Need the answer for question 5 :)

A dog sits 2.6 m from the center of a merry- go-round. a) If the dog undergoes a 1.7 m/s^2 centripetal acceleration, what is the

alexgriva [62]

Answer:

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Explanation

Conceptual analysis :

The dog and the merry-go- round describes a circular motion, then, the following formulas apply :

$a_{c} =\frac{v^{2} }{r}$ Formula (1)

v = ω *r Formula (2)

Where:

$a_{c}$ : Centripetal acceleration(m/s²)

v: linear speed or tangential (m/s)

r : radius of the circle (m)

ω : angular speed ( rad/s)

Data

r= 2.6 m

$a_{c}$ = 1.7 m/s²

Problem develpment

a) We replace data in the formula 1 to calculate the dog's linear speed(v):

$a_{c} =\frac{v^{2} }{r}$

$1.7 =\frac{v^{2} }{2.6}$

$v^{2} =1.7*2.6 = 4.42$

$v=(\sqrt{4.42})\frac{m}{s}$

v= 2.1 m/s

b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).

v = ω *r

2.1 = ω *2.6

ω = 2.1/2.6

ω = 0.807 rad/s

6 0

2 years ago

A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the

SCORPION-xisa [38]

Answer:

Magnitude of the net force acting on the kayak = 39.61 N

Explanation:

Considering motion of kayak:-

Initial velocity, u = 0 m/s

Distance , s = 0.40 m

Final velocity, v = 0.65 m/s

We have equation of motion v² = u² + 2as

Substituting

v² = u² + 2as

0.65² = 0² + 2 x a x 0.4

a = 0.53 m/s²

We have force, F = ma

Mass, m = 75 kg

F = ma = 75 x 0.53 = 39.61 N

Magnitude of the net force acting on the kayak = 39.61 N

6 0

2 years ago

If you subtract vector 3.7 cm at 45° North of East from vector 4.5 cm at 57° West of North using a scale drawing, what is the re

Gelneren [198K]

The resultant vector is 5.2 cm at a direction of 12⁰ west of north.

<h3>Resultant of the two vectors</h3>

The resultant of the two vectors is calculated as follows;

R = a² + b² - 2ab cos(θ)

where;

θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
a is the first vector
b is the second vector

R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)

R² = 27.02

R = 5.2 cm

<h3>Direction of the vector</h3>

θ = 90 - 78⁰

θ = 12⁰

Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.

Learn more about resultant vector here: brainly.com/question/28047791

#SPJ1

3 0

1 year ago

A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.

REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}}$

$= \frac{g \sin \theta}{1 +\frac{2}{3}}$

$= \frac{3}{5} g \sin \theta$

= 0.6 x 9.81 x sin ( 29.7)

$= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0

2 years ago

If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener

inessss [21]

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

3 0

2 years ago