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aleksandr82 [10.1K]
2 years ago
9

A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.

Physics
1 answer:
ipn [44]2 years ago
4 0

A. The acceleration of the ball while it is in flight?

magnitude is 0 m/s² (magnitude is zero)

B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)

C. The initial velocity of the ball 8.036 m/s upward

D. The maximum height reached by the ball is 3.29 m

<h3>A. How to determine the acceleration in the flight</h3>

Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.

<h3>B. How to determine the velocity at maximum height</h3>

At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.

<h3>C. How to determine the initial velocity</h3>
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) = 0 m/s
  • Time of flight (T) = 1.64 s
  • Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
  • Initial velocity (u) =?

v = u - gt (since the ball is going against gravity)

0 = u - (9.8 × 0.82)

0 = u - 8.036

Collect like terms

u = 0 + 8.036

u = 8.036 m/s upward

<h3>D. How to determine the maximum height reached by the ball</h3>
  • Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h)

h = ½gt²

h = ½ × 9.8 × 0.82²

h = 3.29 m

Learn more about motion under gravity:

brainly.com/question/20385439

#SPJ1

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Answer:

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Explanation:

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A piece of aluminum foil has a known surface density of
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That's about 2.4 cm .

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5 0
3 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

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A sprinter generates a constant force of 52 N as he runs 100 m. How much work did he do?
LUCKY_DIMON [66]
You just multiply these two numbers. It's 5200J, or 5.2kJ
3 0
4 years ago
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