Answer:
98N and 147N
Explanation:
We have the following information:
We can find the static fricton force as follow,
Where N is the normal force (mg)
Static friction force at 147N is greater than the force applied hence body does not move.
Which statement is TRUE about white light?
2 answers:
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1) Distance: 9 m, displacement: +5 m
2) The average speed is 3 m/s
3) The average velocity is +2 m/s
Explanation:
1)
We start by reminding the definitions of distance and displacement:
- Distance: it is the total length of the path covered by an object, regardless of the direction of motion
- Displacement: is the distance in a straight line between the initial position and final position of motion
Let's now analyze the motion represented in the figure:
Initial position:
Intermediate position:
Final position:
Therefore, the distance covered is
While the displacement is:
2)
The average speed is a scalar quantity defined as
where
d is the distance travelled
t is the time taken
For the object in this problem,
d = 6 m
t = 2 s
Substituting, we find the average speed:
3)
The average velocity is a vector quantity defined as
where
d is the displacement
t is the time interval
For the object in this problem,
d = +12 m
t = 6 s
Substituting, we find the average velocity:
Where the positive sign + indicates the direction of the velocity vector.
Learn more about average speed and average velocity:
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Answer:
r₂ = 4 r
Explanation:
For this exercise let's use Newton's second law with the magnetic force
F = q v x B
bold letters indicate vectors, the magnitude of this expression is
F = q v B sin θ
in this case we assume that the angle is 90º between the speed and the magnetic field.
If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal
a = v² / r
let's use Newton's second law
F = ma
q v B = m v² / r
r =
Let's apply this expression to our case.
Proton 1
r = \frac{qB_1}{mv_1}
Proton 2
r₂ =
in the exercise indicate some relationships between the two protons
* v₁ = 2 v₂
v₂ = v₁ / 2
* B₂ = 2B₁
we substitute
r₂ =
r₂ = 4
r₂ = 4 r