A turtle crawls at a constant speed, moving 4.77 m in 19.5 s. What was the turtle's velocity?

2 answers:

8 0

The turtles average speed during that time was 0.245 meter per second. We have no way to describe its velocity because you haven't told us anything about the direction it crawled.

topjm [15]4 years ago

5 0

I think that the velocity is 0.245m per second

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an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b

Kay [80]

Answer:

333.3 m

Explanation:

Given

$m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2$

Potential energy = $\frac{2}{3}\ of\ Kinetic\ energy$ ......Equation(1)

We know that

Potential energy=mgh

Kinetic energy = $\frac{1}{2} mv^{2}$

Now From the Equation(1)

$mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m$

3 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$