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3 years ago

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a firework rock is burning at the rate of 14g/s all the products of combustion are ejected in one direction at the speed of 210m

/s what is the greatest height the rockect can have if it is moving vertically upward Give your answer in newtons

1 answer:

Naddika [18.5K]3 years ago

5 0

SO MY NAME IS ROHITH AND I LIKE MM

Explanation:

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You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

3 0

3 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

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3 years ago

Which of the following is not a vector quantity?

baherus [9]

Answer: The answer is density

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3 years ago

It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

$m$ = mass of lead bullet

$c$ = specific heat of lead = 128 J/(kg °C)

$L$ = Latent heat of fusion of lead = 24500 J/kg

$T_{i}$ = initial temperature = 27.4 °C

$T_{f}$ = final temperature = melting point of lead = 327.5 °C

$v$ = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

$(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}$

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3 years ago

How many significant digits are in the measurement 143,000

tangare [24]

6 significant digits

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3 years ago

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