Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
An object at rest does not move and an object in motion does not change its velocity, unless an external force acts upon it
Explanation:
This statement is also known as Newton's first law, or law of inertia.
It states that the state of motion of an object can be changed only if there is an external force (different from zero) acting on it: therefore
- If an object is at rest, it will remain at rest if there is no force acting on it
- If an object is moving, it will continue moving at constant velocity if there is no force acting on it
This phenomenon can be also understood by looking at Newton's second law:
F = ma
where
F is the net force on an object
m is the mass
a is the acceleration
If the net force is zero, F = 0, the acceleration of the object is also zero, a = 0: therefore, the velocity of the object does not change, and it will continue moving at the same velocity (which can be zero, if the object was at rest).
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Answer:
The electric potential will be "259.695 volt".
Explanation:
In the given question, the figure is not provided. Below is the attached figure given.
Given:
Now,
At point P, the electric potential will be:
⇒ 
By putting values, we get
⇒ ![=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]](https://tex.z-dn.net/?f=%3D9%5Ctimes%2010%5E9%20%5B%5Cfrac%7B6.39%5Ctimes%2010%5E%7B-9%7D%7D%7B0.40%7D%20%2B%5Cfrac%7B3.22%5Ctimes%2010%5E%7B-9%7D%7D%7B0.25%7D%20%5D)
⇒ 
