Answer:
1.55 m
Explanation:
The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:
As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.
Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:
The distance between both points, is just the difference between the radius of both spheres, as follows:
r₂ - r₁ = 1.55 m
Answer:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards
</em>
Explanation:
<u>Vertical Throw
</u>
When an object is thrown upwards, it describes a special type of motion ruled only by gravity.
When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.
The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.
With all these concepts in mind, we state that:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards </em>
The other options are not correct because:
A. The acceleration is never upwards
B. The acceleration is never 0
C. Both times are equal
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
The imaginary second level is 60 dB more intense than the real level of the caller.
60 dB means a multiplication of 10⁶ = <em>1 million.</em>
That's how many equally-talented callers it would take to be 60 dB louder than him.
