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.
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
V^2= 2* g * s
V^2 = 2 * 9.8 * 50 = 980
v = square root (980)=...... m/s
Answer:
Charge of the alpha particle
Explanation: