Answer:
The speed of the plank relative to the ice is:
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
(1)
Where:
- m(g) is the mass of the girl
- m(p) is the mass of the plank
- v(g) is the speed of the girl
- v(p) is the speed of the plank
Now, as we have relative velocities, we have:
(2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
I hope it helps you!
The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.
The vertical component of the force on the apple as it falls trough the air is given as;
∑Fy = 0
Fₙ - W = 0
Fₙ = W
where;
- <em>Fₙ is the frictional force on the apple acting upwards</em>
- <em>W is the weight of the apple acting downwards</em>
The free-body diagram of the apple is represented as follows;
↑ Fₙ
Ο
↓ W
Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
Learn more here:brainly.com/question/18770265
Answer:
(A) 1.43secs
(B) -2.50m/s^2
Explanation:
A commuter backs her car out of her garage with an acceleration of 1.40m/s^2
(A) When the speed is 2.00m/s then, the time can be calculated as follows
t= Vf-Vo/a
The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0
= 2.00-0/1.40
= 2.00/1.40
= 1.43secs
(B) The deceleration when the time is 0.800secs can be calculated as follows
a= Vf-Vo/t
= 0-2.00/0.800
= -2.00/0.800
= -2.50m/s^2
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
