As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?

Answers

Answer 1
Answer:

Answer:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t²    .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s


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A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N

Answers

Answer:

1) The volume of helium in the ballon when is fully inflated is 492.8070 m³

2) The magnitude of the force of gravity (with no people) is 869.3119 N

Explanation:

Given:

ρair = density of air = 1.28 kg/m³

ρhelium = density of helium = 0.18 kg/m³

R = radius of balloon = 4.9 m

mtotal = 121 kg

Vtotal = 0.073 m³

mp = average mass per person = 73 kg

Vp = 0.077 m³

g = gravity = 9.8 m/s²

Questions:

1) What is the volume of helium in the balloon when fully inflated, Vhelium = ?

2) What is the magnitude of the force of gravity on the entire system (but with no people), Fg = ?

1) The volume of helium in the ballon when is fully inflated

V_(helium) =(4)/(3) \pi R^(3) =(4)/(3) \pi *4.9^(3) =492.8070m^(3)

2) First, you need to calculate the mass of helium

m_(helium) =\rho _(helium) *V_(helium) =0.18*492.8070=88.7053kg

The magnitude of the force of gravity (with no people)

F_(g) =m_(helium) *g=88.7053*9.8=869.3119N

A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer:

5.78971 m

Explanation:

P_1 = Initial pressure = 0.873 atm

P_2 = Final pressure = 0.0282 atm

V_1 = Initial volume

V_2 = Final volume

r_1 = Initial radius = 16.2 m

r_2 = Final radius

Volume is given by

(4)/(3)\pi r^3

From the ideal gas law we have the relation

(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m

The radius of balloon at lift off is 5.78971 m

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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If you see Sagittarius high in your night sky on June 20 and today is your birthday, what is your zodiac constellation?

Answers

Answer:

Gemini

Explanation:

Zodiac Constellation or Sun sign is the constellation in which the Sun resides on the date of birth of a person. Throughout the year the Sun crosses across 12 constellations thus there are 12 Sun-signs. Though astronomically the Sun crosses across 13 constellations so there should be 13 zodiacs but most of the astrologers do not accept this. On the date of June 20, the Sagittarius which is a summer constellation and a zodiac can be seen high up in the sky in the night time. At this time the Sun will be in a constellation which is almost opposite to Sagittarius in the celestial sphere. That constellation is Gemini. Thus for a person born on 20 June, zodiac will be Gemini.

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

E_i=(mv^2)/(2)+mgh

Here , initial velocity is zero .

Therefore , E_i=mgh=40\ J

Now , final energy of the system :

E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

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The energy provided each hour by heat to the turbine in an electric power plant is 9.0×10^12 J. If 5.4 × 10^12 J of energy is exhausted each hour from the engine as heat, what is the efficiency of this heat engine?

Answers

Answer:

60 %

Explanation:

Efficiency is defined as the ratio of output power  to the input power.

Input energy each hour = 9 x 10^12 J

Output energy each hour = 5.4 x 10^12 J

Efficiency = Output energy per hour / input energy per hour

Efficiency = (5.4 x 10^12) / ( 9 x 10^12) = 5.4 / 9 = 0.6

Efficiency in percentage form = 0.6 x 100 = 60 %

Final answer:

The efficiency of a heat engine is calculated using the formula (Energy Input - Energy Output) / Energy Input. Given the figures, the efficiency of the engine is 40%, indicating that 40% of the input energy is converted into work.

Explanation:

The efficiency of a heat engine is determined by the ratio of work output to energy input. In the given scenario, the turbine in an electric power plant is supplied with 9.0 x 10^12 joules of energy, and 5.4 x 10^12 joules of energy is expelled as heat per hour. We can calculate the efficiency using the equation:

Efficiency = (Energy Input - Energy Output) / Energy Input

By substituting the given values, Efficiency = (9.0 x 10^12 J - 5.4 x 10^12 J) / 9.0 x 10^12 J = 0.4 or 40%

This means the heat engine of the power plant has a 40% efficiency, meaning 40% of the energy input is converted into work while 60% is discarded as waste heat.

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Some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?A. Visible light waves
B. Ultraviolet light waves
C. Infrared Waves
D. Microwaves

Answers

The heat emitted from anything is carried in the form of infrared waves. (C)

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