Answer:
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.
<h3>What is negative angular acceleration?</h3>
- A particle that has a negative angular velocity rotates counterclockwise.
- Negative angular acceleration () is a "push" that is hence counterclockwise.
- The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
- For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
- If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
- It is known that a change in the linear acceleration correlates to a change in the linear velocity.
Let t be the time taken to stop.
ω = 0 rad/s
Use the first equation of motion for rotational motion
ω = ωo + α t
0 = 10.5 - 2 x t
t = 5.25 second
To learn more about angular acceleration, refer to:
brainly.com/question/21278452
#SPJ4
Answer:
0.76 rad/s^2
Explanation:
First, we convert the original and final velocity from rev/s to rad/s:


Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

Now, we just use the formula:


Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151