Since the object is dropped from some height so its initial speed must be zero
acceleration of the object is due to gravity
so we can use kinematics to find the time it will take to drop by x = 22 m
Now the speed after 2.12 s will be given as
so above is the speed and time
Answer:
the person on the boat is moving 15mph on the boat and throws the ball 10mph. you add that together its 25mph. the other person is standing on land so their is no extra speed.
Explanation:
its common
Answer:
v= 13 m/s
Explanation:
Velocity is defined as the derivative of displacement with respect to time
v= ds/dt
Known data
s(t) = 5t + 2t² : distance that the ball has rolled after t seconds
vi= 5 m/s : initial velocity
t= 2 s
Problem develoment
s(t) = 5t + 2t²
v= ds/dt= 5 + 4t : velocity of the ball in function of the time
We replace t =2 s in the equation of velocity
v= 5 + 4(2)
v= 13 m/s : velocity after 2 seconds
Answer:
a) Total mass form, density and axis of rotation location are True
b) I = m r²
Explanation:
a) The moment of inertia is the inertia of the rotational movement is defined as
I = ∫ r² dm
Where r is the distance from the pivot point and m the difference in body mass
In general, mass is expressed through density
ρ = m / V
dm = ρ dV
From these two equations we can see that the moment of inertia depends on mass, density and distance
Let's examine the statements, the moment of inertia depends on
- Linear speed False
- Acceleration angular False
- Total mass form True
- density True
- axis of rotation location True
b) we calculate the moment of inertia of a particle
For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is
I = m r²
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
