Force required to accelerate 10 kg object to 5.9 m/s/s ?
Mass = 10 kg
Acceleration = 5.9 m/s^2
Force = Mass * Acceleration
Force = 10 kg * 5.9 m/s^2
Force = 59 kg m /s^2 = 59 N
Answer:
1408.685 KN/C
Explanation:
Given:
R = 0.45 m
σ = 175 μC/m²
P is located a distance a = 0.75 m
k = 8.99*10^9
- The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:
part a)
Electric Field strength at point P: a = 0.75 m
part b)
Since, R >> a, we can approximate a / R = 0 ,
Hence, E simplified relation becomes:
E = σ / 2*e_o
part c)
Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:
Electric Field strength due to point charge is:
E = k*δ*pi*R^2 / a^2
Since, R << a, Surface area = δ*pi
Hence,
E = (k*δ*pi/a^2)
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
Potassium hydroxide (KOH) and
hydrochloric acid (HCl) react in a beaker. They form potassium chloride (KCl)
and water (H2O). This type of reaction is a double replacement reaction wherein
the two principal reactants exchange after the reaction. The Cl will combine to
the K forming KCl and the H will add to the OH forming H2O.
