Answer:
Approximately
(given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude
is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is
. Apply the unit conversion
:
.
An electric field of magnitude
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Answer
4.8 N
If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.
The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.
The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
pls thank me and brainliest me
Answer:
Newtons first law
Explanation:
object in rest stays at rest
object in motion stays in motion
Answer:

Explanation:
As we know that tension force in the string will be equal to the centripetal force on the string
so we will have

now we have

now we have


now when string length is 0.896 m and its speed is 71.5 m/s then we will have


