<u>Answer:</u>
15.97 N force is tending to pull Rover forward
<u>Explanation:</u>
The woman pulls on the leash with a force of 20.0 N at an angle of 37° above the horizontal. The arrangement is shown in the given figure,
We nee to find the pulling force P. The 20.0 N force has two components, 20.0 cos 37 in horizontal direction and 20.0 sin 37 in vertical direction.
The horizontal component is equal to pulling force P, which will pull Rover forward/
So, P = 20.0 cos 37 = 15.97 N
15.97 N force is tending to pull Rover forward.
Let
upthrust = T
weight = W = mg
Air resistance = F
When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)
Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)
Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
1) metal
Even though metalloids are also conductors of heat and electricity, malleable they are not as good as metals.
Metals are very good conductors of electricity and heat. They are also very hard to touch. Noble gases and non metals are the exact opposite in physical and chemical properties. Metals readily react with oxygen.
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
