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Dominik [7]
3 years ago
13

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

tions every 10 seconds. what is the centripetal acceleration and velocity felt by the ball?
Physics
1 answer:
andrezito [222]3 years ago
4 0

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

f = \frac{5}{10} = 0.5 Hz

now we know that angular frequency is given as

\omega = 2\pi f

\omega = 2(\pi)(0.5)

\omega = \pi rad/s

Now centripetal acceleration is given as

a_c = \omega^2 R

a_c = \pi^2 (0.7)

a_c = 6.9 m/s^2

now the velocity of the ball at this angular frequency is given as

v = R\omega

v = 0.7 (\pi)

v = 2.2 m/s

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The half-life of a certain element is 100 days. How many half-lives will it be before only one-eighth of this element remains?
Natalka [10]

Answer:

3

Explanation:

The half-life is the time it takes for the amount of radioactive isotope to halve. Therefore, we have:

- After 1 half-life, only 1/2 of the element will be left

- After 2 half-lives, only 1/4 of the element will be left

- After 3 half-lives, only 1/8 of the element will be left

So, it will take 3 half-lives for the element to become 1/8 of its original amount.

Mathematically, this can be also verified by using the equation

\frac{N(t)}{N_0}=(\frac{1}{2})^\frac{t}{\tau_{1/2}}

where

N(t) is the amount of the element left at time t

N0 is the initial amount of the element

\tau_{1/2} is the half-life

Substituting t=3\tau_{1/2} (3 half-lives), we find

\frac{N(t)}{N_0}=(\frac{1}{2})^3=\frac{1}{8}

7 0
3 years ago
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
3 years ago
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It would be 300 cm/h
and 5 cm/m
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How is amplitude related to loudness
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Which of the following is not a part of the appendicular skeleton
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