Answer:
Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a time t
Explanation:
Here in the question it is mentioned that a toy car has an initial acceleration of 2m/s² across a horizontal surface so we can say that it is acted upon by an external force
Assuming that the acceleration is constant and the reason for this assumption is there at the last
The major difference between an open system and closed system is in case of open system there will be transfer of matter and in case of closed system there will be no change in matter of the system
If acceleration is constant in case of closed system we can expect the speed of the car after a time t by using the formula
s = u×t + 0·5×a×t²
where s is the distance travelled
t is the time taken to travel that distance
u is the initial velocity
a is the acceleration of that system
But in case of open system as there will be a change of mass there will be a change in velocity of the system so in this case we cannot expect the speed of the car after a time t
And if the acceleration is not constant then we cannot say that the toy car is an open system or closed system, that is why we are assuming that the acceleration of the toy car is constant
The scientist who discovered the two elements radium and polonium would be Marie Curie in 1898.
Hope this helps!
It contains no large maria
The ball will take 2.551 seconds to reach its peak position.
<h3>How much time will the ball take to land?</h3>
We must know how long the balls are in the air before we can predict where they will fall. It will take 2 seconds for both balls to touch the ground.
<h3>How quickly does a ball drop?</h3>
The falling ball travels a distance of d = 12 9.8 (m/s2) t2, with a speed of v = 9.8 (m/s2) t as a function of time. The ball travels 4.9 m in a second. The falling ball's velocity is v = -9.8 (m/s2) t j, and its position is r = (4.9 m - 12 9.8 (m/s2) t2) j as a function of time.
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(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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