Answer:
a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) The total distance traveled by the baseball was 108.7 m.
Explanation:
a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:
Where:
: is the final height =?
: is the initial height = 1 m
: is the initial vertical velocity = v₀sin(45)
v₀: is the initial velocity = 32.5 m/s
g: is the gravity = 9.81 m/s²
t: is the time
First, we need to find the time by using the following equation:
Now, the height is:
Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) To find the distance traveled by the baseball first we need to find the time of flight:
By solving the above quadratic equation we have:
t = 4.73 s
Finally, with that time we can find the distance traveled by the baseball:
Hence, the total distance traveled by the baseball was 108.7 m.
I hope it helps you!
Answer:
what is heat and transfer
Answer:
L/D= 112
Explanation:
Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.
Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.
Lift increases proportionally to the square of the speed.
The solutions to the question is the file attached to this explanation.
Lift,L= qC(l). S---------------------------(1).
and,
Drag,D = qC(d).S ----------------------(2).
Hence, Lift to drag ratio,L/D= C(l)/C(d).
Therefore, we have to compute various angle of attack.(check attached file)...
Then, (L/D) will then be equal to 112.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as
=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)
