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4 years ago

11

You can painlessly wade into a pool, but doing a belly flop off of the high diving board hurts because of

1 answer:

Blizzard [7]4 years ago

7 0

You can painlessly wade into a pool, but doing a belly flop off of the high diving board hurts because of the cohesion of the water molecules.

<u>Explanation:</u>

Water is represented as H20 where two hydrogen atoms are bonded to one oxygen atom. Due to force of attraction between the water molecules they stick together providing a greater force when an object or a body acts upon them.

Hence moving slowly in water or dipping slowly will not produce any opposite form whereas flipping does as it tries to separate the body between the water molecules when we fall with a certain amount of force.

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A rock is thrown horizontally out of a window with a velocity of 20.0 m/s. If the window is 8.50m above the ground, how far

vladimir1956 [14]

The distance from the base of the building the rock will land is 26.4 m

<h3>Data obtained from the question </h3>

Horizontal velocity (u) = 20 m/s
Height (h) = 8.50 m
Distance (s) =?

<h3>Determination of the time to reach the ground </h3>

Height (h) = 8.50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?

h = ½gt²

8.5 = ½ × 9.8 × t²

8.5 = 4.9 × t²

Divide both side by 4.9

t² = 8.5 / 4.9

Take the square root of both side

t = √(8.5 / 4.9)

t = 1.32 s

<h3>How to determine the distance </h3>

Horizontal velocity (u) = 20 m/s
Time (t) = 1.32 s
Distance (s) =?

s = ut

s = 20 × 1.32

s = 26.4 m

Learn more about motion under gravity:

brainly.com/question/22719691

6 0

3 years ago

Noah Formula is in an airplane which is flying at constant speed in a circular course of radius 6500 m, circling O’Hare‘s airpla

algol [13]

Answer: this one looks hard

Explanation:

3 0

4 years ago

4) The mass of Pluto is 1.31 x 1022 kg and its radius is 1.15 x 106 m. What is the acceleration of

Elina [12.6K]

Answer:

0.661 m/s²

Explanation:

g = MG / r²

g = (1.31×10²² kg) (6.67×10⁻¹¹ m³/kg/s²) / (1.15×10⁶ m)²

g = 6.61×10⁻¹ m/s²

g = 0.661 m/s²

3 0

3 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

5 0

3 years ago