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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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Explain why streets and highways have speed limits and not velocity limits.
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We know that speed is a scalar unit while velocity is a vector. The signs are mostly concerned with the magnitude that the car should travel in, the direction is not as important. 

Hope I helped :) 
4 0
3 years ago
How does rhyolite form?
never [62]
Hello There!

I'm pretty sure it is when rhyolitic magma is released.

Hope This Helps You!
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6 0
4 years ago
Read 2 more answers
9. Calculate the distance (in km) that Charlie runs if he maintains the average
Karo-lina-s [1.5K]

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$t=1 hour

As we know the formula to evaluate the total distance d when the average velocity and time is given;

v=\frac{d}{t}

d=v \times t

d=8 \times 1

d=8 k m

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

5 0
3 years ago
An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Substitute the values then we get

-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

Hence, the magnification of apple=0.33

5 0
3 years ago
As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis
klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

d = v_0t + \frac{1}{2}at^2

The block starts from rest, so:

d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2

Now, we can do a summation of forces of the block using Newton's Second Law:

F = ma = m_bg - T

mb = mass of the block

T = tension of string

Solve for tension:

T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N

Now, we can do a summation of torques for the wheel:

\Sigma \tau = rF\\\\\Sigma\tau = rT

Rewrite:

I\alpha = rT

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2

Now, plug in the values into the equation:

I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}

8 0
3 years ago
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