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kodGreya [7K]
3 years ago
9

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

is 8.5 cm at the middle of the short side. What is the centripetal acceleration of the laces on the football? (Enter the magnitude in m/s2.)
Physics
1 answer:
faltersainse [42]3 years ago
6 0

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.

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Misha Larkins [42]
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3 years ago
What do you need to verify in a patient's profile? A. Name, history of illness, medication, dose, strength, frequency B. Name, a
Lubov Fominskaja [6]
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3 0
4 years ago
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Two wheels roll side-by-side without sliding, at the same speed. The radius of wheel 2 is one-half (1/2) the radius of wheel 1.
Lemur [1.5K]

Answer:

w'=(1/2)w

Explanation:

In order to calculate the angular velocity of the second wheel, you use the following formula:

\omega=\frac{v}{r}      (1)

v: speed of the wheel 1 = speed of the wheel 2

r: radius of the wheel 1

For the second wheel you have:

r'=2r

You replace this value of r' in the following equation:

\omega'=\frac{v}{r'}=\frac{v}{2r}=\frac{1}{2}\frac{v}{r}=\frac{1}{2}\omega\\\\\omega'=\frac{1}{2}\omega

The angular velocity of the second wheel is one half of the angular velocity of the first wheel

6 0
3 years ago
New attempt is in progress. Some of the new entries may impact the last attempt grading. Incorrect. The hammer throw is a track-
Morgarella [4.7K]

This question involves the concepts of centripetal force, range of projectile and projectile motion.

The magnitude of centripetal force is "2812.8 N".

First, we will find the velocity of the ball by using the formula of the range of the projectile.

R = \frac{v^2Sin2\theta}{g}

where,

R = range of projectile = 86.75 m

v = speed = ?

θ = launch angle = 47.9°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

86.75\ m = \frac{(v)^2Sin(2)(47.9^o)}{9.81\ m/s^2}\\\\v = \sqrt{\frac{(86.75\ m)(9.81\ m/s^2)}{Sin95.8^o}}

v = 29.25 m/s

Now, we will use the formula to find out the centripetal force:

F_c = \frac{mv^2}{r}

where,

F_c = Centripetal Force = ?

m = mass of the ball = 7.3 kg

v = speed = 29.25 m/s

r = radius = 2.22 m

Therefore,

F_c = \frac{(7.3\ kg)(29.25\ m/s)^2}{2.22\ m}

<u>Fc = 2812.8 N = 2.812 KN</u>

<u />

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

5 0
3 years ago
A hydrogen atom has a radius of 2.5 x 10-11 m<br> Determine the radius of a magnesium atom.
baherus [9]

Answer:

R = 1.5* 10^{-10}m

Explanation:

Given

r = 2.5 * 10^{-11}m -- radius of hydrogen atom

<em>See attachment</em>

Required

Determine the radius of magnesium atom (R)

From the attachment, the ratio of a hydrogen atom to a magnesium atom is:

Ratio = 6mm : 36mm

Simplify

Ratio =1 : 6

Represent the radius as ratio:

Ratio = r : R

Substitute r = 2.5 * 10^{-11}m

Ratio = 2.5 * 10^{-11}m : R

Equate both ratios

2.5 * 10^{-11}m : R = 1 : 6

Express as fraction

\frac{2.5 * 10^{-11}m}{R} = \frac{1}{6}

Cross Multiply

R * 1 = 2.5 * 10^{-11}m * 6

R * 1 = 2.5 * 6* 10^{-11}m

R * 1 = 15* 10^{-11}m

R = 15* 10^{-11}m

R = 1.5*10* 10^{-11}m

R = 1.5* 10^{1-11}m

R = 1.5* 10^{-10}m

Hence, the radius of the magnesium atom is: 1.5* 10^{-10}m

6 0
3 years ago
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