Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
Answer:
Explanation:
We are given that
We have to find the exit temperature.
By steady energy flow equation
Substitute the values
Answer:
P.E = 0.068 J = 68 mJ
Explanation:
First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = -9.8 m/s² (negative sign due to upward motion)
h = height attained by the ball toy = ?
Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)
Vi = Initial Velocity = 3 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²
h = (9 m²/s²)/(19.6 m/s²)
h = 0.46 m
Now, the gravitational potential energy of ball at its peak is given by the following formula:
P.E = mgh
P.E = (0.015 kg)(9.8 m/s²)(0.46 m)
<u>P.E = 0.068 J = 68 mJ</u>
Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as
Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²
