Answer:
<em>2.78m/s²</em>
Explanation:
Complete question:
<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>
According to Newton's second law of motion:
Where:
is the coefficient of friction
g is the acceleration due to gravity
Fm is the moving force acting on the body
Ff is the frictional force
m is the mass of the box
a is the acceleration'
Given
Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:
9.8sin30 - 0.25(9.8)cos30 = ax
9.8(0.5) - 0.25(9.8)(0.866) = ax
4.9 - 2.1217 = ax
ax = 2.78m/s²
<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>
Velocity is defined as Distance divided by Time.
In other words, V = D/T.
Now that we have our formula, we can solve.
Let's plug in the numbers we have.
We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.
Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.
Your velocity is 80 m/s [E]
I hope this helps!
Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface, 
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :
So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
1, 2 and 4 apply. im not sure that 2 ALWAYS applies though.
Answer:
D. No, because the student needs to know the direction that the force is applied
Explanation:
The change in momentum depends on the direction of the force as well as its magnitude. Since the graph only supplies force magnitude information, it is insufficient to allow the student to calculate the change in momentum.
