Answer:
about 19.6° and 73.2°
Explanation:
The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...
y = -4.9(x/s·sec(α))² +x·tan(α)
where s is the launch speed in meters per second.
We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):
-13.6111·tan(α)² +50·tan(α) -16.0511 = 0
This has solutions ...
tan(α) = 0.355408 or 3.31806
The corresponding angles are ...
α = 19.5656° or 73.2282°
The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.
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I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.
Explanation:
The linear analog of angle is angle itself.
The linear analog of angular velocity is linear velocity.
ω is angular velocity, therefore linear velocity is given by v
∴ for linear velocity,
for angular velocity,
The linear analog of angular acceleration is acceleration.
α is angular acceleration whereas as a is linear acceleration.
∴ for linear acceleration, v = u + a.t
for angular acceleration,
The linear analog of moment of inertia is mass.
I is moment of inertia and m is mass,
∴ for linear analog, F = m.a
for angular analog, τ - I.α
Answer:
650.65 K or 377.5°C
Explanation:
Area = A = 10 m²
Thickness of wall = L = 2.5 cm = 2.5×10⁻² m
Inner surface temperature of wall = = 415°C = 688.15 K
Outer surface temperature of wall =
Heat loss through the wall = 3 kW = 3×10³ W
Thermal conductivity of wall = k = 0.2 W/m K
Assumptions made here as follows
- There is not heat generation in the wall itself
- The heat conduction is one dimensional
- Heat flow follows steady state
- The material has same properties in all directions i.e., it is homogeneous.
Considering the above assumptions we use the following formula
∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C
Answer:
18N & 360W
Explanation:
Here,
Velocity of car (v) = 72 ∗ 1000 / 60∗60 = 20m/s
weight of car = 1200N
We have,
μr = Fs / R
Fs = μr ∗ R = 0.015 ∗ 1200 = 18N
Power developed by the engine = 18 ∗ 20 =360W
<h2>Hope this helps!!!</h2>
The answer would be no. Both will have same charge, in both caes charge will reside on surface. In case of solid sphere it will be distributed evenly throughout. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.