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1 year ago

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]1 year 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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A box is moving along the x-axis and its position varies in time according to the expression:

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Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

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3 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

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3 years ago

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marysya [2.9K]

Answer:

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So,

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Therefore, milk has a higher mass.

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Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

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Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

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Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

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3 years ago