PLEASE MARK BRAINLIEST KIND PERSON :)

<u><em>~Luis~</em></u>

julsineya [31]3 years ago
3 0

Answer:

20 meters per second

Explanation:

If an object accelerates for 2 seconds, and accelerates by 10 meters per second, then that objects speed will be 20 meters per second, assuming hat there are no other factors involved.

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An earthquake creates a type of wave that shakes the ground. If a large earthquake occurs in Greece, how can the waves be felt a
Cloud [144]
The sound waves travel through the sea which creates movement which can be felt all the way in Italy in the sea- like a domino effect
7 0
3 years ago
Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the
polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

\rm \theta is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\  d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm

The number of lines per mm is found as;

\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

5 0
2 years ago
The speed of sound waves
neonofarm [45]
B. Is faster in solids than liquids would be the correct answer because the molecules in solids are much closer and can pass along energy faster and more effectively.
4 0
3 years ago
A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho
Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0
3 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
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