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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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You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
iren [92.7K]

Answer:

angle minimum   θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

     fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

     

the friction force formula is

     fr = μ N

     fr = μ W

we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

      μ sin θ - cos θ + ½ cos θ= 0

         

       μ sin θ - ½ cos θ = 0

       sin θ / cos θ = 1/2 μ

       tan θ = 1/2 μ

       θ = tan⁻¹ (1 / 2μ)

       θ = tan⁻¹ (1 (2 0.57))

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7 0
3 years ago
At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.
prohojiy [21]

Answer:

θ = 66.90°

Explanation:

we know that

I= \frac{I_0}{2}cos^2\theta

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

1= \frac{13}{2}cos^2\theta

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5 0
3 years ago
A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V
sladkih [1.3K]

Answer:

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

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Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F

Initial charge across the parallel plates:

Q₁ = CV

Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C

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Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

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                              = 0.924 nC

Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

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