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Artemon [7]
3 years ago
12

A student throws a rock upwards. The rock reaches a maximum height 2.4 seconds after it was released.

Physics
1 answer:
Allisa [31]3 years ago
7 0

Answer:

23.52 m/s

Explanation:

The following data were obtained from the question:

Time taken (t) to reach the maximum height = 2.4 s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)

was thrown as follow:

v = u – gt (since the rock is going against gravity)

0 = u – (9.8 × 2.4)

0 = u – 23.52

Collect like terms

0 + 23.52 = u

u = 23.52 m/s

Therefore, the rock was thrown at a velocity of 23.52 m/s.

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Answer:

The absolute value of the angular deviation is 39.3°.

Explanation:

Given that,

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Distance = 0.9 cm

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Put the value into the formula

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B=24.4\times10^{-6}\ T

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Using formula of direction

\theta=\tan^{-1}(\dfrac{B_{E}}{B_{w}})

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Hence, The absolute value of the angular deviation is 39.3°.

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3 years ago
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