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anyanavicka [17]
3 years ago
13

A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What

was the initial vertical velocity of the golf ball?At what time did the golf ball hit the ground?What was the initial horizontal velocity of the golf ball? What was the final horizontal velocity of the golf ball?
Physics
1 answer:
Ratling [72]3 years ago
4 0

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

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Mandible.

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s
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Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is g_m=1.81\ m/s^2 .

Weight of watermelon in earth , W=44\ N .

Acceleration due to gravity at the surface of earth is g=9.81\ m/s^2 .

We know , weight is given by :

W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg

Therefore , mass at the surface of Jupiter's moon Io is :

W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N

Hence , this is the required solution .

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What is true weightlessness ?​
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Answer:

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2 years ago
Five hundred million years ago, basaltic lava flowed in an area now known as Monticello, the historic home of Thomas Jefferson.
aev [14]

Explanation:

               Igneous -  metamorphic -  sedimentary

A rock cycle provides the cyclic transformation of one rock type to another in nature.

There are three main types of rock involved in the rock cycle;

  • igneous rocks are derived from the cooling and solidification of molten magma
  • metamorphic rocks are changed rocks subjected to intense pressure and temperature
  • sedimentary rocks are derived from rock sediments that have been lithified.

The history of the rock in Monticello begins with igneous rock formation. Basalt is an igneous rock that forms from the cooling and solidification of molten magma. Under intense pressure and temperature regimes, they are changed to metamorphic rocks.

Agents of denudation such as wind, water and glacier weathers the rock and disintegrates it. They are then carried into basins where they are deposited. Here they form sedimentary rock.

The process still goes on as the sedimentary rock gets taken into depth, they can either melt to form igneous rock or be changed to metamorphic rocks.

learn more:

metamorphic process brainly.com/question/869769

sedimentary rocks brainly.com/question/9131992

#learnwithBrainly

5 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
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