Answer:
- See the graph attached with the vectors.
Explanation:
The exercise is to draw the corresponding vectors:
Every one is the multiplication of a scalar by a vector.
The result of multiplying a scalar by a vector is a vector with the same direction of the original vector enlarged in a factor equal to the scalar magntitude.
Thus, in a graph the resulting vector is represented with a parallel arrow and pointing in the same direction as the original vector but with a length equal to the original length multiplied by the magnitude of the scalar.
For instance, the vector 
is represented with an arrow in the same direction of 
and with twice its length.
The figure attached contains the five requested vectors using the procedure explained above.
Answer: D.
Explanation: Just took the quiz.
The change in potential energy of the proton is 5.6 x 
Joule
<h3>
What is a Uniform Electric Field ?</h3>
A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.
Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
- The distance d = 10 cm = 0.1 m
- Electric field E = 3.5 KN/C
- Proton charge q = 1.6 x
C
The Work done = Fd
but F = Eq
Recall that Electric field E = F/q = V/d
Where V = potential difference.
Let us first calculate the V
E = V/d
V = Ed
Substitute all the parameters into the formula above
V = 3.5 × 10³ × 0.1
V = 350 v
from F/q = V/d
make F the subject of formula and substitute it in work formula
F = Vq/d
W.D = Vq/d x d
W.D = Vq
Substitute all the parameters into the formula above
W.D = 350 x 1.6 x
W.D = 5.6 x J
Work done = Energy = Potential Energy
Therefore, the change in potential energy of the proton is 5.6 x <em> Joule</em>
<em />
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Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
Acceleration of the ball is 
Explanation:
The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:
where
F is the net force
m is the mass
a is the acceleration
For the ball in this problem, we have
m = 0.50 kg (mass)
F = 25 N (force)
thereofre, the acceleration of the ball is
Learn more about Newton's second law:
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