Which of the following are true for an object in projectile motion?
1 answer:
The correct options are: B, C and D.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
(acceleration of gravity)
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
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To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:
Travel north 2mph and west to 1mph, then,
The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to
Therefore the total travel time for the man is 1.15hour.
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ = ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀ ) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀ ) Q / (1.25R)²
E = (1 /4πε₀ ) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀ ) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B