Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>
Answer:
Chemicals
Explanation:
Sanitizing method that are uses chlorine, iodine, and quaternary ammonium is <u>chemicals</u>.
Extra Info : -
Sanitizing is also achieved through the use of chemical compounds capable of destroying disease causing bacteria. Common sanitizers are chlorine (bleach), iodine, and quaternary ammonium. Chemical sanitizers have found widespread acceptance in the food service industry.
Answer:
a) t = 1.6 s
b) d = 4.9 m
c) v = 16 m/s
d) θ = 79°
Explanation:
time of fall
t = √(2h/g) = √(2(12)/9.8) = 1.5649... s
d = vt = 3.1(1.56) = 4.8512...
vertical velocity vy = at = 9.8(1.56) = 15.336... m/s
v = √(15.336² + 3.1²) = 15.6464... m/s
θ = arctan(15.336/3.1) = 78.5724...°
Answer: The coefficient of kinetic friction is μ = 0.6
Explanation:
For an object of mass M, the weight is:
W = M*g
where g is the gravitational acceleration: g = 9.8m/s^2
And the friction force between this object and the surface can be written as:
F = W*μ
where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)
In this case, the weight is:
W = 20N
And the friction force is:
F = 12N
Replacing these values in the equation for the friction force we get:
12N = 20N*μ
(12N/20N) = μ = 0.6
The coefficient of kinetic friction is μ = 0.6
Answer:
Light is shown through air on a diamond n=2.42) and it partially reflects and reflects.
Based on the incident angle of 62.5 shown what is the angle of reflection of the light
question of options
21.5
