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MatroZZZ [7]
2 years ago
6

A 0.05 kg ball is attached to the end of a 0.5 m long string. The ball is spun around a circle with a period of 0.20 s. What is

the tension force in the string? ​
Physics
1 answer:
Gekata [30.6K]2 years ago
4 0

The ball spins once around the circle in 0.20 s, meaning it travels a distance equal to its circumference in that time, giving it a linear speed of

<em>v</em> = (2<em>π</em> (0.5 m)) / (0.20 s) = 5<em>π</em> m/s ≈ 15.708 m/s

Use this compute the magnitude of the centripetal acceleration <em>a</em> :

<em>a</em> = <em>v</em>²/ (0.5 m) = 50<em>π</em>² m/s² ≈ 493.48 m/s²

Use Newton's second law to compute the mangitude of the tension <em>F</em> in the string:

<em>F</em> = (0.05 kg) <em>a</em> = 5/2 <em>π</em>² N ≈ 24.674 N ≈ 25 N

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FM radio ________________. a. had a somewhat shorter range than AM radio, but better sound quality. b. was widely adopted in the
svetlana [45]

Answer:

(A) FM Radio had a somewhat shorter ranger than AM radio, but better sound quality.

Explanation:

FM Radio was invented in 1933 by Edwin Armstrong who was an American engineer. FM stands for frequency modulation and AM stands for Amplitude Modulation.

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6 0
3 years ago
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
An airplane travels from st louis to portland, oregon in 4.33 hours. if the distance traveled is 2,742 kilometers, what is the a
Makovka662 [10]

Average speed = (total distance covered) / (total time to cover the distance)

                       =  (2,742 km)  /  (4.33 hours)

                       =  (2,742 / 4.33)    km/hr

                       =      633 km/hr        (rounded)
4 0
3 years ago
Read 2 more answers
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