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3 years ago

Why should you use control while during an experiment?

Physics

1 answer:

Sergio039 [100]3 years ago

8 0

To provide a greater certainty that the observed results are not by chance.

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

4 0

3 years ago

I do not know where to start.

irakobra [83]

Underline the words then eliminate the ones that arent part of the problem!

4 0

2 years ago

A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an

AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0

2 years ago

In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separatio

Keith_Richards [23]

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

$y=\frac{m\lambda D}{d}$

m: order of the fringes

lambda: wavelength 500nm

D: distance to the screen 5 m

d: separation of the slits 1mm=1*10^{-3}m

With the formula you can calculate the separation of two adjacent slits:

$\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm$

hence, the aswer is 0.25cm

5 0

2 years ago

Answer them all please

Amanda [17]

Answer:

respiratory = nose

circulatory = heart

digestive = mouth

nervous = brain

5 0

2 years ago