Answer:F=0.0882kg
Explanation:
The period it takes to make one revolution is 1.5 seconds / revolutions,
v = r * (change in angle / change in time)
the change in angle is 2pi, for one whole revolution. the time is 1.5 second per revolution, and the radius is 0.1.
v = (2 * pi * 0.1 cm * / 1.5second
v = 0.42m/s
a=v^2/r
a=0.42^2 /0.1 =1.764m/s2
F=ma
F=0.05*1.764
F=0.0882kg
Answer:
Well depending on the ramp's texture or shape probably the second one
Explanation:
Answer:
P₁ = 219.3 Pa
Explanation:
This fluid mechanics problem, we can use that the pressure is distributed with the same value throughout the system, which is Pascal's principle.
Let's use the subinidce1 for the small diameter and the subscript 2 for the larger diameter.
P₁ = P₂
pressure is defined by
P = F / A
we subtitute
F₁ / A₁ = F₂ / A₂
F₁ = F₂ A₁ / A₂
the area in a circle is
A = π r² = π d² / 4
we substitute
F₁ = F₂ (d₁ / d₂)²
we calculate
F₁ = 17640 (2/32)²
F₁ = 68.9 N
Having the force to be applied we can find the air pressure on the small plunger
P₁ = F₁ / A₁
P₁ = F₁ 4 / π d₁²
let's calculate
P₁ = 68.9 4 / (π 0.02²)
P₁ = 219.3 Pa
The new pressure P2 is 2.48 atmosphere.
<u>Explanation:</u>
Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.
By using Boyles's law,
pressure is inversely proportional to volume,
P1 V1 = P2 V2
where P1, V1 represents the first pressure and volume,
P2, V2 represents the second pressure and volume
P2 = (P1 V1) / V2
= (1.75 
8.8) / 6.2
P2 = 2.48 atmosphere.
Answer:
11.09 m/s
Explanation:
Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.
The parameters given are:
Initial velocity U = ?
Final velocity V = 9.6 m/s
Acceleration due to gravity g = 9.8m/s^2
Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0
Using third equation of motion
V^2 = U^2 - 2gH
0 = U^2 - 2 × 9.8H
U^2 = 19.6H ........ (1)
Using the formula again for one fourth of its maximum height
9.6^2 = U^2 - 2 × 9.8 × H/4
92.16 = U^2 - 19.6/4H
92.16 = U^2 - 4.9H
U^2 = 92.16 + 4.9H ...... (2)
Substitute U^2 in equation (1) into equation (2)
19.6H = 92.16 + 4.9H
Collect the like terms
19.6H - 4.9H = 92.16
14.7H = 92.16
H = 92.16/14.7
H = 6.269
Substitute H into equation 2
U^2 = 92.16 + 4.9( 6.269)
U^2 = 92.16 + 30.72
U^2 = 122.88
U = 11.09 m/s
Therefore, the initial velocity of the object is 11.09 m/s
