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**

What is a light year

If gear a rotates with a constant angular acceleration of aa = 90 rad>s2, starting from rest, determine the time required for gear d to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear d to attain this angular velocity. Gears a, b, c, and d have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:Find the magnitude and direction of the electric field.

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?C)What would the swimmer's drop time be if the bridge were twice as high?

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

If gear a rotates with a constant angular acceleration of aa = 90 rad>s2, starting from rest, determine the time required for gear d to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear d to attain this angular velocity. Gears a, b, c, and d have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:Find the magnitude and direction of the electric field.

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?C)What would the swimmer's drop time be if the bridge were twice as high?

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

**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

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

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

**Answer:**

5.78971 m

**Explanation:**

= Initial pressure = 0.873 atm

= Final pressure = 0.0282 atm

= Initial volume

= Final volume

= Initial radius = 16.2 m

= Final radius

Volume is given by

From the ideal gas law we have the relation

**The radius of balloon at lift off is 5.78971 m**

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.

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.

#SPJ3

**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.

**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 :

Here , initial velocity is zero .

Therefore ,

Now , final energy of the system :

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

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

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.

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.

#SPJ12

**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 %

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.

The **efficiency** of a heat engine is determined by the ratio of work output to **energy input. I**n 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.

#SPJ3

B. Ultraviolet light waves

C. Infrared Waves

D. Microwaves

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