The lord of the greeks answer d
The ball will take 2.551 seconds to reach its peak position.
<h3>How much time will the ball take to land?</h3>
We must know how long the balls are in the air before we can predict where they will fall. It will take 2 seconds for both balls to touch the ground.
<h3>How quickly does a ball drop?</h3>
The falling ball travels a distance of d = 12 9.8 (m/s2) t2, with a speed of v = 9.8 (m/s2) t as a function of time. The ball travels 4.9 m in a second. The falling ball's velocity is v = -9.8 (m/s2) t j, and its position is r = (4.9 m - 12 9.8 (m/s2) t2) j as a function of time.
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On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
Answer:
F = 147,78*10⁻⁹ [N]
Explanation:
By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .
The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is
tan β = 0,5/0,7
tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰
cos β = 0,81
d = √ (0,5)² + (0,7)² d1stance between charges
d = √0,25 + 0,49
d = √0,74 m
d = 0,86 m
Now Foce between two charges is:
F = K* q₁*q₂/ d² (1)
Where K = 9*10⁹ N*m²/C²
q₁ = 2,5* 10⁻⁹C
q₂ = 3,0*10⁻⁹C
d² = 0,74 m²
Plugging these values in (1)
F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²
F = 91,21 * 10⁻⁹ [N]
And Fx = F*cos β
Fx = 91,21 * 10⁻⁹ *0,81
Fx =73,89*10⁻⁹ [N]
Then total force acting on charge located at x = 0,7 m is:
F = 2* Fx
F = 2*73,89*10⁻⁹ [N]
F = 147,78*10⁻⁹ [N]
