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1 year ago

A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer

Physics

1 answer:

koban [17]1 year ago

5 0

Answer:

the velocity of the swimmer at the bottom of the slide is 6.26 m/s

Explanation:

The computation of the velocity of the swimmer at the bottom of the slide is given below:

v = √2gh

= √2 × 9.8 × 2

= 6.26 m/s

Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s

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Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

2 years ago

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu

Anit [1.1K]

A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N

Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2

</span>

8 0

2 years ago

Read 2 more answers

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$ , $\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$ :

#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y \$ and $L$ .

$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

8 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b

Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0

2 years ago