Answer:
Explanation:
a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.
Answer:
v = 91.8 Km / h
Explanation:
We must start this exercise at the end, let's look for the lighter car acceleration (B), for this we use Newton's second law
fr = m a
a = fr / m
fr = μ N
N-W = 0
Let's replace
a = μ m g / m
a = μ g
a = 0.8 9.8
a = 7.84 m / s²
As the car B after the crash reached an initial velocity vo₀₂ and at the end of the fine speed zero, let us use kinematics
v² = v₀₂² - 2 a x
0 = v₀₂² - 2 a x
v₀₂ = 2 a x₂
v₀₂ = √ (2 7.84 26)
v₀₂ = 20.19 m / s
Let's perform the same procedure for car A, the acceleration is the same as it does not depend on the mass of the vehicles
v₀₁ = √ 2 a x₁
v₀₁ = √ (2 7.84 19)
v₀₂ = 17.36 m / s
Now let's use moment conservation, where the system is the two vehicles
Initial before crash.
p₀ = M v₁ + 0
After the crash
= M v₀₁ + m v₀₂
p₀ =
M v₁ = M v₀₁ + m v₀₂
v₁ = v₀₁ + m / M v₀₂
v₁ = 17.36 + 1841/3000 20.19
v₁ = 20.75 m / s
This is the speed of car 1 (A) just before the crash, now let's look for the speed when I apply the brakes the initial speed (v)
v₁² = v² - 2 a x₁
v = √ (v₁² + 2 a x₁)
v = √ (20.75² + 2 7.84 14)
v = 25.50 m / s
v = 25.50 m / s (1km / 1000m) (3600s / 1h)
v = 91.8 Km / h
Given Information:
Electric field = E = 180,000 N/C
Distance = r = 1.50 cm = 0.015 m
Required Information:
Charge = q = ?
Answer:
charge = q = 4.5x10⁻⁹ C
Explanation:
The electric field is given by
E = kq/r²
Where k is the coulomb constant k = 9x10⁹ Nm²/C²
q = Er²/k
q = 180,000*(0.015)²/9x10⁹
q = 4.5x10⁻⁹ C
Therefore, the charge is 4.5 nC