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Ostrovityanka [42]

3 years ago

14

A golf ball is dropped from rest from a height of 9.50m. It hits the pavement, then bounces back up, rising just 5.70m before fa

lling back down again. A boy then catches the ball on the way down when it is 1.20m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

1 answer:

marta [7]3 years ago

4 0

Answer:

Explanation:

Given

height from which ball is dropped $h_1=9.5\ m$

and it rises to a height of $h_2=5.7\ m$

A person caught the ball in mid-way i.e. at a height of $h_0=1.2\ m$ from pavement

time taken to reach the bottom is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

here initial velocity is zero i.e.

$h_1=0+\frac{1}{2}gt_1^2$

$t_1=\sqrt{\frac{2h_1}{g}}$

Similarly to reach of height $h_2$ time taken is $t_2$

$t_2=\sqrt{\frac{2h_2}{g}}$

The ball is caught in mid-way i.e. ball travel a distance of $h_3=5.7-1.2=4.5\ m$

time taken is $t_3=\sqrt{\frac{2h_3}{g}}$

total time taken $t=t_1+t_2+t_3$

$t=\sqrt{\frac{2h_1}{g}}+\sqrt{\frac{2h_2}{g}}+\sqrt{\frac{2h_3}{g}}$

$t=\sqrt{\frac{2\cdot 9.5}{9.8}}+\sqrt{\frac{2\cdot 5.7}{9.8}}+\sqrt{\frac{2\cdot 4.5}{9.8}}$

$t=3.429\ s$

total time =3.429 s

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bazaltina [42]

Answer:

1845.26 ?

Explanation:

18.46 × 99.96= 1845.2616 = 1845.26

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6 0

3 years ago

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

∑F is net force
m is mass of the third child
a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

the wagon
the children outside the wagon

<h3>Free body diagram</h3>

→ → Ф ←

1st child friction wagon 2nd child

<h3>Acceleration of the child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0

2 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

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3 years ago

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

Alexxx [7]

Answer:

$Q_{2}=1200cm^{3}/s$

Explanation:

Given data

Q₁=200cm³/s

We know that:

$F=n\frac{vA}{l}\\$

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

$Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s$

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2 years ago

For every O ion, two Na ions are needed to balance charges. Which ratio represents the relationship between Na and O? 1 Na:1 O 1

quester [9]

For every O ion, two Na ions are needed to balance charges. Which ratio represents the relationship between Na and O?

2 Na:1 O

6 0

3 years ago

Read 2 more answers