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STALIN [3.7K]
3 years ago
7

What is the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head? The radius of the cir

cle is 1.8 meters, and the ball is traveling at 9.2 meters/second.
Physics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer:

Centripetal acceleration of boy on a string is 47.02 m/s²

Explanation:

It is given that,

Radius of the circle, r = 1.8 m

Speed of the ball, v = 9.2 m/s

Centripetal acceleration is given by :

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

a_c=\dfrac{(9.2)^2}{1.8}

a_c=47.02\ m/s^2

Hence, the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head is 47.02 m/s²

Leya [2.2K]3 years ago
4 0
Well the centripetal acceleration would be 9.82 squared. That is the correct answer. I hope this helped you. Have a great day! :)
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A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli
vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

x=V_{o} cos\theta t (1)

y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2} (2)

Where:

x=0.25 m is the horizontal displacement of the pencil

V_{o} is the pencil's initial velocity

\theta=0\° since we are told the pencil rolls <u>horizontally</u> before falling

t is the time since the pencil falls until it hits the ground

y_{o}=1 m  is the initial height of the pencil

y=0  is the final height of the pencil (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity, always acting vertically downwards

Begining with (1):

x=V_{o} cos(0\°) t (3)

x=V_{o}t (4)

Finding t from (2):

0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2} (5)

t=\sqrt{\frac{-2y_{o}}{g}} (6)

Substituting (6) in (4):

x=V_{o}\sqrt{\frac{-2y_{o}}{g}} (7)

Isolating V_{o}:

V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}} (8)

V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}} (9)

Finally:

V_{o}=0.55 m/s

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3 years ago
A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio
vlabodo [156]

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

F= qVB

Replacing

F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})

F=9.36*10^{-19}

c) Given the second Newton's Law by

F_g = 1.67*10^{-27}*9.81

F_g = 1.64*10^{-26}

Given the electric force by,

F_e = 1.6*10^{-19}*1.5*10^2

F_e = 2.4*10^{-17}N

F=9.36*10^{-19}N

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vladimir2022 [97]
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What is current electricity? write down its use in our daily like ​.
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8 0
2 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

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Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

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3 years ago
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