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2 years ago

A tennis ball rolls off the edge of a table. the table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Physics

1 answer:

k0ka [10]2 years ago

3 0

Answer:

v = 0.363 m/s

Explanation:

Given that,

The table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Here, u = 0 (at rest) for initial vertical velocity as it rolls off the edge of a table.

Let t is the time to fall from the vertical height. So,

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 0.55}{9.8}} \\\\t=0.33\ s$

It can be assumed to find the initial horizontal velocity of the tennis ball. It can be given by :

$v_x=\dfrac{x}{t}\\\\v_x=\dfrac{0.12}{0.33}\\\\v_x=0.363\ m/s$

Hence, the initial horizontal velocity is 0.363 m/s.

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A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l = 4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

$F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s$

Let $\omega$ is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

$v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s$

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0

2 years ago

¿Que es un átomo???????

disa [49]

Un átomo es una porción material menor de un elemento químico que interviene en las reacciones químicas y posee las propiedades características de dicho elemento.

5 0

2 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

2 years ago

Find the radius Rrigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×

Zinaida [17]

Answer:

r = 1.61 x 10^{11} m

Explanation:

energy radiated (H) = 2.7 x 10^31 W

surface temperature (T) = 11,000 k

assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}

we can find the radius of the star from the equation below

H = A x ε x σ x T^{4}

where area (A) = 4 x π x r^{2} (assuming it is a sphere)

therefore the equation becomes

H = 4 x π x r^{2} x ε x σ x T^{4}

2.7 x 10^31 = 4 x π x r^{2} x 1 x 5.67 x 10^{-8} x (11,000)^{4}

r = $\sqrt{\frac{2.7 x 10^31}{4 x π x 1 x 5.67 x 10^{-8} x (11,000)^{4}} }$

r = 1.61 x 10^{11} m

4 0

2 years ago

The net force acting on an object that is in equilibrium is...?

Ratling [72]

Answer: If the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton.

Magnitude: 3.4 N

Direction: 161 deg

HOPE THIS HELPS

8 0

1 year ago