By definition it is known that force equals mass by acceleration. In other words F = m * a. To find the acceleration, you must clear the formula mentioned. Therefore, for a force of 190.08N and a mass of 28 Kg, we have that the acceleration is a = F / m = (190.08) / (28) = 6.79 m / s ^ 2
<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a.
Thus, d = [(0^2)-(15^2)]/(2*-7)
d = [0-(225)]/(-14)
d = 225/14
d = 16.0714 m
With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
Answer: The height (position) of the ball and the acceleration due gravity
Explanation:
In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.
In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy
will be:
Where:
is the mass of the ball
is the acceleration due gravity (assuming the ball is on the Earth surface)
is the height (position) of the ball respect to a given point
Note the value of the gravitational potential energy is directly proportional to the height.
Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:

Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²