Answer:
The answer is 3,064x
Explanation:
When the collision happens, the momentum of the first car is applied to the both of them.
So we can calculate the force that acts on both cars as:
- The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/s
- The acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s
- Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.
Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:
- F = (2020+2940) x 5.78 = 28,684
The friction force acts in the opposite direction and if they stop after moving 2.12 meters;
- Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.
- F - Ffriction = m x V
- 28,684 - μ x 49,600 = 4960 x 5.78
- μ = 3,064x
Answer:
v = Δs/Δt
and use this calcultor"Velocity Calculator | Definition | Formula - Omni Calculator"if u have problems to slove.cos
Explanation:
Answer:
D
Explanation:
as one burns then all stop simple
please mark me as brainlist
Total flour was 5 pounds
Ramsey divided it in 3 bowls equally
it means now
3 bowls has = 5 pounds of flour
to get in 1 bowl we we can divide by 3 both sides
3/3 bowl has = 5/3 pounds of flour
1 bowl has = 5/3 pounds
Write each force in component form:
<em>v </em>₁ : 50 N due east → (50 N) <em>i</em>
<em>v</em> ₂ : 80 N at N 45° E → (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )
The resultant force is the sum of these two vectors:
<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>
Its magnitude is
|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N
and has direction <em>θ</em> such that
tan(<em>θ</em>) = (56.5 N) / (106.5 N) → <em>θ</em> ≈ 28.0°
i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)
