What is the maximum density of water

kogti [31]

Pair production is a direct conversion of radiant energy to matter. It is one of the principal ways in which high-energy gamma rays are absorbed in matter.

6 0

3 years ago

How long will it take a ball to roll 10 meters along the floor at a speed of 0.5 m/s?

IRINA_888 [86]

Answer:

20 seconds.

Explanation:

The following data were obtained from the question:

Distance = 10 m

Speed = 0.5 m/s

Time =...?

The speed of an object is simply defined as the distance travelled by the object per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can obtain the time taken for the ball to travel a distance of 10 m as shown below:

Distance = 10 m

Speed = 0.5 m/s

Time =...?

Speed = Distance /time

0.5 = 10/time

Cross multiply

0.5 × time = 10

Divide both side by 0.5

Time = 10/0.5

Time = 20 secs.

Therefore, it will take 20 seconds for the ball to travel a distance of 10 m.

6 0

3 years ago

Assuming that each of the following objects is a typical example of its class, rank them by increasing density.

inysia [295]

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

Molecular cloud- They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between 10 to 50 Kelvin. Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

Interstellar cloud- They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around just 100 Kelvin while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

1 Msun star- These stars have temperature anywhere between 5300 and 6000 Kelvin. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.

1 Msun protostar- protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between 2000-3000 kelvin and are accompanied by dust usually.

Intercloud gas- These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around 8000 Kelvin.

8 0

3 years ago

A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar

jasenka [17]

Answer:

Part 1) Voltage in secondary windings is 61.08 Volts

Part 2) Current in secondary windings is 0.53 Amperes

Explanation:

The potential developed in the primary and secondary winding of a transformer are related as

$\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

where

Np no of turns in primary coil

Ns no of turns in secondary coil

Vp Voltage of turns in primary coil

Vs Voltage of turns in secondary coil

Applying values in the formula we get

$\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V$

Part 2)

Using Ohm's law the current is given by

$I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A$

5 0

3 years ago

An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

3 years ago

What is the maximum density of water​

What is the maximum density of water