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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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The sales price of a product is $2 per unit; the variable cost is $1 per unit; and fixed costs total $1,000. how many units must
irakobra [83]

Answer:

<u><em>1000 units for breakeven</em></u>

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

Variable costs would be $1x and fixed costs are $1000.

Total costs are = $1x + $1000

At breakeven: Sales = Costs

Sales =m Costs

$2x = $1x + $1000

$1x = $1000

x = 1000 units.

At 1000 units the sales are equal to the costs ("breakeven").

4 0
1 year ago
A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.
Lilit [14]

Answer:

Explanation:

a. The source of centripetal force on the car is  (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c.  Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

6 0
3 years ago
A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the
aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

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\theta= \frac{\omega_f-\omega_i}{2}\times t

putting values

\theta= \frac{0-2}{2}\times5 = 5 rad

8 0
3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
DanielleElmas [232]

Answer:

\alpha =-2.2669642\times^{-10}rad/s^2

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Angular velocity is related to the period by \omega=\frac{2\pi}{T}

Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000&#10;s

T_i=0.0786s

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\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
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What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?​
stellarik [79]

Answer:

Valence electrons are outer shell electrons with an atom and can participate in the formation of chemical bonds. In single covalent bonds, typically both atoms in the bond contribute one valence electron in order to form a shared pair. The ground state of an atom is the lowest energy state of the atom.

8 0
3 years ago
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