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docker41 [41]
3 years ago
13

If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.52 m from the base of the

slide, determine the height h (in m) of the water slide.
Physics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

1.80=0+\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{1.80\times2}{9.8}

t=\sqrt{\dfrac{1.80\times2}{9.8}}

t=0.606\ sec

We need to calculate the velocity

Using formula of velocity

v = \dfrac{d}{t}

Put the value into the formula

v=\dfrac{2.52}{0.606}

v=4.15\ m/s

We need to calculate height

Using conservation of energy

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

Put the value into the formula

h=\dfrac{4.15^2}{2\times9.8}

h=0.878\ m

Hence, The height of the water slide is 0.878 m.

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Instead, suppose that it was fired upward at 60◦ with respect to a horizontal line. Then its horizontal component of velocity is
larisa [96]

Answer:

50 m/s

Explanation:

Angle = 60 degree

Horizontal component of velocity = 50 m/s

A projectile motion is the motion of an object in two dimensions under the influence of gravity.

In this case, the object has no acceleration along horizontal direction, it has acceleration in vertical direction which is equal to the acceleration due to gravity of earth.

When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

5 0
3 years ago
A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo
cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration (a_c) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0
2 years ago
How does the end point differ from the equivalence <br>point of a titration?​
Gwar [14]

<u>Answer:</u>

<em>Equivalence point and end point are terminologies in pH titrations and they are not the same. </em>

<u>Explanation:</u>

In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.

At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation.  The nature of the final solution determines the <em>pH at equivalence point. </em>

<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution. </em>

3 0
3 years ago
What is the acceleration of a 1000kg car subject to a 550N net force?
igomit [66]

Answer:

a=550÷1000

a=0.55m/s²

5 0
3 years ago
PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?
Naily [24]

Answer: the frequency is every 27.322 days

6 0
3 years ago
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