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2 years ago

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A unique type of basketball is played on the planet zarth. During the game, a player flies above the basket and drops the ball i

n from a height of 10 m. If the ball takes 5.0 s to fall, find the acceleration due to gravity on zarth.

2 answers:

pickupchik [31]2 years ago

8 0

Answer:

The acceleration due to gravity on zarth = 0.8 $m/s^2$

Explanation:

The acceleration due to gravity on earth = 9.8 $m/s^2$

We know that time taken(t) by an object to fall a distance (d) is given by

t = $\sqrt{ (2d/g)}$

We are given the height from which the ball is dropped = 10m and the time taken by the ball to fall = 5.0s

So, substituting the given values

d = height of dropping the ball = 10m, t = 5.0 s

i.e. time taken for fall ,in the equation,

$t^2 = 2d/g$

$g= 20/5^2 = 20/25 = 0.8$

Therefore (g) = 0.8 $m/s^2$ which gives the acceleration due to gravity on planet Zarth.

Murljashka [212]2 years ago

6 0

Answer:

$0.8 m/s^2$

Explanation:

The motion of an object in free fall is a uniformly accelerated motion, so we can analyze it using the following suvat equation:

$s=ut+\frac{1}{2}gt^2$

where, taking downward as positive direction:

s is the vertical displacement

u is the initial velocity

t is the time of flight

g is the acceleration of gravity

For the ball dropped in this problem,

s = 10 m

t = 5.0 s

u = 0

Therefore, solving the equation for g, we find the acceleration due to gravity:

$g=\frac{2s}{t^2}=\frac{2(10)}{5^2}=0.8 m/s^2$

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Answer:

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3 years ago

A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At

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Answer:

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(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

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Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

$Q_{1}=Q_{2}$

$v_{1}A_{1}=v_{2}A_{2}$

$v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})$

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$A=\pi\times \dfrac{d^2}{4}$

$v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})$

Put the value into the formula

$v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}$

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The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}$

Where, $h_{1}=h_{2}$

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2$

$P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)$

Put the value into the formula

$P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)$

$P_{1}=188081.702\ Pa$

$P=1.86\ atm$

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

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