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vekshin1
2 years ago
8

A 90 kilogram diver jumped from a cliff. If it takes 2.26 seconds for the diver to hit the water, how high is the cliff?

Physics
2 answers:
Tasya [4]2 years ago
8 0

The height of the cliff will be 25.05 m. The height of the cliff is obtained by the Newton equation of motion.

<h3>What is height?</h3>

The vertical distance between the object's top and bottom is defined as height. It is measured in centimeters, inches, meters, and other units.

The given data in the problem is;

Mass of diver(m)=90 kilogram

Time period(t)=2.26 sec

Initial velocity (u)=0 m/sec

The height attained by the cliff is;

\rm H = ut + \frac{1}{2}gt^2 \\\\ H=\frac{1}{2}gt^2\\\\ H=  \frac{1}{2}(9.81)(2.26)^2\\\\ H= \frac{1}{2}  \times 9.81 \times 5.1076 \\\\ H=25.05 \ m

Hence, the height of the cliff will be 25.05 m.

To learn more about the height, refer to the link;

brainly.com/question/10726356

#SPJ2

Artist 52 [7]2 years ago
3 0

Answer:

3.27 meters

Explanation:

mark brainliest please

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(i) • there is force applied to an objects

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(ii) workdone = force x distance

= 23 x 34

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8 0
3 years ago
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The law of reflection states that if the angle of incidence is 38 degrees, the angle of reflection is ___ degrees.
Tasya [4]

Answer:

38

Explanation:

because the law of reflection states angle of incidence is equal to the angle of reflection

4 0
3 years ago
A ship is travelling due east at 30 km/hr and a boy runs across the deck
dsp73

Answer:

Vr = 20 [km/h]

Explanation:

In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.

Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).

V_{r}=30-10\\V_{r}=20 [km/h]

8 0
3 years ago
A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

5 0
3 years ago
A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?
serious [3.7K]

Answer :  The wavelength of photon is, 4.63\times 10^{2}nm

Explanation : Given,

Energy of photon = 4.29\times 10^{-19}J

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm

Conversion used : 1nm=10^{-9}m

Therefore, the wavelength of photon is, 4.63\times 10^{2}nm

6 0
3 years ago
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