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Sauron [17]
3 years ago
10

Children slide down a frictionless water slide that ends at a height of 1.80 m above the pool. If a child starts from rest at po

int A and lands in the water at point B, a horizontal distance L = 2.42 m from the base of the slide, determine the height h of the water slide.
Physics
1 answer:
irina [24]3 years ago
4 0

Answer:0.813 m

Explanation:

Given

height at which slide ends is h=1.8 m

Horizontal distance L=2.42 m moved by child from slide base to pool

At slide end Child has only horizontal velocity

horizontal velocity is given by u=\sqrt{2gH}

where H=height of slide

time taken by child to complete 1.8 m height

t=\sqrt{\frac{2h}{g}}

t=\sqrt{\frac{2\times 1.8}{9.8}}

t=0.606 s

Horizontal distance moved in this time

L=2.42=u\times t

2.42=\sqrt{2gH}\times 0.606

\sqrt{2gH}=3.99

H=0.813 m      

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A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?
Sauron [17]

The velocity of the ball is 12.5 m/s

Explanation:

The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

v=\frac{d}{t}

First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

L=2\pi r = 2\pi(0.750)=4.7 m

The ball makes 25.0 revolutions, so a total distance of

d=(25.0)L=(25.0)(4.7)=117.5 m

In a time of

t = 9.37 s

So, its velocity is

v=\frac{117.5}{9.37}=12.5 m/s

Learn more about velocity here:

brainly.com/question/5248528

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3 0
3 years ago
Read 2 more answers
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

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5 0
1 year ago
Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher
siniylev [52]

Answer:

v = 4.44 \times 10^5 m/s

Explanation:

By Einstein's Equation of photoelectric effect we know that

h\nu = W + \frac{1}{2}mv^2

here we know that

h\nu = energy of the photons incident on the metal

W = minimum energy required to remove photons from metal

\frac{1}{2}mv^2 = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

W = \frac{hc}{351 nm}

here we know that

hc = 1242 eV-nm

now we have

W = \frac{1242}{351} = 3.54 eV

now by energy equation above when photon of 303 nm incident on the surface

\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2

4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2

(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2

8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2

v = 4.44 \times 10^5 m/s

6 0
3 years ago
A gas has an initial volume of 212 cm^3 at a temperature of 293 K and a pressure of 0.98 atm. What is the final pressure of the
MatroZZZ [7]
For this we use general equation for gases. Our variables represent:

p- pressure
v-volume
t- temperature

P1V1/T1 = P2V2/T2

in this equation we know:
P1,V1 and T1, T2 and V2. 
We have one equation and 1 unknown variable.

P2 = T2P1V1/T1V2 = 1.1atm
8 0
3 years ago
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
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