Hope this helps a little
initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)
A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.
Answer: Option A
<u>Explanation:</u>
The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,
Given that mass of the canoe = 20 kg and its speed =1 m/s
As we know that the Kinetic energy has the formula,

Therefore, substituting the value into the equation, we get,
= 40 J
The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.
All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.
If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.
Answer:
TDR means Timeout Detection and Recovery.
Explanation:
TDR is a feature of the Windows operating system which detects response problems from a graphics card, and recovers to a functional desktop by resetting the card. If the operating system does not receive a response from a graphics card within a certain amount of time (default is 2 seconds), the operating system resets the graphics card.
For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as
F= 618.9 N
<h3>What is the centripetal
force?</h3>
Generally, the equation for the angular speed is mathematically given as
w = v/R
Therefore
w= 4.7/1.8
w= 2.611 rad/s
Where total momentum
Tm= 642.96 + 272.32
Tm= 915.28
and total inertia
Ti= 184 + 246.24
Ti= 430.24
In conclusion, centripetal force
F= mrw^2
F = m*R*w2^2
F = 76*1.8*2.127^2
F= 618.9 N
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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?