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vaieri [72.5K]
3 years ago
6

Calculate the period of a ball tied to a string of length 0.468 m making 3.8 revolutions every second. Answer in units of s. You

r answer must be within ± 2.0%
Physics
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

0.26315 s

Explanation:

The frequency of the ball tied to a string system is 3.8 rev/s.

That means in one second the ball will complete 3.8 revolutions.

The time period will be the reciprocal of this frequency

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.8}\\\Rightarrow T=0.26315\ s

The time period is 0.26315 s

It can be also solved in the following way

1\ s=3.8\ rev\\\Rightarrow 1\ rev=\dfrac{1}{3.8}\ s\\\Rightarrow 1\ rev=0.26315\ s

The time period is 0.26315 s

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Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
Which of the following could be described as "Energy cannot be created or destroyed ,only converted into other forms"?
Kryger [21]

Answer:

C

Explanation:

This is right because that's how u describe it

5 0
3 years ago
the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angl
Cerrena [4.2K]

Answer:

(a) 1.054 m/s²

(b) 1.404 m/s²

Explanation:

0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ))  - μk·m·g·(1 - sin(θ))  = m·a

Which gives;

0.5·g·cos(θ) - μ·g·(1 - sin(θ)   = a

Where:

m = Mass of the of the block

μ = Coefficient of friction

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block

θ = Angle of elevation of the block = 20°

Therefore;

0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°)  - μk×9.81×(1 - sin(20°) = a

(a) When the static friction μs = 0.610  and the dynamic friction μk = 0.500, we have;

0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°)  - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²

(b) When the static friction μs = 0.400  and the dynamic friction μk = 0.300, we have;

0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°)  - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².

3 0
3 years ago
WILL GIVE BRAINLIEST!
Ulleksa [173]

Answer:0kgm/s

Explanation:

Momentum before collision=momentum after collision

Since the momentum of the two blocks have positive sign, it means they are moving in thesame direction

Therefore we use the formula

Momentum (A)+momentum (B)=Momentum (A)+momentum (B)

25+35=60+momentum (B)

60=60+momentum (B)

Momentum (B)=60-60

Momentum (B)=0kgm/s

6 0
3 years ago
How are instantaneous speed and average speed related for an object with constant speed?
Gala2k [10]
In the case of an object with constant speed, the instantaneous speed and average speed are equal.
7 0
3 years ago
Read 2 more answers
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