Ask question

Ask question

All categories

katovenus [111]

3 years ago

5

A reckless coyote engineer straps on a pair of ice skates and attaches a rocket capable of producing 5.0 × 10 3 N of thrust to h

is back. Together, the coyote and the rocket have a mass of 120 kg. If the coyote starts at rest on level, frictionless ice and bends over such that the rocket thrust is directed parallel to the ice, what is his final speed if the rocket burns for 5.0 s?

1 answer:

Marianna [84]3 years ago

3 0

Answer:

208.33 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

a = Acceleration

m = Mass of coyote and the rocket = 120 kg

F = Force of thrust produced by the rockets = $5\times 10^3\ N$

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{5\times 10^3}{120}\\\Rightarrow a=\frac{125}{3}\ m/s^2$

$v=u+at\\\Rightarrow v=0+\frac{125}{3}\times 5\\\Rightarrow v=208.33\ m/s$

Velocity of the rocket at 5 seconds is 208.33 m/s

You might be interested in

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

Describe what it means if a chemical or physical property is periodic

vesna_86 [32]

It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.

6 0

3 years ago

Which statement describes a question that can be answered by a scientific

goblinko [34]

It’s d! hope I can help

4 0

3 years ago

Read 2 more answers

a girl of mass 40 kg wearing high heel of radius 3mm. find the pressure exerted by her on the ground. [answer the question fast]

kakasveta [241]

You should calculate 40 kg and the radius 3mm.

7 0

3 years ago

If there was no frictional force acting on the ball in question 2 what would happen to the player? Use inertia and Newton’s 1st

laila [671]

Newwton's law of inertia states that an object will not be able to move unless force is applied to it

6 0

3 years ago