Answer:
2. [B] = [L]/[T] and [C] = [L]/[T]
Explanation:
I assume you mean this:
A = B² + 2B⁴/C²
Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.
B² = [L]²/[T]²
B = [L]/[T]
B⁴/C² = [L]²/[T]²
C²/B⁴ = [T]²/[L]²
C² = B⁴ [T]²/[L]²
C² = ([L]/[T])⁴ [T]²/[L]²
C² = [L]²/[T]²
C = [L]/[T]
Notice we ignore the 2 coefficient, which is unitless.
1.potential energy 2.kinetic energy 3.electrical energy 4.electrical energy
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s
Answer:
d = 44.64 m
Explanation:
Given that,
Net force acting on the car, F = -8750 N
The mass of the car, m = 1250 kg
Initial speed of the car, u = 25 m/s
Final speed, v = 0 (it stops)
The formula for the net force is :
F = ma
a is acceleration of the car
Let d be the breaking distance. It can be calculated using third equation of motion as :
So, the required distance covered by the car is 44.64 m.
