Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,
![F= ma](https://tex.z-dn.net/?f=F%3D%20ma)
And we know as well that
![a= \frac{v}{t}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bv%7D%7Bt%7D)
Replacing the aceleration value in the equation force we have,
![F= \frac{mv}{t}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7Bmv%7D%7Bt%7D)
Substituting our values we have,
![F= \frac{(0.060)(55)}{4*10^{-3}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B%280.060%29%2855%29%7D%7B4%2A10%5E%7B-3%7D%7D)
![F=825N](https://tex.z-dn.net/?f=F%3D825N)
The weight of the person is then,
![W = mg](https://tex.z-dn.net/?f=W%20%3D%20mg)
![W = (60)(9.8) = 558N](https://tex.z-dn.net/?f=W%20%3D%20%2860%29%289.8%29%20%3D%20558N)
<em>We can conclude that force on the ball is large to lift the ball</em>
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
Answer and Explanation:
The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.
Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.
When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.
The part of the light that continues on into the new medium is <em>transmitted</em>
light. Its forward progress at any point in its journey is <em>transmission</em>.
Its direction usually changes as it crosses the boundary. The bending is <em>
refraction</em>.
The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light. The process of bouncing is <em>reflection</em>.