Compute the ball's angular speed <em>v</em> :
<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s
Use this to find the magnitude of the radial acceleration <em>a</em> :
<em>a</em> = <em>v </em>²/<em>R</em>
where <em>R</em> is the radius of the circular path. We get
<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²
The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude
∑ <em>F</em> = <em>m</em> <em>a</em>
where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then
<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N
The work done by the electric field is equal to the loss of electric potential energy of the proton in moving from its initial location to its final location:
where
is the proton charge,
and
are the voltages in the final and initial locations. Substituting, we get
The answer is 2. The number 2 does not go into 55 evenly, therefore making it not a factor. All the others are factors. You can solve this using the factor list method.
Factors of 55
<em>1, 55</em>
<em>5, 11</em>
<em>11, 5</em>
<em>55, 1</em>
This works for all problems involving factoring.
<em>H a v e a f a n t a s t i c n i g h t / d a y !</em>
-Dylan (AKA Animus)
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Answer: I am very confused. Where are the answer choices
Explanation: