Answer:
I = 113.014 kg.m^2
m = 2075.56 kg
wf = 3.942 rad/s
Explanation:
Given:
- The constant Force applied F = 300 N
- The radius of the wheel r = 0.33 m
- The angular acceleration α = 0.876 rad / s^2
Find:
(a) What is the moment of inertia of the wheel (in kg · m2)?
(b) What is the mass (in kg) of the wheel?
(c) The wheel starts from rest and the tangential force remains constant over a time period of t= 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
Solution:
- We will apply Newton's second law for the rotational motion of the disc given by:
F*r = I*α
Where, I: The moment of inertia of the cylindrical wheel.
I = F*r / α
I = 300*0.33 / 0.876
I = 113.014 kg.m^2
- Assuming the cylindrical wheel as cylindrical disc with moment inertia given as:
I = 0.5*m*r^2
m = 2*I / r^2
Where, m is the mass of the wheel in kg.
m = 2*113.014 / 0.33^2
m = 2075.56 kg
- The initial angular velocity wi = 0, after time t sec the final angular speed wf can be determined by rotational kinematics equation 1:
wf = wi + α*t
wf = 0 + 0.876*(4.5)
wf = 3.942 rad/s
Answer:
<em>The glider's new speed is 68.90 m/s</em>
Explanation:
<u>Principle Of Conservation Of Mechanical Energy</u>
The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is
The gravitational potential energy when it's at a height h from the zero reference is
The total mechanical energy is
The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.
The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is
Operating in terms of m
Then we know the glider dives to 278 meters and we need to know their final speed, let's call it 
. The final mechanical energy is
Operating and factoring
Both mechanical energies must be the same, so
Simplifying by m and rearranging
Computing
The glider's new speed is 68.90 m/s
<h2>Solution :</h2>
Here ,
• Height of sign post = 30 m
• Distance between signpost and truck = 24 m
Let the
• Top of signpost = A
• Bottom of signpost = B
• The end of truck facing sign post be = C
Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .
Where
• AB = Height of signpost = 30 m
• BC = distance between both = 24 m
• AC = Minimum length of ladder
→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )
→ AC² = (30)² + (24)²
→ AC² = 900 + 576
→ AC² = 1476
→ AC = 38.41875
or AC apx = 38.42
So minimum height of ladder = 38.42
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
Answer:
The formula to calculate velocity in this case:
v = v0 + at
=> a = (v - v0)/t
= (50 - 0)/4
= 50/4 = 12.5 (m/s2)
Hope this helps!
:)
