Two factors that can be used to evaluate ______ are life expectancy and quality of life.

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

If you are in a spaceship that is sitting on the surface of a planet, you feel your weight. How does this compare to the weight

nordsb [41]

Answer:

You will feel more weight if it is accelerating out of the planet.

You will feel less weight if it is accelerating towards the planet.

Explanation:

The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.

When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.

When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.

If the spaceship is free falling at the gravitational acceleration you experience a zero weight

8 0

3 years ago

Most of the mass of an element is found in the?

svetoff [14.1K]

Most of the mass of an element is found in the NUCLEUS

6 0

3 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

3 years ago

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

aliina [53]

Answer:

v_f = 10.38 m / s

Explanation:

For this exercise we can use the relationship between work and kinetic energy

W = ΔK

note that the two quantities are scalars

Work is defined by the relation

W = F. Δx

the bold are vectors. The displacement is

Δx = r_f -r₀

Δx = (11.6 i - 2j) - (4.4 i + 5j)

Δx = (7.2 i - 7 j) m

W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

i.i = j.j = 1

i.j = 0

W = 4 7.2 + 9 7

W = 91.8 J

the initial kinetic energy is

Ko = ½ m vo²

Ko = ½ 2.0 4.0²

Ko = 16 J

we substitute in the initial equation

W = K_f - K₀

K_f = W + K₀

½ m v_f² = W + K₀

v_f² = 2 / m (W + K₀)

v_f² = 2/2 (91.8 + 16)

v_f = √107.8

v_f = 10.38 m / s

3 0

3 years ago