Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Answer: The machine must apply the force over a shorter distance. That's because a machine doesn't change the amount of work and work equals force times distance. Therefore, if force increases, distance must decrease
FALSE
HOPE THIS HELPS
Answer:
After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+
)
Explanation:
I believe it is a part C question.
The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........
Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.
Answer:
10.3 cm³
Explanation:
From the question given above, the following data were obtained:
Original volume (V₁) = 10 cm³
Initial temperature (θ₁) = 20 °C
Final temperature (θ₂) = 50 °C
Cubic expansivity (γ) = 10¯³ K¯¹
Final volume (V₂) =?
γ = V₂ – V₁ / V₁(θ₂ – θ₁)
10¯³ = V₂ – 10 / 10( 50 – 20)
10¯³ = V₂ – 10 / 10(30)
10¯³ = V₂ – 10 / 300
Cross multiply
10¯³ × 300 = V₂ – 10
0.3 = V₂ – 10
Collect like terms
0.3 + 10 = V₂
10.3 = V₂
V₂ = 10.3 cm³
Thus, the volume at 50 °C is 10.3 cm³
