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3 years ago

14

What goes up but never comes down i know the answer but do you?

1 answer:

Nady [450]3 years ago

4 0

Answer:

Our Age :(

Explanation:

we will always age

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What is the difference of Mitosis and Meiosis? Please explain it in easy way.

lesya692 [45]

Answer:

The nuclear division processes of mitosis and meiosis occur during cell division. During mitosis, body cells are divided, while during meiosis, s3x cells are divided. At mitosis, a cell divides once; at meiosis, it divides twice.

Explanation:

3 0

3 years ago

Steam enters an adiabatic turbine at 5 MPa and 4500C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbi

REY [17]

Answer:350.92 KJ/kg

Explanation:

Given the process is reversible adiabatic i.e it is isentropic

$P_1=5 MPa$

$T_1=4500 MPa$

$P_2=1.4 MPa$

From steam table

$h_1=3317.03KJ/kg$

For isentropic process $s_1=s_2$

at $P_2=1.4 MPa$

$s_2=6.82 KJ/kg-k$

$h_2=2966.11 KJ/kg$

Therefore Work output of the turbine per unit mass of steam is = $h_1-h_2$

=3317.03-2966.11

=350.92 KJ/kg

8 0

3 years ago

Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad

alisha [4.7K]

Answer:

a) $g(r) = 4\pi \cdot G \cdot \rho\cdot r$ , b) $g = 4\pi \cdot G \cdot \rho\cdot r_{P}$

Explanation:

a) The acceleration due to gravity inside the planet is:

$dg = G\cdot \frac{\rho \cdot dV}{r^{2}}$

$dg = G\cdot \frac{\rho \cdot dV}{r^{2}}$

$dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}$

$dg = 4\pi\cdot G\cdot \rho \,dr$

$g(r) = 4\pi \cdot G \cdot \rho\cdot r$

b) The acceleration at the surface of the planet is:

$g = 4\pi \cdot G \cdot \rho\cdot r_{P}$

5 0

3 years ago

Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th

irinina [24]

I'm not quite sure what happens to Fay so I didn't finish but hope it helps

5 0

3 years ago

Read 2 more answers

Aluminum wire 1.00 mm in radius is to be used to construct a solenoid 15.0 mm in radius so that the magnitude of the magnetic fi

Leto [7]

To solve this problem we will proceed to use the equations given for the calculation of the resistance, in order to find the radius of the cable. Once the length is found we can find the number of turns of the solenoid and finally the net length of it

The resistance of the wire is

$R= \frac{\rho L}{A}$

$\rho$ = Resistivity

L = Length

A = Cross-sectional Area

That can be also expressed as,

$R = \frac{\rho L}{\pi r^2}$

Rearranging the equation for the length of the wire we have

$L= \frac{R\pi r^2}{\rho}$

$L= \frac{3\Omega \pi(1*10^{-3})}{2.655*10^{-8}\Omega \cdot m}$

$L = 354.8m$

The number of turns of the solenoid is

$N = \frac{L}{2\pi r} \rightarrow$ Denominator is equal to the circumference of the loop

$N = \frac{354.8}{2\pi(15*10^{-3})}$

$N = 3766$

Finally the Length of he solenoid is

$l = N (\phi)$

Where \phi is the diameter of wire

$l = N (2r)$

$l = (3766)(2*(1*10^{-3}))$

$l = 7.532m$

Therefore the length of the solenoid is 7.532m

7 0

4 years ago