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3 years ago

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A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height

it was released from.

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}$

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}$

<u>h = 0.46 m</u>

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A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H

SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

$v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s$

Time taken by the ball to reach the highest point is 2.44 seconds

a)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m$

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s$

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

$v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s$

The ball will hit the ground at 25.2117 m/s

8 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

If wavelength and speed of a wave are 4 m and 332 m/s respectively, calculate its frequency<br>

Furkat [3]

Explanation:

<em>Given </em>

<em>wavelength </em><em>=</em><em> </em><em>4</em><em> </em><em>m</em>

<em>speed </em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>m/</em><em>s</em>

<em>frequency </em><em>=</em><em> </em><em>?</em>

<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>

<em>wavelength</em><em> </em><em>=</em><em> </em><em>speed </em><em>/</em><em> </em><em>frequency </em>

<em>4</em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>/</em><em> </em><em>frequency </em>

<em>frequency </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em>/</em><em>4</em>

<em>Therefore </em><em> </em><em>frequency </em><em>is </em><em>8</em><em>3</em><em> </em><em>Hertz </em><em>.</em>

4 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

4 years ago

Read 2 more answers

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

$F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules$

Answer:196 Joules

I hope it helps

5 0

3 years ago