The magnitude of the electrostatic force is 0.09 N, and the force is attractive
Explanation:
The strength of the electrostatic force acting on a charged particle is given by
where
q is the charge
E is the magnitude of the electric field
For the charge in this problem, we have:
is the electric field
is the charge
Therefore, the force experienced by this charge is
And the negative sign means that the force is attractive: in fact, the charge is negative, while the charge that is producing the field is positive (because the electric field points outward), therefore the negative charge is attracted towards the positive charge that is generating the field.
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Take wire and a nail strip 2 in of wire then wrap it around a nail
Answer:
All parts of a wheel rotating about a fixed axis must have the same angular speed.
All parts of a wheel rotating about a fixed axis must have the same angular acceleration.
Explanation:
In a rigid body in rotation (a rotating wheel) with respect to a fixed axis, each particle on the body turns around the same angle in a certain time interval, having the same angular speed and the same angular acceleration. Otherwise, the wheel will deform, therefore it would not be in equilibrium
Unbalanced forces
When the forces acting on an object do not balance, the resultant force will cause the object to accelerate in the direction of the resultant force.
In other words, a resultant force on a body will cause it to change its velocity. This simply means that unbalanced forces will cause:
acceleration
deceleration
change in direction. The relationship between the resultant force, the mass of the object and the object’s acceleration is:
resultant force (N)
=
mass (kg)
×
acceleration (m/s
2
)
F
=
m
×
a
You will have to be able to state this equation in the examination.
Question
A car weighs 1,000 kg. The resultant force is 5,000 N. Use the Fma triangle to find the acceleration of the car.
The question is incomplete.
Here is the complete question (only part A):
Two children, Ferdinand and Isabella, are playing with
a water hose on a sunny summer day. Isabella is holding the hose
inher hand 1.0 meters above the ground and is trying to
spray Ferdinand, who is standing 10.0 meters away.
Part A
Will Isabella be able to spray Ferdinand
if the water is flowing out of the hose at a constant
speed Vo of 3.5
meters per second? Assume that the hose is pointed parallel to the
ground and take the magnitude of the acceleration due to gravity
to be 9.81 meters per second, per second.
Answer: No. Since the water jet runs 1.58 m before falling into the ground, it cannot spray Ferdinand who is 10.0 m away.
Explanation:
1) Data:
yo = 1.0m
Vox = 3.5 m/s
Voy = 0
x = 10.0 m
g = 9.81 m/s²
2) Type of motion and formulas
Parabolic motion
y = yo + Voy×t - g×t² / 2
x = Vox × t
3) Solution:
i) From y = yo + Voy×t - g×t² / 2 calculate t until the water reaches the ground (y = 0).
From the data yo = 1, and Voy = 0
0 = 1.0 - 4.905t² ⇒ t = √ (1.0 / 4.905) = 0.45 s
ii) Calculate x using t = 0.45s
From the data Vox = 3.5m/s
x = Vox × t = 3.5m/s × 0.45s = 1.58 m
iii) Conclusion: since the water jet runs 1.58 m before falling into the ground, it cannot spray Ferdinand who is 10.0 m away.