Emily throws the ball at 30 degree below the horizontal
so here the speed is 14 m/s and hence we will find its horizontal and vertical components
vertical distance between them
now we will use kinematics in order to find the time taken by the ball to reach at Allison
here acceleration is due to gravity
now we will have
now solving above quadratic equation we have
now in order to find the horizontal distance where ball will fall is given as
here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time
so the distance at which Allison is standing to catch the ball will be 5.33 m
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
Answer:
<h2>
187,500N/m</h2>
Explanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
The value of the spring constant is 187,500N/m
