Ok so the expression that you will be doing is water-water+object. The actual expression is 120-80. The answer would be 40mL. Remember, don't forget your units! :)
Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.
We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s
we can use the formulae for acceleration to calculate the time taken/
(final - initial velocity)/timetaken=10
(30-0)/timetaken=10
timetaken =30/10=3 seconds
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
Answer:
a) In order to catch the ball at the level at which it is thrown in the direction of motion.
b)Speed of the receiver will be 7.52m/s
Explanation:
Calculating range,R= Vo^2Sin2theta/g
R= (20^2×Sin(2×30)/9.8 = 35.35m
Let receiver be(R-20) = 35.35-20= 15.35m
The horizontal component of the ball is:
Vox= Vocostheta= 20× cos30°
Vox= 17.32m/s
Time taken to coverR=35.35m with 17.32m/s will be:
t=R/Vox= 35.35/17.32
t= 2.04seconds
b)Speed required to cover 15.35m at 2.04seconds
Vxreciever= d/t = 15.35/2.04 = 7.52m/s
