The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Answer:
12°F
Explanation:
Calculation for how much subcooling is there in the condenser
Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula
Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature
Let plug in the formula
Amount of Condenser subcooling=120°-108f
Amount of Condenser subcooling=12°F
Therefore the amount of subcooling that is there in the condenser will be 12°F
Answer:
Yes. Inertia keeps the speed maintained though my feet leave the ground.
Explanation:
Inertia is the resistance to the change in position of any object this means this resistance will keep me traveling at 30 km/s relative to the sun. If the person wants to change the position we apply force to do that because inertia is opposing us to not do that. We are always traveling with 30km/s relative to sun due to inertia.
Answer:
Static Friction - acts on objects when they are resting on a surface
Sliding Friction - friction that acts on objects when they are sliding over a surface
Rolling Friction - friction that acts on objects when they are rolling over a surface
Fluid Friction - friction that acts on objects that are moving through a fluid
Explanation:
Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book
, car parked on a hill.
Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.
Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.
Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon, sucking water through a straw, submarine moving through water.
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
