Ask question

Ask question

All categories

3 years ago

10

An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

ate her weight when she reaches an altitude of 6,400 kilometers above the surface of Earth.

1 answer:

PIT_PIT [208]3 years ago

5 0

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

You might be interested in

When air resistance balances the weight of an object that is falling, the velocity _____.

miskamm [114]

Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.

5 0

3 years ago

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta

Tomtit [17]

Answer:

(a) 185 N/m

(b) 3.083 kg

Explanation:

(a)

Using,

E = 1/2ke²....................... Equation 1

Where E = work done to compress the spring, k = spring constant of the spring, e = compression of the spring.

make k the subject of the equation

k = 2E/e²............... Equation 2

Given: E = 3.7 J, e = 0.20 m

Substitute into equation 2

k = 2(3.7)/0.2²

k = 185 N/m.

(b)

Using,

F = ma.............. Equation 2

Where F = force applied to the spring, m = mass attached to the spring, a = acceleration of the spring.

But from hook's law,

F = ke................. Equation 3

substitute equation 3 into equation 2

ke = ma

make m the subject of the equation

m = ke/a................ Equation 4

Given: k = 185 N/m, e = 0.2 m, a = 12 m/s²

Substitute into equation 4

m = 185(0.2)/12

m = 3.083 kg

3 0

3 years ago

Read 2 more answers

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

Your friend says that the second law of thermodynamics can't be true because life itself is a highly ordered system that wouldn'

spayn [35]

This topic is actually quite controversial, but the answer in this case would be C.

Just some food for thought, the 2nd law of thermodynamics entropy of the universe is always increasing, but that doesn't necessarily mean that earth's entropy has to. As long as the net change in entropy of the universe is increasing it doesn't matter if one planet is decreasing a nominal amount. Next, Earth as said is not a closed system and you could argue that the sunlight and energy from the sun is increasing the total energy within the system that is earth meaning that it is increasing in entropy. Next, if you consider increasing entropy as an increase in the number of possible permutations that the universe or parts of the universe can take, then it is completely possible that an ordered planet and life is possible, although rare. This theory explains why there are so many life forms and why entropy is actually increasing when divergent evolution occurs.

8 0

3 years ago

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

Dominik [7]

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

b.

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

4 0

3 years ago