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3 years ago

A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?

1 answer:

8 0

R=ut+gt^2/2
r- displacement (height to find)
u - initial speed (zero)
t - time taken

r=0*5.8 + 10*5.8^2 /2 = 168.2 meters

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A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr

7nadin3 [17]

Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)

7 0

2 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

Can anyone ask me how many protons, neutrons, and electrons I need and just explain where to put them.

Aliun [14]

Answer:

on the first shell (ring) there will be 2 electrons

and on the 2nd shell there will be only one electron

while in the nucleus (the middle of the diagram) there will be 4 neutrons and 3 protons

Explanation:

you can see the picture attached

8 0

3 years ago

A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

6 0

3 years ago

Read 2 more answers

a 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. if the c

velikii [3]

consider the velocity towards the pitcher as positive

m = mass of the baseball = 0.145 kg

v₀ = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = 3 x 10⁻³ sec

F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

3 0

3 years ago