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1 year ago

6

Two rocks are thrown directly upward with the same initial speeds, one on earth and one on our moon, where the acceleration due

to gravity is one-sixth what it is on earth. Part A If the rock on the moon rises to a height H, how high will the rock rise on the earth

1 answer:

emmasim [6.3K]1 year ago

8 0

Answer:

The rock will rise H/6 units high on earth

Explanation:

In order to find the height to which rock rises, we use 3rd equation of motion. The 3rd equation of motion is as follows:

2gh = Vf² - Vi²

h = (Vf² - Vi²)/2g

where,

h = height

Vf = Final Velocity

Vi = Initial Velocity

g = acceleration due to gravity

<u>ON MOON</u>:

On moon:

h = H

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi

g = -(1/6) g (negative sign due to upward motion)

Therefore,

H = (0² - Vi²)/[-(2)(1/6)(g)]

H = 3Vi²/g

H/3 = Vi²/g ------ equation (1)

<u>ON EARTH</u>:

On earth:

h = ?

Vf = 0 m/s (rock stops at highest point for a moment)

Vi = Vi (same initial velocity)

g = - g (negative sign due to upward motion)

Therefore,

h = (0² - Vi²)/(-2g)

h = Vi²/2g

h = (1/2)(Vi²/g)

using equation (1), we get:

h = (1/2)(H/3)

<u>h = H/6</u>

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A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

so height is 69.68 m

6 0

2 years ago

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$

The time taken to fall down is 2.4158 seconds

$h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m$

Height of the buidling is 28.6260196842 m

7 0

1 year ago

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

wolverine [178]

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

6 0

1 year ago

A swimming pool heater has to be able to raise the temperatureof the 40,000 gallons of water in the pool by 10 degress Celsius.H

andrew11 [14]

Answer:b)1770 kWh

Explanation:

Given

volume of water $V=40,000 gallons$

Temperature rise $\Delta T=10^{\circ}C$

$c=4186 J/kg-^{\circ} C$

also 1 kg mass is approximately is 1 gallon

therefore 40,000 gallon is equivalent to 3.8\times 40000 kg

heat Required to raise temperature is

$Q=mc\Delta T$

$Q=3.8\times 40000\times 4186\times 10$

$Q=63,627.2\times 10^5 J$

$Q=63,672.2\times 10^2 KJ$

$Q=1767.42 kWh\approx 1770 kWh$

3 0

1 year ago

The speed of sound in dry air at 20 °C is 343.5 m s-1, and the frequency of the sound from the note C# above middle C on the pia

Nastasia [14]

Answer:

1.23917 m

0.14323 s

Explanation:

v = Speed of sound in dry air at 20 °C = 343.5 m/s

f = Frequency of note C# = 277.2 /s = 277.2 Hz

λ = Wavelength

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343.5}{277.2}\\\Rightarrow \lambda=1.23917\ m$

Wavelength = 1.23917 m

Distance the wave needs to travel is 49.2 m

Time = Distance / Speed

$\text{Time}=\frac{49.2}{343.5}=0.14323\ s$

Time taken for the sound to travel across the concert hall is 0.14323 s

4 0

1 year ago