Answer: Object B
Explanation: Acceleration is directly proportional to force and inversely proportional to mass. It implies that more massive objects accelerates at a slower rate.
For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Answer:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Explanation:
The center of mass of a system of particles (
), measured in meters, is defined by this weighted average:
(1)
Where:
- Mass of the i-th particle, measured in kilograms.
- Location of the i-th particle with respect to origin, measured in meters.
If we know that 
, 
, 
, 
, 
and 
, then the coordinates of the third particle are:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Answer:
Option (D) is correct.
Explanation:
The balloon lands horizontally at a distance of 420 m from a point where it as released.
Velocity of air balloon along +X axis =10 m/s
velocity of ball=4 m/s along + X axis
the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s
time taken= 30 s
The distance traveled is given by d= v t
d= 14 (30)
d= 420 m
Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.
You can’t solve it because you don’t have c in the question
