Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
Answer:
this is a no brainer
Explanation:
As air pressure in an area increases, the density of the gas particles in that area increases.
The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.
<h3>What is Potential and Kinetic energy?</h3>
Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.
mass of rod, mab = 2.4kg
mass of rod, mbc = 4kg
conservation of energy
potential energy at position 1,
V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18
V1 = 11.30112
kinetic energy T1 at position 1 is zero
potential energy at position 2 is zero
K.E at position 2,
= 1/3 *4 * (0.36)²
=0.10368kg m²
= 1/12 *4 * (0.6)²
=0.12kg m²
on putting the values in above equation we get,
T₂ = 1.0667vb²
0 + 11.30112 = 1.0667vb² + 0
vb = 3.2549 m/s
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0.04m²
Explanation:
Given parameters:
Pressure = 250000Pa
Weight = 40000N
Unknown:
Area of each foot = ?
Solution:
Pressure is the force exerted per unit area of a body
Pressure =
To find the area;
Area =
Area = = 0.16m²
The force exerted by all the four feet is 0.16m²
the area of each feet = = 0.04m²
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Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
For 8 shot burst, average recoil force on the gun is :
So, the average recoil force on the gun during that 0.40 s burst is 45 N.