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Lelu [443]
3 years ago
12

A child slides down a frictionless playground slide from a height of 2.1 m above the ground. If she starts with an initial speed

of 4 m/s and the end of the slide is 0.7 m above the ground, her final speed will be: (in m/s)
Physics
1 answer:
baherus [9]3 years ago
8 0

Answer: Final speed will be 6.59 m/s

Explanation:  Initial height of slide is 2.1 m

Final height = 0.7 m

Initial speed, u = 4 m/s

Final speed, v = ??

To find the final speed in this question we can use equation of motion i.e.

v^{2} - u^{2} = 2aS

Here, S is the distance covered by the child on the slide i.e. net height  

S = 2.1 - 0.7 = 1.4 m

and a = g, acceleration due to gravity i.e. 9.8 m/s

Put all the values in the equation,

v^{2} = u^{2} + 2aS

v^{2} = 4^{2} + 2 \times 9.8  \times 1.4

v^{2} = 16 + 27.44 = 43.44

Taking square root both the sides, we get

v = 6.59 m/s

Final speed will be 6.59 m/s

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In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

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for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

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Which gas is the most abundant of the trace gases?
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8 0
3 years ago
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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

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