<em>Answer: D</em>
<em>Given data:</em>
The electric field due to charge q₁ is (E₁) = 1.5 × 10⁵ N/C
The electric field due to charge q₂ is (E₂) = 7.2 × 10⁵ N/C
Determine the net electric field at point P (Enet) = ?
electric field measured in Newtons/coulomb
<em>We know that,</em>
The net electric field (Enet) is vector sum of E₁ and E₂ caused by the charge q₁ and q₂ respectively
Enet = E₁ + E₂
= (1.5 × 10⁵)+ (7.2 × 10⁵)
= 8.7 × 10⁵ Newtons/coulomb
<em>The net electric field at point P is 8.7 × 10⁵ N/C</em>
speed Δv = v₂ = 8.15ms⁻¹
v₁ = 0ms⁻¹
time =t = 5s
acceleration = a = speed / time taken
a = Δv / Δt ( as Δv = v₂ - v₁)
a = 8.15ms⁻¹ / 5 s
a = 1.6ms⁻²
<u><em>Answer:</em></u>
The answer is 1400 J, according to my Physics teacher.
<u><em>Explanation:</em></u>
You need to take into account everything that is listed in the question; it's important to remember that the question is asking about the change in gravitational potential energy of the object-object-Earth system from 0s to 10s, not 0s to 20s. :)
Answer:
0.832 m/s
Explanation:
The work done by the spring W equals the kinetic energy of the object K
The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀ = initial compression = 0.065 m and x₁ = final compression = 0.032 m
The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object
Since W = K,
1/2k(x₀² - x₁²) = 1/2mv²
k(x₀² - x₁²) = mv²
mv² = k(x₀² - x₁²)
v² = [(k/m)(x₀² - x₁²)]
taking square root of both sides, we have
v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),
v = √[(k/m)√(x₀² - x₁²)]
v = ω√(x₀² - x₁²)]
Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have
v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]
v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]
v = 14.7 rad/s × √(0.003201 m²)
v = 14.7 rad/s × 0.056577
v = 0.832 m/s
Answer:
Ensure each member of the audience can see and read the aid.
Explanation:
