In orbit how high is the altitude

How the Law of Conservation of Matter is supported by the experimental demonstrations?

Delvig [45]

The law of conservation of matter states that matter cannot be created nor destroyed. It can only be transformed from one form to another. To state an example where this is shown, let's say a piece of paper is burning. Not having a scientific background, you would say that the matter is being destroyed. But in reality, the paper is simple being transformed to ash, carbon dioxide, and water vapor. Overall, the total mass would still remain the same.

6 0

3 years ago

What are food source of calcium?

bearhunter [10]

Answer:

Milk

Explanation:

It can be milk and other foods similar to milk like ice cream, cheese, yogurt, etc. This is why people say milk is good for your bones cause it has calcium

5 0

4 years ago

Read 2 more answers

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

3 years ago

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

$d = ut + \frac{1}{2}at^2$

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

7 0

3 years ago