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bekas [8.4K]
3 years ago
10

Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball

be 2.0 second later?
Physics
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

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Answer:

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The distance between two adjacent nodes or two adjacent anti-nodes is equivalent to half the wavelength.

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-- First, we have to decide how to handle the two resistors.

The effective resistance of resistors in series is the sum
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