What is the impulse of a 2 kg object that starts at 10 m/s and comes to a stop over 0.4 seconds?
1 answer:
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Answer:
h = 4.04 m
Explanation:
Given that,
Mass of a child, m = 25 kg
The speed of the child at the bottom of the swing is 8.9 m/s
We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,
Put all the values,
So, the child is able to go at a height of 4.04 m.
The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°
cos θ = Adjacent side / Hypotenuse
θ = 47°
Hypotenuse = Length of ladder = 8.5 m
cos 47° = Adjacent side / 8.5
Adjacent side = Initial distance of base of ladder from the building = 5.8 m
Adjacent side 2 = Final distance of base of ladder from the building
Adjacent side 2 = 5.8 - 0.82 = 4.98 m
cos θ = Adjacent side 2 / Hypotenuse
cos θ = 4.98 / 8.5 = 0.59
θ =
( 0.59 )
θ = 53.84°
The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.
Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°
To know more about trigonometric ratios
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