The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²
<h3>What is angular acceleration?</h3>
Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²
It is expressed thus;
α= Δω ÷ Δt
Where α = angular acceleration
Δw = change in angular velocity = velocity ÷ radius
Δ t = change in time
<h3>How to calculate the angular acceleration</h3>
Using the formula:
α= Δω ÷ Δt
v = 5. 33m/s, r = 0. 33m and t = 6.27s
Substitute the values to get the angular velocity
Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s
Substitute the value of Δw into the equation
α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²
Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²
Learn more about angular acceleration here:
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I believe the answer is:
~D. Very dusty irregular galaxy.
Hope this helps!!!
Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
Answer:
Here is something that may help you!!
Explanation:
I found it in a cite (not that I'm plagiarizing, or anything).
Answer:
1. about 1.5 AU
2. about 5 AU
3. about 8 light-years
4. about 100,000 light-years
5. less than 0.01 AU
Explanation:
a. Mars is about 1.5 AU from the Sun.
b. Jupiter is about 5 AU from the Sun.
c. The star Sirius is about 8 light-years from the Sun.
d. The diameter of the Milky Way Galaxy is about 100,000 light-years.
e. The distance from Earth to the Moon is less than 0.01 AU.
Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.